NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

The chapter is divided into five exercises and one miscellaneous exercise. Here is a brief summary of each exercise:

• Exercise 9.1: This exercise asks students to verify whether a given function is a solution of a given differential equation. A general solution is the one where the independent arbitrary constants of the equation are equal to the order of the equation.  So for an equation d2y/dx2 + y = 0, its general solution would be given as y = K Cos x + C sin x,  since it has 2 arbitrary constants K and C which are equal to the order of the equation that is 2. To find the particular solution of a differential equation, the arbitrary constants need to be given particular values. So, in the above example, above if we replace K = C = 1, we get the solution y = cos x + sin x which is termed as the particular solution of the differential equation.

• Exercise 9.2: This exercise asks students to find the order and degree of given differential equations. In this, you would learn how to formulate differential equations given “n” arbitrary constants and differentiate the equation n times over to get the n + 1 equations. When you eliminate the arbitrary constants from these n + 1 equations, you will get the required differential equation. You would use this method to derive a differential equation that represents a family of curves.

• Exercise 9.3: This exercise asks students to solve differential equations of the first order and first degree.

• Exercise 9.4: This exercise asks students to solve differential equations of the first order and higher degree.

• Exercise 9.5: This exercise asks students to solve differential equations of second order and higher degree.

Overall, this chapter is essential for students who want to pursue higher education in Mathematics and Physics. It provides a solid foundation for students to understand differential equations and their applications in real-world problems.

Access NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations

Exercise 9.1

1. Determine order and degree (if defined) of differential equation $\frac{{{\text{d}}^{\text{4}}}\text{y}}{\text{d}{{\text{x}}^{\text{4}}}}\text{+sin}\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)\text{=0}.$

The highest order between the two terms is of $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$ which is four.

The differential equation contains a trigonometric derivative term and is not completely polynomial in its derivative, thus degree is not defined.

2. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{ +5y=0}$.

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{ +5y=0}$.

The highest order term is $\text{y }\!\!'\!\!\text{ }$, thus the order is one.

As the derivative is of completely polynomial nature is and highest power of derivative is of $\text{y }\!\!'\!\!\text{ }$ which is one. Thus degree is one.

3. Determine order and degree (if defined) of differential equation ${{\left( \frac{\text{ds}}{\text{dt}} \right)}^{\text{4}}}\text{+3s}\frac{{{\text{d}}^{\text{2}}}\text{s}}{\text{d}{{\text{t}}^{\text{2}}}}\text{=0}$.

Ans: The given differential equation is ${{\left( \frac{\text{ds}}{\text{dt}} \right)}^{\text{4}}}\text{+3s}\frac{{{\text{d}}^{\text{2}}}s}{\text{d}{{\text{t}}^{\text{2}}}}\text{=0}$.

The highest order term is $\frac{{{\text{d}}^{\text{2}}}\text{s}}{\text{d}{{\text{t}}^{\text{2}}}}$, thus the order is two.

As the derivative is of completely polynomial nature is and highest power of derivative term$\frac{{{\text{d}}^{\text{2}}}\text{s}}{\text{d}{{\text{t}}^{\text{2}}}}$ which is one. Thus the degree is one.

4. Determine order and degree (if defined) of differential equation ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}\text{+cos}\left( \frac{\text{dy}}{\text{dx}} \right)\text{=0}$.

Ans: The given differential equation is ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}\text{+cos}\left( \frac{\text{dy}}{\text{dx}} \right)\text{=0}$.

The highest order term is $\frac{{{\text{d}}^{\text{2}}}y}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order is two.

The differential equation contains a trigonometric derivative term and is not completely polynomial in its derivative, thus degree is not defined.

5. Determine order and degree (if defined) of differential equation ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}}\text{-cos3x+sin3x=0}$.

Ans: The given differential equation is ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}\text{-cos3x+sin3x=0}$.

The highest order term is $\frac{{{\text{d}}^{\text{2}}}y}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order is two.

As the derivative is of completely polynomial nature is and highest power of derivative term $\frac{{{\text{d}}^{\text{2}}}y}{\text{d}{{\text{x}}^{\text{2}}}}$ which is one. Thus degree is one.

6. Determine order and degree (if defined) of differential equation ${{\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)}^{\text{2}}}\text{+}{{\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)}^{\text{3}}}\text{+}{{\left( \text{y }\!\!'\!\!\text{ } \right)}^{\text{4}}}\text{+}{{\text{y}}^{\text{5}}}\text{=0}$.

Ans: The given differential equation is ${{\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)}^{\text{2}}}\text{+}{{\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)}^{\text{3}}}\text{+}{{\left( \text{y }\!\!'\!\!\text{ } \right)}^{\text{4}}}\text{+}{{\text{y}}^{\text{5}}}\text{=0}$.

The highest order term is ${{\left( y''' \right)}^{2}}$, thus the order is three.

The differential equation is of the polynomial form and the power of highest order term $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$ is two, thus the degree is two.

7. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ +2y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +y }\!\!'\!\!\text{ =0}$.

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ +2y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +y }\!\!'\!\!\text{ =0}$.

The highest order derivative in the differential equation is $y'''$. Thus its order is three.

The differential equation is polynomial with the highest order term $y'''$ having a degree one. Thus the degree is one.

8. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{ +y=e }\!\!'\!\!\text{ }$.

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{ +y=e }\!\!'\!\!\text{ }$. Therefore:

$\Rightarrow \text{y }\!\!'\!\!\text{ +y-e }\!\!'\!\!\text{ =0}$ 

The highest order derivative in the differential equation is $\text{y }\!\!'\!\!\text{ }$. Thus its order is one.

The given equation is of polynomial form with the highest order term $\text{y }\!\!'\!\!\text{ }$ with degree one. Thus the degree is one.

9. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +}{{\left( \text{y }\!\!'\!\!\text{ } \right)}^{\text{2}}}\text{+2y=0}$.

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +}{{\left( \text{y }\!\!'\!\!\text{ } \right)}^{\text{2}}}\text{+2y=0}$.

The highest order derivative in the differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$. Thus its order is two.

The given equation is of polynomial form with the highest order term $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$ with highest degree one. Thus the degree is one.

10. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +2y }\!\!'\!\!\text{ +siny=0}$ .

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +2y }\!\!'\!\!\text{ +siny=0}$.

The highest order derivative in the differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$. Thus its order is two.

The given equation is of polynomial form with the highest order term $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$ with the highest degree one. Thus the degree is one.

11. Find the degree of the differential equation ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{3}}}\text{+}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}\text{+sin}\left( \frac{\text{dy}}{\text{dx}} \right)\text{+1=0}$.

(A)$\text{3}$

(B)$\text{2}$

(C)$\text{1}$  

(D)not defined

Ans: The given differential equation is ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{3}}}\text{+}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}\text{+sin}\left( \frac{\text{dy}}{\text{dx}} \right)\text{+1=0}$.

The differential equation is not polynomial in its derivative because of the term $\text{sin}\left( \frac{\text{dy}}{\text{dx}} \right)$ thus its order is not defined.

The correct answer is (D).

12. Find the order of the differential equation $\text{2}{{\text{x}}^{\text{2}}}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-3}\frac{\text{dy}}{\text{dx}}\text{+y=0}$.

(A)$\text{2}$

(B)$\text{1}$

(C)$\text{0}$  

(D)not defined

Ans: The given differential equation is $\text{2}{{\text{x}}^{\text{2}}}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-3}\frac{\text{dy}}{\text{dx}}\text{+y=0}$.

The highest order term of the equation is $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order is two.

The correct answer is (A).

Exercise 9.2

1. Verify the function $\text{y=}{{\text{e}}^{\text{x}}}\text{+1}$ is solution of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ -y=0}$ .

Ans: The given function is $\text{y=}{{\text{e}}^{\text{x}}}+1$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( {{e}^{x}}+1 \right)$ 

$\Rightarrow y'={{e}^{x}}$       ……(1)

Take the derivative of the above equation:

$\frac{d}{dx}\left( y' \right)=\frac{d}{dx}\left( {{e}^{x}} \right)$ 

$\Rightarrow y''={{e}^{x}}$ 

Using result from equation (1):

$y''-y'=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ -y }\!\!'\!\!\text{ =0}$.

 2. Verify the function $\text{y=}{{\text{x}}^{\text{2}}}\text{+2x+C}$ is solution of differential equation $\text{y }\!\!'\!\!\text{ -2x-2=0}$ .

Ans: The given function is $\text{y=}{{\text{x}}^{\text{2}}}\text{+2x+C}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( {{x}^{2}}+2x+C \right)$ 

$\Rightarrow y'=2x+2$

$\Rightarrow y'-2x-2=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{ -2x-2=0}$.

3. Verify the function $\text{y=cos x +C}$ is solution of differential equation $\text{y }\!\!'\!\!\text{ +sin x =0}$ .

Ans: The given function is $\text{y=cos x + C}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( \cos x+C \right)$ 

$\Rightarrow y'=-\sin x$

$\Rightarrow y'+\sin x=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  + sin x = 0}$.

4. Verify the function $\text{y=}\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}$ is solution of differential equation $\text{y }\!\!'\!\!\text{ =}\frac{\text{xy}}{\text{1+}{{\text{x}}^{\text{2}}}}$ .

Ans: The given function is $\text{y = }\sqrt{1+{{x}^{2}}}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( \sqrt{1+{{x}^{2}}} \right)$ 

$y'=\frac{1}{2\sqrt{1+{{x}^{2}}}}\times \frac{d}{dx}\left( 1+{{x}^{2}} \right)$

$y'=\frac{2}{2\sqrt{1+{{x}^{2}}}}$ 

$\Rightarrow y'=\frac{1}{\sqrt{1+{{x}^{2}}}}$ 

Multiply numerator and denominator by $\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}$:

$y'=\frac{1}{\sqrt{1+{{x}^{2}}}}\times \frac{\sqrt{1+{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}}$

Substitute $\text{y  = }\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}$ 

$\Rightarrow y'=\frac{xy}{1+{{x}^{2}}}$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  =}\frac{xy}{1+{{x}^{2}}}$.

5. Verify the function $\text{y=Ax}$ is solution of differential equation $\text{xy }\!\!'\!\!\text{ =y}\left( x\ne 0 \right)$ .

Ans: The given function is $\text{y = Ax}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( Ax \right)$ 

$\Rightarrow y'=A$

Multiply by $\text{x}$ on both side:

$xy'=Ax$ 

Substitute $\text{y = Ax}$:

$\Rightarrow xy'=y$ 

Thus the given function is solution of differential equation $xy'=y\left( x\ne 0 \right)$.

6. Verify the function $\text{y=xsin x}$ is solution of differential equation $\text{xy }\!\!'\!\!\text{ =y+x}\sqrt{{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}}\left( x\ne 0\,\text{and}\,x>y\,\text{or}\,x<-y \right)$ .

Ans: The given function is $\text{y = xsin x}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( x\sin x \right)$ 

$\Rightarrow y'=\sin x\frac{d}{dx}\left( x \right)+x\frac{d}{dx}\left( \sin x \right)$

$\Rightarrow y'=\sin x+x\cos x$ 

Multiply by $\text{x}$ on both side:

$xy'=x\left( \sin x+x\cos x \right)$ 

$xy'=x\sin x+{{x}^{2}}\cos x$ 

Substitute $\text{y = xsin x}$:

$\Rightarrow xy'=y+{{x}^{2}}\cos x$ 

Use $\text{sin x = }\frac{\text{y}}{\text{x}}$ and substitute $\text{cos x}$:

$xy'=y+{{x}^{2}}\sqrt{1-{{\sin }^{2}}x}$ 

$xy'=y+{{x}^{2}}\sqrt{1-{{\left( \frac{y}{x} \right)}^{2}}}$ 

$\Rightarrow xy'=y+x\sqrt{{{y}^{2}}-{{x}^{2}}}$ 

Thus the given function is solution of differential equation $\text{xy }\!\!'\!\!\text{  = y + x}\sqrt{{{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}$.

7. Verify the function $\text{xy=log y + C}$ is solution of differential equation $\text{y }\!\!'\!\!\text{ =}\frac{{{\text{y}}^{\text{2}}}}{\text{1-xy}}\left( \text{xy}\ne \text{1} \right)$ .

Ans: The given function is $\text{xy = log y + C}$ .

Take derivative on both side:

$\frac{d}{dx}\left( xy \right)=\frac{d}{dx}\left( \log y \right)$ 

$\Rightarrow y\frac{d}{dx}\left( x \right)+x\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}$

$\Rightarrow y+xy'=\frac{1}{y}y'$ 

$\Rightarrow {{y}^{2}}+xy\,y'=y'$ 

Shift the $\text{y }\!\!'\!\!\text{ }$ terms on one side and take it common. 

$\Rightarrow \left( xy-1 \right)y'=-{{y}^{2}}$ 

$\Rightarrow y'=\frac{{{y}^{2}}}{1-xy}$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  = }\frac{{{\text{y}}^{\text{2}}}}{\text{1-xy}}$.

8. Verify the function $\text{y-cosy=x}$ is solution of differential equation \[\left( \text{ysin y + cos y + x} \right)\text{y }\!\!'\!\!\text{ = 1}\] .

Ans: The given function is $\text{y - cos y = x}$ .

Take derivative on both side:

$\frac{dy}{dx}-\frac{d}{dx}\left( \cos y \right)=\frac{d}{dx}\left( x \right)$ 

$\Rightarrow y'+y'\sin y=1$

$\Rightarrow y'\left( 1+\sin y \right)=1$ 

$\Rightarrow y'=\frac{1}{1+\sin y}$ 

Multiply by \[\left( \text{ysin y + cos y + x} \right)\] on both side:

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{\left( \text{ysin y + cos y + x} \right)}{1+\sin y}\] 

Substitute $y=\cos y+x$ in the numerator:

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{\left( \text{ysin y + y} \right)}{1+\sin y}\] 

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{y\left( \text{sin y + 1} \right)}{1+\sin y}\] 

$\Rightarrow \left( \text{ysin y + cos y + x} \right)y'=y$ 

Thus the given function is solution of differential equation$\left( \text{ysin y + cos y + x} \right)\text{y }\!\!'\!\!\text{  = y}$.

9. Verify the function $\text{x+y=ta}{{\text{n}}^{-1}}\text{y}$ is solution of differential equation \[{{\text{y}}^{\text{2}}}\text{y }\!\!'\!\!\text{ +}{{\text{y}}^{\text{2}}}\text{+1=0}\] .

Ans: The given function is $\text{x + y = ta}{{\text{n}}^{-1}}y$ .

Take derivative on both side:

$\frac{d}{dx}\left( x+y \right)=\frac{d}{dx}\left( ta{{n}^{-1}}y \right)$ 

$1+y'=\left( \frac{1}{1+{{y}^{2}}} \right)y'$

$\Rightarrow y'\left[ \frac{1}{1+{{y}^{2}}}-1 \right]=1$ 

$\Rightarrow y'\left[ \frac{1-\left( 1+{{y}^{2}} \right)}{1+{{y}^{2}}} \right]=1$ 

$\Rightarrow y'\left[ \frac{-{{y}^{2}}}{1+{{y}^{2}}} \right]=1$ 

$\Rightarrow -{{y}^{2}}y'=1+{{y}^{2}}$ 

$\Rightarrow {{y}^{2}}y'+{{y}^{2}}+1=0$ 

Thus the given function is solution of differential equation ${{y}^{2}}y'+{{y}^{2}}+1=0$.

10. Verify the function $\text{y=}\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}\text{x}\in \left( \text{-a,a} \right)$ is solution of differential equation \[\text{x+y}\frac{\text{dy}}{\text{dx}}\text{=0}\left( \text{y}\ne \text{0} \right)\] .

Ans: The given function is:

\[y\text{ }=\text{ }\sqrt{{{a}^{2}}-{{x}^{2}}}\,x\in \left( -a,a \right)\] .

Take derivative on both side:

$\frac{dy}{dx}=\frac{d}{dx}\left( \sqrt{{{a}^{2}}-{{x}^{2}}} \right)$ 

$\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{{{a}^{2}}-{{x}^{2}}}}\frac{d}{dx}\left( {{a}^{2}}-{{x}^{2}} \right)$

$\Rightarrow \frac{dy}{dx}=\frac{-2x}{2\sqrt{{{a}^{2}}-{{x}^{2}}}}$ 

$\Rightarrow \frac{dy}{dx}=\frac{-x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}$ 

Substitute $\text{y = }\sqrt{{{\text{a}}^{\text{2}}}\text{ - }{{\text{x}}^{\text{2}}}}$ 

$\Rightarrow \frac{dy}{dx}=\frac{-x}{y}$ 

$\Rightarrow x+y\frac{dy}{dx}=0$ 

Thus the given function is solution of differential equation:

$x+y\frac{dy}{dx}=0\left( y\ne 0 \right)$.

11. Find the numbers of arbitrary constants in the general solution of a differential equation of fourth order.

(A)$\text{0}$

(B)$\text{2}$

(C)$\text{3}$  

(D)$\text{4}$ 

Ans: The number of arbitrary constants in the general solution of a differential equation is equal to its order. As the given differential equation is of fourth order, thus it has four arbitrary constants in its solution.

The correct answer is (D).

12. Find the numbers of arbitrary constants in the particular solution of a differential equation of third order.

(A)$\text{3}$

(B)$\text{2}$

(C)$\text{1}$  

(D)$\text{0}$ 

Ans: The particular solution of any differential equation does not have any arbitrary constants. Thus it has zero constants in its solution.

The correct answer is (D).

Exercise 9.3

1. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$.

Use trigonometric half –angle identities to simplify:
$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{\sin }^{2}}\frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}}$ 

$\Rightarrow \frac{dy}{dx}={{\tan }^{2}}\frac{x}{2}$ 

$\Rightarrow \frac{dy}{dx}={{\sec }^{2}}\frac{x}{2}-1$ 

Separate the differentials and integrate:

$\int dy=\int \left( {{\sec }^{2}}\frac{x}{2}-1 \right)dx$ 

$\Rightarrow y=\int {{\sec }^{2}}\frac{x}{2}dx-\int dx$ 

$\Rightarrow y=2\tan \frac{x}{2}-x+C$ 

Thus the general solution of given differential equation is $\text{y = 2tan}\frac{\text{x}}{\text{2}}\text{ - x + C}$.

2. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{{\text{y}}^{\text{2}}}}\left( \text{-2  y  2} \right)$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{{\text{y}}^{\text{2}}}}\left( \text{-2  y  2} \right)$.

Simplify the expression:

$\frac{dy}{dx}=\sqrt{4-{{y}^{2}}}$ 

$\Rightarrow \frac{dy}{\sqrt{4-{{y}^{2}}}}=dx$ 

Use standard integration:

$\int \frac{dy}{\sqrt{4-{{y}^{2}}}}=\int dx$ 

$\Rightarrow {{\sin }^{-1}}\frac{y}{2}=x+C$ 

$\Rightarrow \frac{y}{2}=\sin \left( x+C \right)$ 

$\Rightarrow y=2\sin \left( x+C \right)$ 

Thus the general solution of given differential equation is $\text{y = 2 sin}\left( \text{x + C} \right)$.

3. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{+ y =1}\left( \text{y}\ne \text{1} \right)$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{+ y =1}\left( \text{y}\ne \text{1} \right)$.

Simplify the expression:

$\frac{dy}{dx}+y=1$ 

$\Rightarrow \frac{dy}{dx}=1-y$ 

$\Rightarrow \frac{dy}{1-y}=dx$ 

Use standard integration:

$\int \frac{dy}{1-y}=\int dx$ 

$\Rightarrow -\log \left( 1-y \right)=x+C$ 

$\Rightarrow \log \left( 1-y \right)=-\left( x+C \right)$ 

$\Rightarrow 1-y={{e}^{-\left( x+C \right)}}$ 

$y=1-A{{e}^{-x}}\,\left( A={{e}^{-C}} \right)$ 

Thus the general solution of given differential equation is $\text{y=1- A}{{\text{e}}^{\text{-x}}}\,$.

4. Find the general solution for $\text{se}{{\text{c}}^{\text{2}}}\text{x}\,\text{tany}\,\text{dx+se}{{\text{c}}^{\text{2}}}\text{y}\,\text{tanx}\,\text{dy=0}$.

Ans: The given differential equation is:

${{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy=0$.

Divide both side by $\text{tan x tan y}$:

$\frac{{{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy}{\tan x\tan y}=0$ 

\[\Rightarrow \frac{{{\sec }^{2}}x}{\tan x}dx+\frac{{{\sec }^{2}}y}{\tan y}dy=0\] 

Integrate both side:

$\int \frac{{{\sec }^{2}}x}{\tan x}dx+\int \frac{{{\sec }^{2}}y}{\tan y}dy=0$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=-\int \frac{{{\sec }^{2}}x}{\tan x}dx$       ……(1)

Use a substitution method for integration. Substitute $\text{tan x = u}$:

For integral on RHS:

$\Rightarrow \tan x=u$ 

$\Rightarrow {{\sec }^{2}}xdx=du$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\log u$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\log \left( \tan x \right)$ 

Thus evaluating result form (1):

$\Rightarrow \log \left( \tan y \right)=-\log \left( \tan x \right)+\log \left( C \right)$ 

$\Rightarrow \log \left( \tan y \right)=\log \left( \frac{C}{\tan x} \right)$ 

$\Rightarrow \tan x\tan y=C$ 

Thus the general solution of given differential equation is $\text{tan x tan y = C}$.

5. Find the general solution for $\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{dy-}\left( {{\text{e}}^{\text{x}}}\text{-}{{\text{e}}^{\text{-x}}} \right)\text{dx=0}$ .

Ans: The given differential equation is:

$\left( {{e}^{x}}+{{e}^{-x}} \right)dy-\left( {{e}^{x}}-{{e}^{-x}} \right)dx=0$.

Simplify the expression:

$dy=\left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx$ 

Integrate both side:

$\int dy=\int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx$ ……(1)

Use a substitution method for integration. Substitute \[{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}}\text{=t}\]:

For integral on RHS:

$\Rightarrow {{e}^{x}}+{{e}^{-x}}=t$ 

$\Rightarrow \left( {{e}^{x}}-{{e}^{-x}} \right)dx=dt$ 

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\int \frac{dt}{t}$ s

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\ln t+C$ 

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\log \left( {{e}^{x}}+{{e}^{-x}} \right)+C$ 

Thus evaluating result form (1):

$y=\log \left( {{e}^{x}}+{{e}^{-x}} \right)+C$ 

Thus the general solution of given differential equation is $\text{y=log}\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{+C}$.

6. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\left( \text{1+}{{\text{y}}^{\text{2}}} \right)$.

Ans: The given differential equation is:

$\frac{dy}{dx}=\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)$.

Simplify the expression:

$\frac{dy}{1+{{y}^{2}}}=\left( 1+{{x}^{2}} \right)dx$ 

Integrate both side:

$\int \frac{dy}{1+{{y}^{2}}}=\int \left( 1+{{x}^{2}} \right)dx$

Use standard integration:

${{\tan }^{-1}}y=\int dx+\int {{x}^{2}}dx$ 

$\Rightarrow {{\tan }^{-1}}y=x+\frac{{{x}^{3}}}{3}+C$ 

Thus the general solution of given differential equation is $\text{ta}{{\text{n}}^{\text{-1}}}\text{y=x+}\frac{{{\text{x}}^{\text{3}}}}{\text{3}}\text{+C}$.

7. Find the general solution for $\text{ylog y dx-xdy=0}$.

Ans: The given differential equation is:

$y\log ydx-xdy=0$.

Simplify the expression:

$y\log ydx=xdy$ 

$\Rightarrow \frac{dx}{x}=\frac{dy}{y\log y}$ 

Integrate both side:

$\int \frac{dy}{y\log y}=\int \frac{dx}{x}$ ……(1)

Use substitution method for integration on LHS. Substitute \[\text{log y=t}\]:

$\log y=t$ 

$\Rightarrow \frac{1}{y}dy=dt$ 

$\Rightarrow \int \frac{dy}{y\log y}=\int \frac{dt}{t}$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log t$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log \left( \log y \right)$ 

Evaluating expression (1):

$\log \left( \log y \right)=\log x+\log C$ 

$\Rightarrow \log \left( \log y \right)=\log Cx$ 

$\Rightarrow \log y=Cx$ 

$\Rightarrow y={{e}^{Cx}}$ 

Thus the general solution of given differential equation is $\text{y=}{{\text{e}}^{\text{Cx}}}$.

8. Find the general solution for ${{\text{x}}^{\text{5}}}\frac{\text{dy}}{\text{dx}}\text{=-}{{\text{y}}^{\text{5}}}$.

Ans: The given differential equation is:

${{x}^{5}}\frac{dy}{dx}=-{{y}^{5}}$.

Simplify the expression:

$\frac{dy}{{{y}^{5}}}=-\frac{dx}{{{x}^{5}}}$ 

Integrate both side:

$\int \frac{dy}{{{y}^{5}}}=-\int \frac{dx}{{{x}^{-5}}}$ 

\[\Rightarrow \int {{y}^{-5}}dy=-\int {{x}^{-5}}dx\] 

$\Rightarrow \frac{{{y}^{-5+1}}}{-5+1}=-\frac{{{x}^{-5+1}}}{-5+1}+C$ 

$\Rightarrow \frac{{{y}^{-4}}}{-4}=-\frac{{{x}^{-4}}}{-4}+C$ 

$\Rightarrow {{x}^{-4}}+{{y}^{-4}}=-4C$ 

$\Rightarrow {{x}^{-4}}+{{y}^{-4}}=A\,\,\,\,\left( A=-4C \right)$ 

Thus the general solution of given differential equation is ${{\text{x}}^{\text{-4}}}\text{+}{{\text{y}}^{\text{-4}}}\text{=A}$.

9. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=si}{{\text{n}}^{\text{-1}}}\text{x}$.

Ans: The given differential equation is:

$\frac{dy}{dx}={{\sin }^{-1}}x$.

Simplify the expression:

$dy={{\sin }^{-1}}xdx$ 

Integrate both side:

$\int dy=\int {{\sin }^{-1}}xdx$ 

\[\Rightarrow y=\int 1\times {{\sin }^{-1}}xdx\] 

Use product rule of integration:

$\int {{\sin }^{-1}}xdx={{\sin }^{-1}}x\int dx-\int \left( \frac{1}{\sqrt{1-{{x}^{2}}}}\int dx \right)dx$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x-\int \frac{x}{\sqrt{1-{{x}^{2}}}}dx$ 

Substitute $\text{1-}{{\text{x}}^{\text{2}}}\text{=}{{\text{t}}^{\text{2}}}$ 

$1-{{x}^{2}}={{t}^{2}}$ 

$\Rightarrow -2xdx=2tdt$ 

$\Rightarrow -xdx=tdt$ 

Evaluating the integral:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+\int \frac{tdt}{\sqrt{{{t}^{2}}}}$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+t+C$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$ 

$\Rightarrow y=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$ 

Thus the general solution of given differential equation is $\text{y=xsi}{{\text{n}}^{\text{-1}}}\text{x+}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{+C}$,

10. Find the general solution for ${{\text{e}}^{\text{x}}}\text{tanydx+}\left( \text{1-}{{\text{e}}^{\text{x}}} \right)\text{se}{{\text{c}}^{\text{2}}}\text{ydy=0}$.

Ans: The given differential equation is:

${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$.

Simplify the expression:

$\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=-{{e}^{x}}\tan ydx$ 

$\Rightarrow \frac{{{\sec }^{2}}y}{\tan y}dy=-\frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx$ 

Integrate both side:

$\int \frac{{{\sec }^{2}}y}{\tan y}dy=-\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx$ ……(1)

Substitute $\text{tan y=u}$ 

$\tan y=u$ 

\[\Rightarrow {{\sec }^{2}}y=du\] 

Evaluating the LHS integral of (1):

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\log u$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\log \left( \tan y \right)$ 

Substitute $\text{1-}{{\text{e}}^{x}}\text{=v}$ 

$1-{{e}^{x}}=v$ 

$\Rightarrow -{{e}^{x}}dx=dv$ 

Evaluating the RHS integral of (1):

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\int \frac{dv}{v}$ 

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\log v$ 

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\log \left( 1-{{e}^{x}} \right)$ 

Therefore the integral (1) will be:

$\log \left( \tan y \right)=\log \left( 1-{{e}^{x}} \right)+\log C$ 

$\Rightarrow \log \left( \tan y \right)=\log C\left( 1-{{e}^{x}} \right)$ 

$\Rightarrow \tan y=C\left( 1-{{e}^{x}} \right)$ 

Thus the general solution of given differential equation is $\text{tany=C}\left( \text{1-}{{\text{e}}^{\text{x}}} \right)$,

11. Find the particular solution of $\left( {{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}}\text{+x+1} \right)\frac{\text{dy}}{\text{dx}}\text{=2}{{\text{x}}^{\text{2}}}\text{+x;}\,\,\text{y=1,}\,\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)\frac{dy}{dx}=2{{x}^{2}}+x;\,\,y=1,\,x=0$.

Simplify the expression:

$\frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( {{x}^{3}}+x+{{x}^{2}}+1 \right)}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}$ 

$\Rightarrow dy=\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}dx$ 

Integrate both side:

$\int dy=\int \frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}dx$ ……(1)

Use partial fraction method to simplify the RHS:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$

$\Rightarrow 2{{x}^{2}}+x=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+1 \right)$ 

\[\Rightarrow 2{{x}^{2}}+x=\left( A+B \right){{x}^{2}}+\left( B+C \right)x+\left( A+C \right)\] 

By comparing coefficients:

$A+B=2$ 

$B+C=1$ 

$A+C=0$ 

Solving this we get:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{\left( \frac{1}{2} \right)}{x+1}+\frac{\left( \frac{3}{2} \right)x+\left( -\frac{1}{2} \right)}{{{x}^{2}}+1}$ 

$\Rightarrow \frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{1}{2}\left( \frac{1}{x+1}+\frac{3x-1}{{{x}^{2}}+1} \right)$ 

Rewriting the  integral(1):

$y=\frac{1}{2}\int \left( \frac{1}{x+1}+\frac{3x-1}{{{x}^{2}}+1} \right)dx$ 

\[\Rightarrow y=\frac{1}{2}\int \frac{1}{x+1}dx+\frac{1}{2}\int \frac{3x-1}{{{x}^{2}}+1}dx\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{2}\int \frac{x}{{{x}^{2}}+1}dx-\frac{1}{2}\int \frac{1}{{{x}^{2}}+1}dx\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\int \frac{2x}{{{x}^{2}}+1}dx-\frac{1}{2}{{\tan }^{-1}}x\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\log \left( {{x}^{2}}+1 \right)-\frac{1}{2}{{\tan }^{-1}}x+C\] 

For $y=1$ when $\text{x=0}$. 

\[1=\frac{1}{2}\log \left( 0+1 \right)+\frac{3}{4}\log \left( 0+1 \right)-\frac{1}{2}{{\tan }^{-1}}0+C\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is :

$\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\log \left( {{x}^{2}}+1 \right)-\frac{1}{2}{{\tan }^{-1}}x+1$.

12. Find the particular solution of $\text{x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)\frac{\text{dy}}{\text{dx}}\text{=1}$; $\text{y=0}$ when $\text{x=2}$ to satisfy the given condition.

Ans: The given differential equation is:

$x\left( {{x}^{2}}-1 \right)\frac{dy}{dx}=1$; $y=0$ when $x=2$ 

Simplify the expression:

$x\left( {{x}^{2}}-1 \right)\frac{dy}{dx}=1$ 

$\Rightarrow dy=\frac{dx}{x\left( {{x}^{2}}-1 \right)}$ 

$\Rightarrow dy=\frac{dx}{x\left( x-1 \right)\left( x+1 \right)}$ 

Integrate both side:

$\int dy=\int \frac{dx}{x\left( x-1 \right)\left( x+1 \right)}$ ……(1)

Use partial fraction method to simplify the RHS:

$\frac{1}{x\left( x-1 \right)\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$

$\Rightarrow 1=A\left( {{x}^{2}}-1 \right)+Bx\left( x+1 \right)+Cx\left( x-1 \right)$ 

\[\Rightarrow 1=\left( A+B+C \right){{x}^{2}}+\left( B-C \right)x-A\] 

By comparing coefficients:

$A+B+C=0$ 

$B-C=0$ 

$-A=1$ 

Solving this we get:

$\frac{1}{x\left( x-1 \right)\left( x+1 \right)}=\frac{\left( -1 \right)}{x}+\frac{\left( \frac{1}{2} \right)}{x-1}+\frac{\left( \frac{1}{2} \right)}{x+1}$ 

$\Rightarrow \frac{1}{x\left( x-1 \right)\left( x+1 \right)}=-\frac{1}{x}+\frac{1}{2}\left( \frac{1}{x-1}+\frac{1}{x+1} \right)$ 

Rewriting the  integral(1):

$y=\int \left( -\frac{1}{x}+\frac{1}{2}\left( \frac{1}{x-1}+\frac{1}{x+1} \right) \right)dx$ 

\[\Rightarrow y=-\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{x-1}dx+\frac{1}{2}\int \frac{1}{x+1}dx\] 

\[\Rightarrow y=-\log x+\frac{1}{2}\log \left( x-1 \right)+\frac{1}{2}\log \left( x+1 \right)+\log C\] 

\[\Rightarrow y=-\frac{2}{2}\log x+\frac{1}{2}\log \left( x-1 \right)+\frac{1}{2}\log \left( x+1 \right)+\frac{2}{2}\log C\] 

\[\Rightarrow y=\frac{1}{2}\left( -\log {{x}^{2}}+\log \left( x-1 \right)+\log \left( x+1 \right)+\log {{C}^{2}} \right)\] 

\[\Rightarrow y=\frac{1}{2}\log \left[ \frac{{{C}^{2}}\left( {{x}^{2}}-1 \right)}{{{x}^{2}}} \right]\] 

For $y=0$ when $\text{x=2}$. 

\[0=\frac{1}{2}\log \left[ \frac{{{C}^{2}}\left( {{2}^{2}}-1 \right)}{{{2}^{2}}} \right]\] 

\[\Rightarrow 0=\log \left[ \frac{3{{C}^{2}}}{4} \right]\] 

$\Rightarrow \frac{3{{C}^{2}}}{4}=1$ 

$\Rightarrow {{C}^{2}}=\frac{4}{3}$ 

Thus the required particular solution is :

$y=\frac{1}{2}\log \left[ \frac{4\left( {{x}^{2}}-1 \right)}{3{{x}^{2}}} \right]$.

13. Find the particular solution of $\text{cos}\left( \frac{\text{dy}}{\text{dx}} \right)\text{=a}\,\,\left( \text{a}\in \text{R} \right)$; $\text{y=1}$ when $\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\cos \left( \frac{dy}{dx} \right)=a\,\,\left( a\in R \right)$; $y=1$ when $x=0$

Simplify the expression:

$\cos \left( \frac{dy}{dx} \right)=a\,$ 

$\Rightarrow \frac{dy}{dx}={{\cos }^{-1}}a$ 

$\Rightarrow dy={{\cos }^{-1}}adx$ 

Integrate both side:

$\int dy=\int {{\cos }^{-1}}adx$ 

$\Rightarrow y={{\cos }^{-1}}a\int dx$ 

\[\Rightarrow y=x{{\cos }^{-1}}a+C\] 

For $y=1$ when $\text{x=0}$. 

\[1=0{{\cos }^{-1}}a+C\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is:

\[y=x{{\cos }^{-1}}a+1\].

\[\Rightarrow \frac{y-1}{x}={{\cos }^{-1}}a\]

\[\Rightarrow \cos \left( \frac{y-1}{x} \right)= a\]

14. Find the particular solution of $\frac{\text{dy}}{\text{dx}}\text{=ytanx}$; $\text{y=1}$ when $\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\frac{dy}{dx}=y\tan x$; $y=1$ when $x=0$

Simplify the expression:

$\frac{dy}{dx}=y\tan x$

$\Rightarrow \frac{dy}{y}=\tan xdx$ 

Integrate both side:

$\int \frac{dy}{y}=\int \tan xdx$ 

$\Rightarrow \log y=\log \left( \sec x \right)+\log C$ 

\[\Rightarrow \log y=\log \left( C\sec x \right)\] 

$\Rightarrow y=C\sec x$ 

For $\text{y=1}$ when $\text{x=0}$. 

\[1=C\sec 0\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is :

\[y=\sec x\]. 

15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is $\text{y }\!\!'\!\!\text{ =}{{\text{e}}^{\text{x}}}\text{sin x}$.

Ans: The given differential equation is:

$y'={{e}^{x}}\sin x$

The curve passes through $\left( 0,0 \right)$.

Simplify the expression:

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\sin x$

$\Rightarrow dy={{e}^{x}}\sin xdx$ 

Integrate both side:

$\int dy=\int {{e}^{x}}\sin xdx$ 

Use product rules for integration of RHS. Let:

$I=\int {{e}^{x}}\sin xdx$ 

$\Rightarrow I=\sin x\int {{e}^{x}}dx-\int \left( \cos x\int {{e}^{x}}dx \right)dx$ 

$\Rightarrow I={{e}^{x}}\sin x-\int {{e}^{x}}\cos xdx$ 

$\Rightarrow I={{e}^{x}}\sin x-\left( \cos x\int {{e}^{x}}dx+\int \left( \sin x\int {{e}^{x}}dx \right)dx \right)$ 

$\Rightarrow I={{e}^{x}}\sin x-{{e}^{x}}\cos x-\int \left( {{e}^{x}}\sin x \right)dx$ 

$\Rightarrow I={{e}^{x}}\sin x-{{e}^{x}}\cos x-I$ 

$\Rightarrow I=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x \right)$ 

Thus integral will be:

$y=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x \right)+C$ 

Thus as the curve passes through $\left( \text{0,0} \right)$ 

$0=\frac{{{e}^{0}}}{2}\left( \sin 0-\cos 0 \right)+C$

$\Rightarrow 0=\frac{1}{2}\left( 0-1 \right)+C$ 

$\Rightarrow C=\frac{1}{2}$ 

Thus the equation of the curve will be:

$y=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x \right)+\frac{1}{2}$ 

$\Rightarrow y=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x+1 \right)$ 

16. For the differential equation $\text{xy}\frac{\text{dy}}{\text{dx}}\text{=}\left( \text{x+2} \right)\left( \text{y+2} \right)$ find the solution curve passing through the point $\left( \text{1,-1} \right)$.

Ans: The given differential equation is:

$xy\frac{dy}{dx}=\left( x+2 \right)\left( y+2 \right)$

The curve passes through $\left( 1,-1 \right)$.

Simplify the expression:

\[\Rightarrow \left( \frac{y}{y+2} \right)dy=\frac{\left( x+2 \right)}{x}dx\]

$\Rightarrow \left( 1-\frac{2}{y+2} \right)dy=\frac{\left( x+2 \right)}{x}dx$ 

Integrate both side:

$\int \left( 1-\frac{2}{y+2} \right)dy=\int \frac{\left( x+2 \right)}{x}dx$ 

$\Rightarrow \int dy-2\int \frac{1}{y+2}dy=\int \frac{x}{x}dx+\int \frac{2}{x}dx$ 

$\Rightarrow y-2\log \left( y+2 \right)=x+2\log x+C$ 

$\Rightarrow y-x=2\log \left( y+2 \right)+2\log x+C$ 

$\Rightarrow \Rightarrow y-x=2\log \left[ x\left( y+2 \right) \right]+C$ 

$\Rightarrow y-x=\log \left[ {{x}^{2}}{{\left( y+2 \right)}^{2}} \right]+C$ 

Thus as the curve passes through $\left( \text{1,-1} \right)$ 

$\Rightarrow -1-1=\log \left[ {{\left( 1 \right)}^{2}}{{\left( -1+2 \right)}^{2}} \right]+C$

$\Rightarrow -2=\log 1+C$ 

$\Rightarrow C=-2$ 

Thus the equation of the curve will be:

$y-x=\log \left[ {{x}^{2}}{{\left( y+2 \right)}^{2}} \right]-2$ 

$\Rightarrow y-x+2=\log \left( {{x}^{2}}{{\left( y+2 \right)}^{2}} \right)$ 

17. Find the equation of a curve passing through the point $\left( \text{0,-2} \right)$ given that at any point $\left( \text{x,y} \right)$ on the curve, the product of the slope of its tangent and $\text{y}$-coordinate of the point is equal to the $\text{x}$ -coordinate of the point.

Ans: According to , the equation is given by:

$y\frac{dy}{dx}=x$ 

The curve passes through $\left( 0,-2 \right)$.

Simplify the expression:

\[\Rightarrow ydy=xdx\] 

Integrate both side:

$\int ydy=\int xdx$ 

$\Rightarrow \frac{{{y}^{2}}}{2}=\frac{{{x}^{2}}}{2}+C$ 

$\Rightarrow {{y}^{2}}-{{x}^{2}}=2C$ 

Thus as the curve passes through $\left( \text{0,-2} \right)$ 

$\Rightarrow {{\left( -2 \right)}^{2}}-{{0}^{2}}=2C$

$\Rightarrow 4=2C$ 

$\Rightarrow C=2$ 

Thus the equation of the curve will be:

${{y}^{2}}-{{x}^{2}}=2\left( 2 \right)$ 

$\Rightarrow {{y}^{2}}-{{x}^{2}}=4$ .

18. At any point $\left( \text{x,y} \right)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point$\left( \text{-4,-3} \right)$ . Find the equation of the curve given that it passes through $\left( \text{-2,1} \right)$.

Ans: Let the point of contact of the tangent be $\left( \text{x,y} \right)$.Then the slope of the segment joining point of contact and $\left( \text{-4,-3} \right)$:

$m=\frac{y+3}{x+4}$ 

According to  the for the slope of tangent $\frac{\text{dy}}{\text{dx}}$ it follows:

$\frac{dy}{dx}=2m$ 

$\Rightarrow \frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)$

Simplify the expression:

$\frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)$ 

$\frac{dy}{y+3}=\frac{2}{x+4}dx$ 

Integrate both side:

$\int \frac{dy}{y+3}=\int \frac{2}{x+4}dx$ 

$\Rightarrow \log \left( y+3 \right)=2\log \left( x+4 \right)+\log C$ 

\[\Rightarrow \log \left( y+3 \right)=\log {{\left( x+4 \right)}^{2}}+\log C\] 

\[\Rightarrow \log \left( y+3 \right)=\log C{{\left( x+4 \right)}^{2}}\] 

\[\Rightarrow y+3=C{{\left( x+4 \right)}^{2}}\] 

Thus as the curve passes through $\left( \text{-2,1} \right)$ 

\[1+3=C{{\left( -2+4 \right)}^{2}}\]

$\Rightarrow 4=4C$ 

$\Rightarrow C=1$ 

Thus the equation of the curve will be:

\[y+3={{\left( x+4 \right)}^{2}}\] .

19. The volume of spherical balloons being inflated changes at a constant rate. If initially its radius is $\text{3}$ units and after \[\text{3}\] seconds it is $\text{6}$ units. Find the radius of balloon after $\text{t}$ seconds.

Ans: Let the volume of spherical balloon be $\text{V}$ and its radius $\text{r}$. Let the rate of change of volume be $\text{k}$.

$\frac{dV}{dt}=k$ 

$\Rightarrow \frac{d}{dt}\left( \frac{4}{3}\pi {{r}^{3}} \right)=k$ 

$\Rightarrow \frac{4}{3}\pi \frac{d}{dt}\left( {{r}^{3}} \right)=k$ 

$\Rightarrow \frac{4}{3}\pi \left( 3{{r}^{2}} \right)\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi {{r}^{2}}\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi {{r}^{2}}dr=kdt$ 

Integrate both side:
$\int 4\pi {{r}^{2}}dr=\int kdt$ 

$\Rightarrow 4\pi \int {{r}^{2}}dr=kt+C$ 

$\Rightarrow \frac{4}{3}\pi {{r}^{3}}=kt+C$ 

At initial time, $\text{t=0}$ and $\text{r=3}$:

$\frac{4}{3}\pi {{3}^{3}}=k\left( 0 \right)+C$ 

$\Rightarrow C=36\pi $

At $\text{t=3}$ the radius $\text{r=6}$:

$\frac{4}{3}\pi \left( {{6}^{3}} \right)=k\left( 3 \right)+36\pi $ 

$\Rightarrow 3k=288\pi -36\pi $ 

$\Rightarrow k=84\pi $ 

Thus the radius-time relation can be given by:

$\frac{4}{3}\pi {{r}^{3}}=84\pi t+36\pi$ 

$\Rightarrow {{r}^{3}}=63t+27$ 

$\Rightarrow r={{\left( 63t+27 \right)}^{\frac{1}{3}}}$

The radius of balloon after $\text{t}$ seconds is given by: $r={{\left( 63t+27 \right)}^{\frac{1}{3}}}$ .

.

20. In a bank, principal increases continuously at the rate of $\text{r}$% per year. Find the value of $\text{r}$ if Rs. $\text{100}$doubles itself in $\text{10}$ years $\left( \text{lo}{{\text{g}}_{\text{e}}}\text{2=0}\text{.6931} \right)$.

Ans: Let the principal be $\text{p}$, according to :

$\frac{dp}{dt}=\left( \frac{r}{100} \right)p$ 

Simplify the expression:

$\frac{dp}{p}=\left( \frac{r}{100} \right)dt$ 

Integrate both side:
$\int \frac{dp}{p}=\int \left( \frac{r}{100} \right)dt$ 

$\Rightarrow \log p=\frac{rt}{100}+c$ 

\[\Rightarrow p={{e}^{\frac{rt}{100}+c}}\] 

\[\Rightarrow p=A{{e}^{\frac{rt}{100}}}\,\,\,\left( A={{e}^{c}} \right)\] 

At $\text{t=0}$, $\text{p=100}$:

\[100=A{{e}^{\frac{r\left( 0 \right)}{100}}}\] 

$\Rightarrow A=100$ 

Thus the principle and rate of interest relation:

$p=100{{e}^{\frac{rt}{100}}}$ 

At $\text{t=10}$, $\text{p=2}\times \text{100=200}$:

\[200=100{{e}^{\frac{r\left( 10 \right)}{100}}}\] 

$\Rightarrow 2={{e}^{\frac{r}{10}}}$ 

Take logarithm on both side:

$\log \left( {{e}^{\frac{r}{10}}} \right)=\log \left( 2 \right)$ 

$\Rightarrow \frac{r}{10}=0.6931$ 

$\Rightarrow r=6.931$ 

Thus the rate of interest $\text{r=6}\text{.931 }\!\!%\!\!\text{ }$. 

21. In a bank, principal increases continuously at the rate of $\text{5}$% per year. An amount of Rs $\text{1000}$ is deposited with this bank, how much will it worth after $\text{10}$  years $\left( {{\text{e}}^{\text{0}\text{.5}}}\text{=1}\text{.648} \right)$.

Ans: Let the principal be $\text{p}$, according to  principle increases at the rate of $\text{5}$% per year.

$\frac{dp}{dt}=\left( \frac{5}{100} \right)p$ 

Simplify the expression:

$\frac{dp}{dt}=\frac{p}{20}$ 

$\Rightarrow \frac{dp}{p}=\frac{1}{20}dt$ 

Integrate both side:

$\int \frac{dp}{p}=\int \frac{1}{20}dt$ 

$\Rightarrow \log p=\frac{t}{20}+C$ 

$\Rightarrow p={{e}^{\frac{t}{20}+C}}$ 

$\Rightarrow p=A{{e}^{\frac{t}{20}}}\,\,\left( A={{e}^{C}} \right)$
At $\text{t=0}$, $\text{p=1000}$: 

$1000=A{{e}^{\frac{0}{20}}}$ 

$\Rightarrow A=1000$ 

Thus the relation of principal and time relation:

$\Rightarrow p=1000{{e}^{\frac{t}{20}}}$ 

At $\text{t=10}$: 

$p=1000{{e}^{\frac{10}{20}}}$ 

$\Rightarrow p=1000{{e}^{0.5}}$

$\Rightarrow p=1000\times 1.648$ 

$\Rightarrow p=1648$  

Thus after $\text{10}$ this year the amount will become Rs. $\text{1648}$.

22. In a culture, the bacteria count is $\text{100000}$ . The number is increased by $\text{10 }\!\!%\!\!\text{ }$  in $\text{2}$ hours. In how many hours will the count reach $\text{200000}$, if the rate of growth of bacteria is proportional to the number present?

Ans: Let the number of bacteria be $\text{y}$ at time $\text{t}$. According to :

$\frac{dy}{dt}\propto y$ 

$\frac{dy}{dt}=cy$ 

Here $\text{c}$ is constant.

Simplify the expression:
$\frac{dy}{y}=cdt$ 

Integrate both side:
$\int \frac{dy}{y}=\int cdt$ 

$\Rightarrow \log y=ct+D$ 

$\Rightarrow y={{e}^{ct+D}}$ 

$\Rightarrow y=A{{e}^{ct}}\,\,\left( A={{e}^{D}} \right)$ 

At $\text{t=0}$, $\text{y=100000}$: 

$100000=A{{e}^{c\left( 0 \right)}}$ 

$\Rightarrow A=100000$ 

At $\text{t=2}$, $\text{y=}\frac{11}{10}\left( \text{100000} \right)=110000$: 

$y=100000{{e}^{ct}}$ 

$\Rightarrow 110000=100000{{e}^{c\left( 2 \right)}}$ 

$\Rightarrow {{e}^{2c}}=\frac{11}{10}$ 

$\Rightarrow 2c=\log \left( \frac{11}{10} \right)$ 

$\Rightarrow c=\frac{1}{2}\log \left( \frac{11}{10} \right)$ ……(1)

For $\text{y=200000}$:

$200000=100000{{e}^{ct}}$ 

$\Rightarrow {{e}^{ct}}=2$ 

$\Rightarrow ct=\log 2$ 

$\Rightarrow t=\frac{\log 2}{c}$ 

Back substituting using expression (1):
\[t=\frac{\log 2}{\frac{1}{2}\log \left( \frac{11}{10} \right)}\] 

$\Rightarrow t=\frac{2\log 2}{\log \left( \frac{11}{10} \right)}$ 

Thus time required for bacteria to reach $\text{200000}$ is \[\text{t=}\frac{\text{log2}}{\frac{\text{1}}{\text{2}}\text{log}\left( \frac{\text{11}}{\text{10}} \right)}\] hrs.

23. Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x+y}}}$.

(A)${{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-y}}}\text{=C}$ 

(B) ${{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{y}}}\text{=C}$

(C) ${{\text{e}}^{\text{-x}}}\text{+}{{\text{e}}^{\text{y}}}\text{=C}$

(D) ${{\text{e}}^{\text{-x}}}\text{+}{{\text{e}}^{\text{-y}}}\text{=C}$

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x+y}}}$. Simplify the expression:

$\frac{dy}{dx}={{e}^{x}}{{e}^{y}}$ 

$\Rightarrow \frac{dy}{{{e}^{y}}}={{e}^{x}}dx$ 

$\Rightarrow {{e}^{-y}}dy={{e}^{x}}dx$ 

Integrate both side:

$\int {{e}^{-y}}dy=\int {{e}^{x}}dx$ 

$\Rightarrow -{{e}^{-y}}={{e}^{x}}+D$ 

$\Rightarrow {{e}^{x}}+{{e}^{-y}}=-D$ 

$\Rightarrow {{e}^{x}}+{{e}^{-y}}=C\,\,\left( C=-D \right)$ 

Thus the general solution of given differential equation is $ {{e}^{x}}+{{e}^{-y}}=C$ 

Thus the correct option is (A).

Exercise 9.4

1. Show that, differential equation $\left( {{\text{x}}^{\text{2}}}\text{+xy} \right)\text{dy=}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)\text{dx}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+xy}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+xy}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\left( \lambda x \right)}^{2}}+{{\left( \lambda y \right)}^{2}}}{{{\left( \lambda x \right)}^{2}}+\left( \lambda x \right)\left( \lambda y \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\lambda }^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)}{{{\lambda }^{2}}\left( {{x}^{2}}+xy \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+xy}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{{{x}^{2}}+{{\left( vx \right)}^{2}}}{{{x}^{2}}+x\left( vx \right)}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left( 1+{{v}^{2}} \right)}{{{x}^{2}}\left( 1+v \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{1+{{v}^{2}}}{1+v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1+{{v}^{2}}}{1+v}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1+{{v}^{2}}-v-{{v}^{2}}}{1+v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-v}{1+v}$ 

Separate the differentials:

$\frac{1+v}{1-v}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1+v}{1-v}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{2-\left( 1-v \right)}{1-v}dv=\log x-\log k$ 

$\Rightarrow \int \frac{2}{1-v}dv-\int \frac{1-v}{1-v}dv=\log \frac{x}{k}$ 

$\Rightarrow -2\log \left( 1-v \right)-\int dv=\log \frac{x}{k}$ 

$\Rightarrow -2\log \left( 1-v \right)-v=\log \frac{x}{k}$ 

$\Rightarrow v=-\log \frac{x}{k}-2\log \left( 1-v \right)$ 

$\Rightarrow v=\log \left( \frac{k}{x{{\left( 1-v \right)}^{2}}} \right)$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{y}{x}=\log \left( \frac{k}{x{{\left( 1-\frac{y}{x} \right)}^{2}}} \right)$ 

$\Rightarrow \frac{y}{x}=\log \left( \frac{kx}{{{\left( x-y \right)}^{2}}} \right)$ 

$\Rightarrow {{e}^{\frac{y}{x}}}=\frac{kx}{{{\left( x-y \right)}^{2}}}$

$\Rightarrow {{\left( x-y \right)}^{2}}=kx{{e}^{-\frac{y}{x}}}$  

 The solution of the given differential equation ${{\left( \text{x-y} \right)}^{\text{2}}}\text{=kx}{{\text{e}}^{\text{-}\frac{\text{y}}{\text{x}}}}$ .

2. Show that, differential equation $\text{y }\!\!'\!\!\text{ =}\frac{\text{x+y}}{\text{x}}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=\frac{x+y}{x}$ 

Checking for homogeneity:

.$F\left( x,y \right)=\frac{x+y}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda x+\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( x+y \right)}{\lambda \left( x \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{x+y}{x}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{x+\left( vx \right)}{x}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( 1+v \right)}{x}$ 

$\Rightarrow v+x\frac{dv}{dx}=1+v$ 

$\Rightarrow x\frac{dv}{dx}=1$ 

Separate the differentials:

$dv=\frac{dx}{x}$ 

Integrate both side:

$\int dv=\int \frac{dx}{x}$ 

$\Rightarrow \int dv=\log x+\log k$ 

$\Rightarrow v=\log kx$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{y}{x}=\log kx$ 

$\Rightarrow y=x\log kx$  

 The solution of the given differential equation $\text{y=x log kx}$ 

3. Show that, differential equation $\left( \text{x-y} \right)\text{dy-}\left( \text{x+y} \right)\text{dx=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=\frac{x+y}{x-y}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{x+y}{x-y}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( x+y \right)}{\lambda \left( x-y \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{x+y}{x-y}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{x+\left( vx \right)}{x-vx}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( 1+v \right)}{x\left( 1-v \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{1+v}{1-v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v$

$\Rightarrow x\frac{dv}{dx}=\frac{1+v-v+{{v}^{2}}}{1-v}$  

$\Rightarrow x\frac{dv}{dx}=\frac{1+{{v}^{2}}}{1-v}$ 

Separate the differentials:

$\frac{1-v}{1+{{v}^{2}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1-v}{1+{{v}^{2}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{1-v}{1+{{v}^{2}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{1}{1+{{v}^{2}}}dv-\int \frac{v}{1+{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow {{\tan }^{-1}}v-\frac{1}{2}\int \frac{2v}{1+{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow {{\tan }^{-1}}v-\frac{1}{2}\log \left( 1+{{v}^{2}} \right)=\log x+C$ 

$\Rightarrow {{\tan }^{-1}}v=\log x+\frac{1}{2}\log \left( 1+{{v}^{2}} \right)+C$ 

$\Rightarrow {{\tan }^{-1}}v=\frac{1}{2}\log {{x}^{2}}+\frac{1}{2}\log \left( 1+{{v}^{2}} \right)+C$ 

$\Rightarrow {{\tan }^{-1}}v=\frac{1}{2}\log \left[ {{x}^{2}}\left( 1+{{v}^{2}} \right) \right]+C$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow {{\tan }^{-1}}\frac{y}{x}=\frac{1}{2}\log \left[ {{x}^{2}}\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right) \right]+C$ 

$\Rightarrow {{\tan }^{-1}}\frac{y}{x}=\frac{1}{2}\log \left( {{x}^{2}}+{{y}^{2}} \right)+C$  

The solution of the given differential equation $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{y}}{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{log}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)\text{+C}$.

4. Show that, differential equation $\left( {{\text{x}}^{2}}\text{-}{{\text{y}}^{2}} \right)\text{dx+2xy dy=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=-\frac{{{x}^{2}}-{{y}^{2}}}{2xy}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{{{x}^{2}}-{{y}^{2}}}{2xy}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{\left( \lambda x \right)}^{2}}-{{\left( \lambda y \right)}^{2}}}{2\left( \lambda x \right)\left( \lambda y \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{2}}{{x}^{2}}-{{\lambda }^{2}}{{y}^{2}}}{2{{\lambda }^{2}}xy}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\lambda }^{2}}\left( 2xy \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{x}^{2}}-{{y}^{2}}}{2xy}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=-\frac{{{x}^{2}}-{{\left( vx \right)}^{2}}}{2x\left( vx \right)}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left( {{v}^{2}}-1 \right)}{{{x}^{2}}\left( 2v \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{{{v}^{2}}-1}{2v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{{{v}^{2}}-1}{2v}-v$

$\Rightarrow x\frac{dv}{dx}=\frac{{{v}^{2}}-1-2{{v}^{2}}}{2v}$  

$\Rightarrow x\frac{dv}{dx}=-\frac{1+{{v}^{2}}}{2v}$ 

Separate the differentials:

$\frac{2v}{1+{{v}^{2}}}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \frac{2v}{1+{{v}^{2}}}dv=-\int \frac{dx}{x}$ 

$\Rightarrow \log \left( 1+{{v}^{2}} \right)=-\log x+C$ 

$\Rightarrow \log \left( 1+{{v}^{2}} \right)+\log x=C$ 

$\Rightarrow \log \left[ x\left( 1+{{v}^{2}} \right) \right]=C$

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \log \left[ x\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right) \right]=C$ 

$\Rightarrow \left( \frac{{{x}^{2}}+{{y}^{2}}}{x} \right)=k\,\,k={{e}^{C}}$  

$\Rightarrow {{x}^{2}}+{{y}^{2}}=kx$ 

 The solution of the given differential equation ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=kx}$.

5. Show that, differential equation ${{\text{x}}^{\text{2}}}\frac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{2}}}\text{-2}{{\text{y}}^{\text{2}}}\text{+xy}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

\[\frac{dy}{dx}=\frac{{{x}^{2}}-2{{y}^{2}}+xy}{{{x}^{2}}}\] 

Checking for homogeneity:

$F\left( x,y \right)=\frac{{{x}^{2}}-2{{y}^{2}}+xy}{{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\lambda }^{2}}{{x}^{2}}-2{{\lambda }^{2}}{{y}^{2}}+\left( \lambda x \right)\left( \lambda y \right)}{{{\lambda }^{2}}{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\lambda }^{2}}\left( {{x}^{2}}-2{{y}^{2}}+xy \right)}{{{\lambda }^{2}}\left( {{x}^{2}} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{x}^{2}}-2{{y}^{2}}+xy}{{{x}^{2}}}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{{{x}^{2}}-2{{\left( vx \right)}^{2}}+x\left( vx \right)}{{{x}^{2}}}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left( 1-2{{v}^{2}}+v \right)}{{{x}^{2}}}$ 

$\Rightarrow v+x\frac{dv}{dx}=1-2{{v}^{2}}+v$ 

$\Rightarrow x\frac{dv}{dx}=1-2{{v}^{2}}$ 

Separate the differentials:

$\frac{1}{1-2{{v}^{2}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1}{1-2{{v}^{2}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \frac{1}{2}\int \frac{1}{\frac{1}{2}-{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow \frac{1}{2}\int \frac{1}{{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}-{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow \frac{1}{2}\left( \frac{1}{2\times \frac{1}{\sqrt{2}}} \right)\log \frac{\left| \frac{1}{\sqrt{2}}+v \right|}{\left| \frac{1}{\sqrt{2}}-v \right|}=\log x+C$

$\Rightarrow \frac{1}{2\sqrt{2}}\log \frac{\left| 1+\sqrt{2}v \right|}{\left| 1-\sqrt{2}v \right|}=\log x+C$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

$\Rightarrow \frac{1}{2\sqrt{2}}\log \frac{\left| 1+\sqrt{2}\left( \frac{y}{x} \right) \right|}{\left| 1-\sqrt{2}\left( \frac{y}{x} \right) \right|}=\log x+C$ 

$\Rightarrow \frac{1}{2\sqrt{2}}\log \frac{\left| x+\sqrt{2}y \right|}{\left| x-\sqrt{2}y \right|}=\log x+C$ 

 The solution of the given differential equation $\frac{\text{1}}{\text{2}\sqrt{\text{2}}}\text{log}\frac{\left| \text{x+}\sqrt{\text{2}}\text{y} \right|}{\left| \text{x-}\sqrt{\text{2}}\text{y} \right|}\text{=logx+C}$.

6. Show that, differential equation $\text{xdy-ydx=}\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}}\text{dx}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$xdy=\sqrt{{{x}^{2}}+{{y}^{2}}}dx+ydx$ 

$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\sqrt{{{\lambda }^{2}}{{x}^{2}}+{{\lambda }^{2}}{{y}^{2}}}+\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\sqrt{{{\lambda }^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)}+\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( \sqrt{{{x}^{2}}+{{y}^{2}}}+y \right)}{\lambda \left( x \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{\sqrt{{{x}^{2}}+{{\left( vx \right)}^{2}}}+vx}{x}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( \sqrt{1+{{v}^{2}}}+v \right)}{x}$ 

$\Rightarrow v+x\frac{dv}{dx}=\sqrt{1+{{v}^{2}}}+v$ 

$\Rightarrow x\frac{dv}{dx}=\sqrt{1+{{v}^{2}}}$ 

Separate the differentials:

$\frac{1}{\sqrt{1+{{v}^{2}}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1}{\sqrt{1+{{v}^{2}}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \log \left| v+\sqrt{1+{{v}^{2}}} \right|=\log x+\log C$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

$\Rightarrow \log \left| \frac{y}{x}+\sqrt{1+\frac{{{y}^{2}}}{{{x}^{2}}}} \right|=\log Cx$ 

$\Rightarrow \log \left| \frac{y+\sqrt{{{y}^{2}}+{{x}^{2}}}}{x} \right|=\log Cx$ 

$y+\sqrt{{{y}^{2}}+{{x}^{2}}}=C{{x}^{2}}$ 

7. Show that, differential equation $\left\{ \text{xcos}\left( \frac{\text{y}}{\text{x}} \right)\text{+ysin}\left( \frac{\text{y}}{\text{x}} \right) \right\}\text{ydx=}\left\{ \text{ysin}\left( \frac{\text{y}}{\text{x}} \right)\text{-xcos}\left( \frac{\text{y}}{\text{x}} \right) \right\}\text{xdy}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}ydx=\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}xdy$ 

$\frac{dy}{dx}=\frac{\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\left\{ \lambda x\cos \left( \frac{\lambda y}{\lambda x} \right)+y\sin \left( \frac{\lambda y}{\lambda x} \right) \right\}\lambda y}{\left\{ \lambda y\sin \left( \frac{\lambda y}{\lambda x} \right)-\lambda x\cos \left( \frac{\lambda y}{\lambda x} \right) \right\}\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\lambda }^{2}}\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{{{\lambda }^{2}}\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{\left\{ x\cos \left( \frac{xv}{x} \right)+vx\sin \left( \frac{vx}{x} \right) \right\}vx}{\left\{ vx\sin \left( \frac{vx}{x} \right)-x\cos \left( \frac{vx}{x} \right) \right\}x}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left\{ \cos v+v\sin v \right\}v}{{{x}^{2}}\left\{ v\sin v-\cos v \right\}}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v}{v\sin v-\cos v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v}{v\sin v-\cos v}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v-{{v}^{2}}\sin v+v\cos v}{v\sin v-\cos v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}$ 

Separate the differentials:

$\frac{v\sin v-\cos v}{v\cos v}dv=2\frac{dx}{x}$ 

Integrate both side:

$\int \frac{v\sin v-\cos v}{v\cos v}dv=2\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{v\sin v}{v\cos v}dv-\int \frac{\cos v}{v\cos v}dv=2\int \frac{dx}{x}$

$\Rightarrow \int \tan vdv-\int \frac{1}{v}dv=2\log \left| x \right|+\log C$ 

$\Rightarrow \log \left| \sec v \right|-\log \left| v \right|=\log C{{\left| x \right|}^{2}}$

$\Rightarrow \log \frac{\left| \sec v \right|}{\left| v \right|}=\log C{{\left| x \right|}^{2}}$ 

$\Rightarrow \sec v=Cv{{x}^{2}}$   

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

$\Rightarrow \sec \frac{y}{x}=C\left( \frac{y}{x} \right){{x}^{2}}$ 

$\Rightarrow \cos \frac{y}{x}=\frac{k}{xy}\,\,k=\frac{1}{C}$ 

The solution of the given differential equation $\text{cos}\frac{\text{y}}{\text{x}}\text{=}\frac{\text{k}}{\text{xy}}$.

8. Show that, differential equation $\text{x}\frac{\text{dy}}{\text{dx}}\text{-y+xsin}\left( \frac{\text{y}}{\text{x}} \right)\text{=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$x\frac{dy}{dx}-y+x\sin \left( \frac{y}{x} \right)=0$ 

$\frac{dy}{dx}=\frac{y-x\sin \left( \frac{y}{x} \right)}{x}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{y-x\sin \left( \frac{y}{x} \right)}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y-\lambda x\sin \left( \frac{\lambda y}{\lambda x} \right)}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( y-x\sin \left( \frac{y}{x} \right) \right)}{\lambda x}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{y-x\sin \left( \frac{y}{x} \right)}{x}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{vx-x\sin \left( \frac{vx}{x} \right)}{x}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( v-\sin v \right)}{x}$ 

$\Rightarrow v+x\frac{dv}{dx}=v-\sin v$ 

$\Rightarrow x\frac{dv}{dx}=-\sin v$ 

Separate the differentials:

$\frac{1}{\sin v}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \text{cosec}\,vdv=-\int \frac{dx}{x}$ 

$\Rightarrow \log \left| \text{cosec}\,v-\cot v \right|=-\log x+\log C$

$\Rightarrow \log \left| \text{cosec}\,v-\cot v \right|=\log \frac{C}{x}$ 

$\Rightarrow \text{cosec}\,v-\cot v=\frac{C}{x}$

$\Rightarrow \frac{1}{\sin v}-\frac{\cos v}{\sin v}=\frac{C}{x}$ 

$\Rightarrow 1-\cos v=\frac{C}{x}\sin v$   

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow 1-\cos \frac{y}{x}=\frac{C}{x}\sin \frac{y}{x}$ 

$\Rightarrow x\left( 1-\cos \frac{y}{x} \right)=C\sin \left( \frac{y}{x} \right)$ 

The solution of the given differential equation $\text{x}\left( \text{1-cos}\frac{\text{y}}{\text{x}} \right)\text{=Csin}\left( \frac{\text{y}}{\text{x}} \right)$.

9. Show that, differential equation $\text{ydx+xlog}\left( \frac{\text{y}}{\text{x}} \right)\text{dy-2xdy=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$ydx=2xdy-x\log \left( \frac{y}{x} \right)dy$ 

$\frac{dy}{dx}=\frac{y}{2x-x\log \left( \frac{y}{x} \right)}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{y}{2x-x\log \left( \frac{y}{x} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y}{2\lambda x-\lambda x\log \left( \frac{\lambda y}{\lambda x} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y}{\lambda \left( 2x-x\log \left( \frac{y}{x} \right) \right)}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{y}{2x-x\log \left( \frac{y}{x} \right)}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{vx}{2x-x\log \left( \frac{vx}{x} \right)}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{xv}{x\left( 2-\log \left( v \right) \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{v}{2-\log \left( v \right)}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v}{2-\log v}-v$

$\Rightarrow x\frac{dv}{dx}=\frac{v-2v+v\log v}{2-\log v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v\log v-v}{2-\log v}$  

Separate the differentials:

$\frac{2-\log v}{v\log v-v}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{2-\log v}{v\log v-v}dv=\int \frac{dx}{x}$ 

$\int \frac{1+1-\log v}{v\left( \log v-v \right)}dv=\log x+\log C$

$\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv+\int \frac{1-\log v}{v\left( \log v-1 \right)}dv=\log x+\log C$ 

$\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv-\int \frac{1}{v}dv=\log Cx$ ……(1)

Solving :

$\int \frac{1}{v\left( \log v-1 \right)}dv$

Substitute $\text{log v-1=t}$:

$\log v-1=t$   

$\Rightarrow \frac{1}{v}dv=dt$   

Thus the integral will be:

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\int \frac{dt}{t}\] 

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\log t\] 

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\log \left( \log v-1 \right)\] 

Using above result for solving (1):

$\Rightarrow \log \left( \log v-1 \right)-\log v=\log Cx$ 

$\Rightarrow \log \frac{\log v-1}{v}=\log Cx$ 

$\Rightarrow \frac{\log v-1}{v}=Cx$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{\log \frac{y}{x}-1}{\frac{y}{x}}=Cx$ 

$\Rightarrow \log \frac{y}{x}-1=Cx\left( \frac{y}{x} \right)$ 

$\Rightarrow \log \frac{y}{x}-1=Cy$ 

 The solution of the given differential equation $\text{log}\frac{\text{y}}{\text{x}}\text{-1=Cy}$.

10. Show that, differential equation $\left( \text{1+}{{\text{e}}^{\frac{\text{x}}{\text{y}}}} \right)\text{dx+}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\left( \text{1-}\frac{\text{x}}{\text{y}} \right)\text{dy=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\left( 1+{{e}^{\frac{x}{y}}} \right)dx=-{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)dy$ 

$\frac{dx}{dy}=-\frac{{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)}{\left( 1+{{e}^{\frac{x}{y}}} \right)}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)}{\left( 1+{{e}^{\frac{x}{y}}} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{e}^{\frac{\lambda x}{\lambda y}}}\left( 1-\frac{\lambda x}{\lambda y} \right)}{\left( 1+{{e}^{\frac{\lambda x}{\lambda y}}} \right)}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)}{\left( 1+{{e}^{\frac{x}{y}}} \right)}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{x=vy}$:

\[\frac{d\left( vy \right)}{dy}=-\frac{{{e}^{\frac{vy}{y}}}\left( 1-\frac{vy}{y} \right)}{1+{{e}^{\frac{vy}{y}}}}\] 

$\Rightarrow v\frac{dy}{dy}+y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}$ 

$\Rightarrow v+y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}$ 

$\Rightarrow y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}-v$

$\Rightarrow y\frac{dv}{dy}=\frac{-{{e}^{v}}+v{{e}^{v}}-v-v{{e}^{v}}}{1+{{e}^{v}}}$ 

$\Rightarrow y\frac{dv}{dy}=\frac{-\left( {{e}^{v}}+v \right)}{1+{{e}^{v}}}$  

Separate the differentials:

$\frac{1+{{e}^{v}}}{{{e}^{v}}+v}dv=-\frac{dy}{y}$ 

Integrate both side:

$\int \frac{1+{{e}^{v}}}{{{e}^{v}}+v}dv=-\int \frac{dy}{y}$ 

\[\int \frac{{{e}^{v}}+1}{{{e}^{v}}+v}dv=-\log y+\log C\] ……(1)

Solving the LHS integral. Substitute ${{\text{e}}^{\text{v}}}\text{+v=t}$:

\[{{e}^{v}}+v=t\] 

$\Rightarrow \left( {{e}^{v}}+1 \right)dv=dt$ 

Solving the expression (1):

\[\Rightarrow \int \frac{1}{t}dt=\log \frac{C}{y}\] 

\[\Rightarrow \log \left( t \right)=\log \frac{C}{y}\]

\[\Rightarrow \log \left( {{e}^{v}}+v \right)=\log \frac{C}{y}\] 

\[\Rightarrow {{e}^{v}}+v=\frac{C}{y}\] 

Back substitute $\text{v=}\frac{\text{x}}{\text{y}}$:

\[\Rightarrow {{e}^{\frac{x}{y}}}+\frac{x}{y}=\frac{C}{y}\] 

\[\Rightarrow y{{e}^{\frac{x}{y}}}+x=C\] 

The solution of the given differential equation \[\text{y}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{+x=C}\].

11. For the differential equation $\left( \text{x+y} \right)\text{dy+}\left( \text{x-y} \right)\text{dx=0}$.Find the particular solution for the condition $\text{y=1}$ when $\text{x=1}$.

Ans: Given differential equation is:

$\left( x+y \right)dy+\left( x-y \right)dx=0$

$\Rightarrow \frac{dy}{dx}=-\frac{x-y}{x+y}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{x-y}{x+y}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\lambda x-\lambda y}{\lambda x+\lambda y}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\lambda \left( x-y \right)}{\lambda \left( x+y \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{x-y}{x+y}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=-\frac{x-\left( vx \right)}{x+\left( vx \right)}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( v-1 \right)}{x\left( v+1 \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{v-1}{v+1}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v-1}{v+1}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v-1-{{v}^{2}}-v}{v+1}$ 

$\Rightarrow x\frac{dv}{dx}=-\frac{1+{{v}^{2}}}{v+1}$ 

Separate the differentials:

$\frac{v+1}{1+{{v}^{2}}}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \frac{v+1}{1+{{v}^{2}}}dv=-\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{v}{1+{{v}^{2}}}dv+\int \frac{1}{1+{{v}^{2}}}dv=-\log x+k$ 

$\Rightarrow \frac{1}{2}\log \left( 1+{{v}^{2}} \right)+{{\tan }^{-1}}v+\log x=k$ 

$\Rightarrow \frac{1}{2}\log \left[ x\left( 1+{{v}^{2}} \right) \right]+{{\tan }^{-1}}v=k$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{1}{2}\log \left[ x\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right) \right]+{{\tan }^{-1}}\frac{y}{x}=k$ 

$\Rightarrow \frac{1}{2}\log \left[ \frac{{{x}^{2}}+{{y}^{2}}}{x} \right]+{{\tan }^{-1}}\frac{y}{x}=k$  

 Now $\text{y=1}$ and $\text{x=1}$:

$\Rightarrow \frac{1}{2}\log \left[ \frac{{{1}^{2}}+{{1}^{2}}}{1} \right]+{{\tan }^{-1}}\frac{1}{1}=k$  

$k=\frac{1}{2}\log 2+\frac{\pi }{4}$ 

The required particular solution:

$\Rightarrow \frac{1}{2}\log \left[ \frac{{{x}^{2}}+{{y}^{2}}}{x} \right]+{{\tan }^{-1}}\frac{y}{x}=\frac{1}{2}\log 2+\frac{\pi }{4}$ 

12. For the differential equation ${{\text{x}}^{2}}\text{dy+}\left( \text{xy+}{{\text{y}}^{2}} \right)\text{dx=0}$.Find the particular solution for the condition $\text{y=1}$ when $\text{x=1}$.

Ans: Given differential equation is:

${{x}^{2}}dy+\left( xy+{{y}^{2}} \right)dx=0$

$\Rightarrow \frac{dy}{dx}=-\frac{xy+{{y}^{2}}}{{{x}^{2}}}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{xy+{{y}^{2}}}{{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\left( \lambda x \right)\left( \lambda y \right)+{{\lambda }^{2}}{{y}^{2}}}{{{\lambda }^{2}}{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{2}}\left( xy+{{y}^{2}} \right)}{{{\lambda }^{2}}{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{xy+{{y}^{2}}}{{{x}^{2}}}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=-\frac{x\left( vx \right)+{{\left( vx \right)}^{2}}}{{{x}^{2}}}$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=-\frac{v{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{{{x}^{2}}}\] 

$\Rightarrow v+x\frac{dv}{dx}=-\frac{{{x}^{2}}\left( v+{{v}^{2}} \right)}{{{x}^{2}}}$ 

$\Rightarrow v+x\frac{dv}{dx}=-v-{{v}^{2}}$ 

$\Rightarrow x\frac{dv}{dx}=-{{v}^{2}}-2v$ 

Separate the differentials:

$\frac{1}{{{v}^{2}}+2v}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1}{{{v}^{2}}+2v}dv=-\int \frac{dx}{x}$ 

$\Rightarrow \frac{1}{2}\int \frac{v+2-v}{v\left( v+2 \right)}dv=-\log x+\log C$ 

$\Rightarrow \frac{1}{2}\int \frac{v+2}{v\left( v+2 \right)}dv-\frac{1}{2}\int \frac{v}{v\left( v+2 \right)}dv=-\log x+\log C$ 

$\Rightarrow \frac{1}{2}\int \frac{1}{v}dv-\frac{1}{2}\int \frac{1}{v+2}dv=-\log x+\log C$ 

$\Rightarrow \frac{1}{2}\log v-\frac{1}{2}\log \left( v+2 \right)=\log \frac{C}{x}$ 

$\Rightarrow \frac{1}{2}\log \frac{v}{v+2}=\log \frac{C}{x}$ 

\[\Rightarrow \frac{v}{v+2}={{\left( \frac{C}{x} \right)}^{2}}\] 

\[\Rightarrow \frac{v}{v+2}=\frac{{{C}^{2}}}{{{x}^{2}}}\] 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

\[\Rightarrow \frac{\frac{y}{x}}{\frac{y}{x}+2}=\frac{{{C}^{2}}}{{{x}^{2}}}\] 

\[\Rightarrow \frac{{{x}^{2}}y}{y+2x}={{C}^{2}}\]  

 Now $\text{y=1}$ and $\text{x=1}$:

\[\Rightarrow \frac{{{1}^{2}}\left( 1 \right)}{1+2\left( 1 \right)}={{C}^{2}}\]  

\[\Rightarrow {{C}^{2}}=\frac{1}{3}\] 

The required particular solution:

\[\Rightarrow \frac{{{x}^{2}}y}{y+2x}=\frac{1}{3}\] 

\[\Rightarrow y+2x=3{{x}^{2}}y\] 

13. For the differential equation $\left[ \text{xsi}{{\text{n}}^{\text{2}}}\left( \frac{\text{x}}{\text{y}} \right)\text{-y} \right]\text{dx+xdy=0}$.Find the particular solution for the condition $\text{y=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ when $\text{x=1}$.

Ans: Given differential equation is:

$\left[ x{{\sin }^{2}}\left( \frac{x}{y} \right)-y \right]dx+xdy=0$

$\Rightarrow \frac{dy}{dx}=-\frac{\left[ x{{\sin }^{2}}\left( \frac{x}{y} \right)-y \right]}{x}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{x{{\sin }^{2}}\left( \frac{x}{y} \right)-y}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\lambda x{{\sin }^{2}}\left( \frac{\lambda x}{\lambda y} \right)-\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\lambda \left( x{{\sin }^{2}}\left( \frac{y}{x} \right)-y \right)}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{x{{\sin }^{2}}\left( \frac{y}{x} \right)-y}{x}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=-\frac{x{{\sin }^{2}}\left( \frac{vx}{x} \right)-vx}{x}$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{-x{{\sin }^{2}}\left( v \right)+vx}{x}\] 

$\Rightarrow v+x\frac{dv}{dx}=-{{\sin }^{2}}v+v$ 

$\Rightarrow x\frac{dv}{dx}=-{{\sin }^{2}}v$ 

Separate the differentials:

$\text{cose}{{\text{c}}^{2}}vdv=-\frac{dx}{x}$ 

Integrate both side:

$\int \text{cose}{{\text{c}}^{2}}vdv=-\int \frac{dx}{x}$ 

$\Rightarrow -\cot v=-\log x-\log C$ 

$\Rightarrow \cot v=\log Cx$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \cot \frac{y}{x}=\log Cx$ 

 Now $\text{y=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ and $\text{x=1}$:

$\cot \frac{\frac{\pi }{4}}{1}=\log C\left( 1 \right)$  

\[\Rightarrow \log C=\cot \frac{\pi }{4}\] 

\[\Rightarrow \log C=1\] 

\[\Rightarrow C=e\] 

The required particular solution:

\[\Rightarrow \cot \frac{y}{x}=\log \left| ex \right|\] 

14. For the differential equation $\frac{\text{dy}}{\text{dx}}\text{-}\frac{\text{y}}{\text{x}}\text{+cosec}\left( \frac{\text{y}}{\text{x}} \right)\text{=0}$.Find the particular solution for the condition $\text{y=0}$ when $\text{x=1}$.

Ans: Given differential equation is:

$\frac{dy}{dx}-\frac{y}{x}+\text{cosec}\left( \frac{y}{x} \right)=0$

\[\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\text{cosec}\left( \frac{y}{x} \right)\] 

Checking for homogeneity:

$F\left( x,y \right)=\frac{y}{x}-\text{cosec}\left( \frac{y}{x} \right)$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y}{\lambda x}-\text{cosec}\left( \frac{\lambda y}{\lambda x} \right)$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{y}{x}-\text{cosec}\left( \frac{y}{x} \right)$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=v-\text{cosec}\left( v \right)$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=v-\text{cosec}\left( v \right)\] 

$\Rightarrow v+x\frac{dv}{dx}=v-\text{cosec}\left( v \right)$ 

$\Rightarrow x\frac{dv}{dx}=-\text{cosec}\left( v \right)$ 

Separate the differentials:

$\sin vdv=-\frac{dx}{x}$ 

Integrate both side:

$\int \sin vdv=-\int \frac{dx}{x}$ 

\[\Rightarrow -\cos v=-\log \left| x \right|-\log C\] 

\[\cos v=\log \left| Cx \right|\] 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \cos \frac{y}{x}=\log \left| Cx \right|$ 

 Now $\text{y=0}$ and $\text{x=1}$:

$\Rightarrow \cos \frac{0}{1}=\log \left| C1 \right|$  

\[\Rightarrow \log C=1\] 

\[\Rightarrow C=e\] 

The required particular solution:

\[\Rightarrow \cos \frac{y}{x}=\log \left| ex \right|\].

15. For the differential equation $\text{2xy+}{{\text{y}}^{\text{2}}}\text{-2}{{\text{x}}^{\text{2}}}\frac{\text{dy}}{\text{dx}}\text{=0}$.Find the particular solution for the condition $\text{y=2}$ when $\text{x=1}$.

Ans: Given differential equation is:

$2xy+{{y}^{2}}-2{{x}^{2}}\frac{dy}{dx}=0$

\[\Rightarrow \frac{dy}{dx}=\frac{2xy+{{y}^{2}}}{2{{x}^{2}}}\] 

Checking for homogeneity:

$F\left( x,y \right)=\frac{2xy+{{y}^{2}}}{2{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{2\left( \lambda x \right)\left( \lambda y \right)+{{\lambda }^{2}}{{y}^{2}}}{2{{\lambda }^{2}}{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{2xy+{{y}^{2}}}{2{{x}^{2}}}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{2x\left( vx \right)+{{\left( vx \right)}^{2}}}{2{{x}^{2}}}$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{2v+{{v}^{2}}}{2}\] 

$\Rightarrow v+x\frac{dv}{dx}=v+\frac{{{v}^{2}}}{2}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{{{v}^{2}}}{2}$ 

Separate the differentials:

$\frac{dv}{{{v}^{2}}}=\frac{1}{2}\left( \frac{dx}{x} \right)$ 

Integrate both side:

$2\int \frac{dv}{{{v}^{2}}}=\int \left( \frac{dx}{x} \right)$ 

\[\Rightarrow \frac{{{v}^{-2+1}}}{-2+1}=\log \left| x \right|+C\] 

\[-\frac{2}{v}=\log \left| x \right|+C\] 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow -\frac{2x}{y}=\log \left| x \right|+C$ 

 Now $\text{y=2}$ and $\text{x=1}$:

$\Rightarrow -\frac{2\left( 1 \right)}{2}=\log \left| 1 \right|+C$  

$\Rightarrow C=-1$ 

The required particular solution:

$\Rightarrow -\frac{2x}{y}=\log \left| x \right|-1$.

$\Rightarrow y=\frac{2x}{1-\log \left| x \right|}\,\,\left( x\ne 0,e \right)$ 

16. What substitution should be used for solving homogeneous differential equation $\frac{\text{dx}}{\text{dy}}\text{=h}\left( \frac{\text{x}}{\text{y}} \right)$.

A.$\text{y=vx}$ 

B. $\text{v=yx}$

C. $\text{x=vy}$

D. $\text{x=v}$

Ans: The required substitution will be: 

$\frac{x}{y}=v$ 

$\Rightarrow x=vy$ 

The correct answer is (C).

17. Which of the following equation is homogeneous:

1. $\left( \text{4x+6y+5} \right)\text{dy-}\left( \text{3y+2x+4} \right)\text{dx=0}$ 

2. $\left( \text{xy} \right)\text{dx-}\left( {{\text{x}}^{3}}\text{+}{{\text{y}}^{3}} \right)\text{dy=0}$ 

3. $\left( {{\text{x}}^{\text{3}}}\text{+2}{{\text{y}}^{\text{2}}} \right)\text{dx+2xydy=0}$ 

4. ${{\text{y}}^{\text{2}}}\text{dx+}\left( {{\text{x}}^{\text{2}}}\text{-xy-}{{\text{y}}^{\text{2}}} \right)\text{dy=0}$ 

Ans: For option (A):

$F\left( x,y \right)=\frac{3y+2x+4}{4x+6y+5}$

$F\left( \lambda x,\lambda y \right)=\frac{3\lambda y+2\lambda x+4}{4\lambda x+6\lambda y+5}$ 

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  

For option (B):

$F\left( x,y \right)=\frac{xy}{{{x}^{3}}+{{y}^{3}}}$

$F\left( \lambda x,\lambda y \right)=\frac{\left( \lambda x \right)\left( \lambda y \right)}{{{\left( \lambda x \right)}^{3}}+{{\left( \lambda y \right)}^{3}}}$ 

$F\left( \lambda x,\lambda y \right)=\frac{xy}{\lambda \left( x+y \right)}$ 

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  

For option (C):

$F\left( x,y \right)=-\frac{{{x}^{3}}+2{{y}^{2}}}{2xy}$

$F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{3}}{{x}^{3}}+2{{\lambda }^{2}}{{y}^{2}}}{2\left( \lambda x \right)\left( \lambda y \right)}$ 

$F\left( \lambda x,\lambda y \right)=-\frac{\lambda {{x}^{3}}+2{{y}^{2}}}{2xy}$ 

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  

For option (D):

$F\left( x,y \right)=-\frac{{{y}^{2}}}{{{x}^{2}}-xy-{{y}^{2}}}$

$F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{2}}{{y}^{2}}}{{{\lambda }^{2}}{{x}^{2}}-\left( \lambda x \right)\left( \lambda y \right)-{{\lambda }^{2}}{{y}^{2}}}$ 

$F\left( \lambda x,\lambda y \right)=-\frac{{{y}^{2}}}{{{x}^{2}}-xy-{{y}^{2}}}$ 

$F\left( \lambda x,\lambda y \right)=F\left( x,y \right)$  

Thus the correct answer is option (D).

Exercise 9.5

1. Find the general solution for the differential equation $\frac{\text{dy}}{\text{dx}}\text{+2y=sinx}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+2y=\sin x$

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=2$ 

$Q=\sin x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int 2dx}}$ 

$\Rightarrow I.F={{e}^{2x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$y{{e}^{2x}}=\int \sin x\left( {{e}^{2x}} \right)dx+C$ 

\[\Rightarrow y{{e}^{2x}}=I+C\,\,\,\left( I=\int \sin x\left( {{e}^{2x}} \right)dx \right)\] ……(1)

\[I=\int \sin x\left( {{e}^{2x}} \right)dx\] 

$\Rightarrow I=\left( \sin x \right)\int {{e}^{2x}}dx-\int \left( \left( \sin x \right)'\int {{e}^{2x}}dx \right)dx$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\int \left( \cos x\left( \frac{{{e}^{2x}}}{2} \right) \right)dx$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \cos x\int {{e}^{2x}}dx-\int \left( \left( \cos x \right)'\left( \int {{e}^{2x}}dx \right) \right)dx \right]$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \frac{{{e}^{2x}}}{2}\cos x+\frac{1}{2}\int {{e}^{2x}}\left( \sin x \right)dx \right]$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \frac{{{e}^{2x}}}{2}\cos x+\frac{1}{2}I \right]$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{{{e}^{2x}}}{4}\cos x-\frac{1}{4}I$ 

$\Rightarrow \frac{5}{4}I=\frac{{{e}^{2x}}}{2}\sin x-\frac{{{e}^{2x}}}{4}\cos x$ 

\[\Rightarrow I=\frac{2{{e}^{2x}}}{5}\sin x-\frac{{{e}^{2x}}}{5}\cos x\] 

\[\Rightarrow I=\frac{{{e}^{2x}}}{5}\left[ 2\sin x-\cos x \right]\] 

Back substituting $\text{I}$ in expression (1):

\[\Rightarrow y{{e}^{2x}}=\frac{{{e}^{2x}}}{5}\left[ 2\sin x-\cos x \right]+C\,\] 

\[\Rightarrow y=\frac{1}{5}\left( 2\sin x-\cos x \right)+C{{e}^{-2x}}\] 

The general solution for given differential equation is \[\text{y=}\frac{\text{1}}{\text{5}}\left( \text{2sinx-cosx} \right)\text{+C}{{\text{e}}^{\text{-2x}}}\].

2. Find the general solution for the differential equation $\frac{\text{dy}}{\text{dx}}\text{+3y=}{{\text{e}}^{\text{-2x}}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+3y={{e}^{-2x}}$

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=3$ 

$Q={{e}^{-2x}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int 3dx}}$ 

$\Rightarrow I.F={{e}^{3x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y{{e}^{3x}}=\int {{e}^{-2x}}\left( {{e}^{3x}} \right)dx+C$ 

$\Rightarrow y{{e}^{3x}}=\int {{e}^{-2x+3x}}dx+C$ 

$\Rightarrow y{{e}^{3x}}=\int {{e}^{x}}dx+C$ 

$\Rightarrow y{{e}^{3x}}={{e}^{x}}+C$ 

$\Rightarrow y={{e}^{-2x}}+C{{e}^{-3x}}$ 

The general solution for given differential equation is $\text{y=}{{\text{e}}^{\text{-2x}}}\text{+C}{{\text{e}}^{\text{-3x}}}$.

3. Find the general solution for the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\frac{\text{y}}{\text{x}}\text{=}{{\text{x}}^{\text{2}}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+\frac{y}{x}={{x}^{2}}$

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{1}{x}$ 

$Q={{x}^{2}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{x}dx}}$ 

$\Rightarrow I.F={{e}^{\log x}}$

$\Rightarrow I.F=x$  

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow yx=\int {{x}^{2}}\left( x \right)dx+C$ 

$\Rightarrow xy=\int {{x}^{3}}dx+C$ 

$\Rightarrow xy=\frac{{{x}^{3+1}}}{3+1}+C$ 

$\Rightarrow xy=\frac{{{x}^{4}}}{4}+C$ 

The general solution for given differential equation is $\text{xy=}\frac{{{\text{x}}^{\text{4}}}}{\text{4}}\text{+C}$.

4. Find the general solution for the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\left( \text{secx} \right)\text{y=tanx}\,\,\left( \text{0}\le \text{x}\le \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$.

Ans: The given differential equation is:

$\frac{dy}{dx}+\left( \sec x \right)y=\tan x$

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\sec x$ 

$Q=\tan x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \sec xdx}}$ 

$\Rightarrow I.F={{e}^{\log \left( \sec x+\tan x \right)}}$

$\Rightarrow I.F=\left( \sec x+\tan x \right)$  

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\int \tan x\left( \sec x+\tan x \right)dx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\int \tan x\sec xdx+\int {{\tan }^{2}}xdx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\int \left( {{\sec }^{2}}x-1 \right)dx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\int {{\sec }^{2}}xdx-\int dx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\tan x-x+C$ 

The general solution for given differential equation is $\text{y}\left( \text{secx+tanx} \right)\text{=secx+tanx-x+C}$.

5. Find the general solution for the differential equation $\text{co}{{\text{s}}^{\text{2}}}\text{x}\frac{\text{dy}}{\text{dx}}\text{+y=tanx}\,\,\left( \text{0}\le \text{x}\le \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$.

Ans: The given differential equation is:

${{\cos }^{2}}x\frac{dy}{dx}+y=\tan x\,$

$\frac{dy}{dx}+\left( {{\sec }^{2}}x \right)y={{\sec }^{2}}x\tan x\,$ 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p={{\sec }^{2}}x$ 

$Q={{\sec }^{2}}x\tan x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int {{\sec }^{2}}xdx}}$ 

$\Rightarrow I.F={{e}^{\tan x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y{{e}^{\tan x}}=\int {{e}^{\tan x}}\left( {{\sec }^{2}}x\tan x \right)dx+C$ 

$\Rightarrow y{{e}^{\tan x}}=I+C\,\,\left( I=\int {{e}^{\tan x}}\left( {{\sec }^{2}}x\tan x \right)dx \right)$ ……(1)

Solving the integral $\text{I}$:

$I=\int {{e}^{\tan x}}\left( {{\sec }^{2}}x\tan x \right)dx$ 

Substitute $\text{tanx=t}$ :
$\tan x=t$ 

$\Rightarrow {{\sec }^{2}}xdx=dt$ 

$\Rightarrow I=\int {{e}^{t}}tdt$ 

$\Rightarrow I=t\int {{e}^{t}}dt-\int \left( \left( t \right)'\int {{e}^{t}}dt \right)dt$ 

$\Rightarrow I=t{{e}^{t}}-\int \left( {{e}^{t}} \right)dt$ 

$\Rightarrow I=t{{e}^{t}}-{{e}^{t}}$ 

Back substitute $\text{t}$:

$I=\tan x{{e}^{\tan x}}-{{e}^{\tan x}}$ 

Back substitute $\text{I}$ in expression (1):

$\Rightarrow y{{e}^{\tan x}}=I+C\,\,$ 

$\Rightarrow y{{e}^{\tan x}}=\tan x{{e}^{\tan x}}-{{e}^{\tan x}}+C\,\,$ 

The general solution for given differential equation is $\text{y}{{\text{e}}^{\text{tanx}}}\text{=tanx}{{\text{e}}^{\text{tanx}}}\text{-}{{\text{e}}^{\text{tanx}}}\text{+C}\,\,$.

6. Find the general solution for the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}\text{+2y=}{{\text{x}}^{2}}\text{logx}$.

Ans: The given differential equation is:

$x\frac{dy}{dx}+2y={{x}^{2}}\log x$

\[\frac{dy}{dx}+\frac{2}{x}y=x\log x\] 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{2}{x}$ 

$Q=x\log x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{2}{x}dx}}$ 

$\Rightarrow I.F={{e}^{2\log x}}$

$\Rightarrow I.F={{e}^{\log {{x}^{2}}}}$ 

$\Rightarrow I.F={{x}^{2}}$  

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y{{x}^{2}}=\int {{x}^{2}}\left( x\log x \right)dx+C$ 

\[\Rightarrow y{{x}^{2}}=\int {{x}^{3}}\log xdx+C\] 

\[\Rightarrow y{{x}^{2}}=I+C\,\,\left( I=\int {{x}^{3}}\log xdx \right)\] ……(1)

Solving the integral $\text{I}$:

\[I=\int {{x}^{3}}\log xdx\] 

\[\Rightarrow I=\log x\int {{x}^{3}}dx-\int \left( \left( \log x \right)'\int {{x}^{3}}dx \right)dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\int \left( \frac{1}{x}\left( \frac{{{x}^{4}}}{4} \right) \right)dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{1}{4}\int {{x}^{3}}dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{1}{4}\left( \frac{{{x}^{4}}}{4} \right)\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{{{x}^{4}}}{16}\] 

Back substitute $\text{I}$ in expression (1):

\[\Rightarrow y{{x}^{2}}=I+C\] 

\[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\frac{{{x}^{4}}}{16}+C\] 

\[\Rightarrow y=\frac{{{x}^{2}}}{16}\left( 4\log x-1 \right)+C{{x}^{-2}}\] 

The general solution for given differential equation is \[\text{y=}\frac{{{\text{x}}^{\text{2}}}}{\text{16}}\left( \text{4logx-1} \right)\text{+C}{{\text{x}}^{\text{-2}}}\].

7. Find the general solution for the differential equation $\text{xlogx}\frac{\text{dy}}{\text{dx}}\text{+y=}\frac{\text{2}}{\text{x}}\text{logx}$.

Ans: The given differential equation is:

$x\log x\frac{dy}{dx}+y=\frac{2}{x}\log x$

$\frac{dy}{dx}+\frac{y}{x\log x}=\frac{2}{{{x}^{2}}}$ 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{1}{x\log x}$ 

$Q=\frac{2}{{{x}^{2}}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{x\log x}dx}}$ 

Substitute $\text{logx=t}$:
$\log x=t$ 

$\Rightarrow \frac{1}{x}dx=dt$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{t}dt}}$

$\Rightarrow I.F={{e}^{\log t}}$ 

$\Rightarrow I.F=t$

$\Rightarrow I.F=\log x$   

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\log x=\int \frac{2}{{{x}^{2}}}\left( \log x \right)dx+C$ 

$\Rightarrow y\log x=I+C\,\,\left( I=\int \frac{2}{{{x}^{2}}}\left( \log x \right)dx \right)$ ……(1)

Solving the integral $\text{I}$:

$I=\int \frac{2}{{{x}^{2}}}\left( \log x \right)dx$ 

\[I=2\left[ \log x\int \frac{1}{{{x}^{2}}}dx-\int \left( \left( \log x \right)'\int \frac{1}{{{x}^{2}}}dx \right)dx \right]\] 

\[I=2\left[ \log x\left( \frac{-1}{x} \right)-\int \left( \frac{1}{x}\left( \frac{-1}{x} \right) \right)dx \right]\] 

\[I=2\left[ -\frac{\log x}{x}+\int \left( \frac{1}{{{x}^{2}}} \right)dx \right]\] 

\[I=2\left[ -\frac{\log x}{x}-\frac{1}{x} \right]\] 

Back substitute $\text{I}$ in expression (1):

$y\log x=I+C$ 

\[\Rightarrow y\log x=2\left[ -\frac{\log x}{x}-\frac{1}{x} \right]+C\] 

\[\Rightarrow y\log x=-\frac{2}{x}\left( \log x+1 \right)+C\] 

The general solution for given differential equation is \[\text{ylogx=-}\frac{\text{2}}{\text{x}}\left( \text{logx+1} \right)\text{+C}\].

8. Find the general solution for the differential equation $\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\text{dy+2xydx=cotxdx}\,\,\left( \text{x}\ne \text{0} \right)$.

Ans: The given differential equation is:

$\left( 1+{{x}^{2}} \right)dy+2xydx=\cot xdx\,\,\left( x\ne 0 \right)$

$\Rightarrow \frac{dy}{dx}+\frac{2xy}{1+{{x}^{2}}}=\frac{\cot x}{1+{{x}^{2}}}$ 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{2x}{1+{{x}^{2}}}$ 

$Q=\frac{\cot x}{1+{{x}^{2}}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{2x}{1+{{x}^{2}}}dx}}$ 

Substitute $\text{logx=t}$:
$1+{{x}^{2}}=t$ 

$\Rightarrow 2xdx=dt$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{t}dt}}$

$\Rightarrow I.F={{e}^{\log t}}$ 

$\Rightarrow I.F=t$

$\Rightarrow I.F=1+{{x}^{2}}$   

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \frac{\cot x}{1+{{x}^{2}}}\left( 1+{{x}^{2}} \right)dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \cot xdx+C$

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\log \left| \sin x \right|+C$ 

The general solution for given differential equation is $\text{y}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\text{=log}\left| \text{sinx} \right|\text{+C}$.

9. Find the general solution for the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}\text{+y-x+xycotx=0}\,\,\left( \text{x}\ne \text{0} \right)$.

Ans: The given differential equation is:

$x\frac{dy}{dx}+y-x+xy\cot x=0\,$

$\Rightarrow \frac{dy}{dx}+\frac{y}{x}-1+y\cot x=0$ 

$\Rightarrow \frac{dy}{dx}+\left( \frac{1}{x}+\cot x \right)y=1$ 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\left( \frac{1}{x}+\cot x \right)$ 

$Q=1$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \left( \frac{1}{x}+\cot x \right)dx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{x}dx+\int \cot xdx}}$

$\Rightarrow I.F={{e}^{\log x+\log \left( \sin x \right)}}$ 

$\Rightarrow I.F={{e}^{\log \left( x\sin x \right)}}$

$\Rightarrow I.F=x\sin x$   

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( x\sin x \right)=\int \left( 1 \right)\left( x\sin x \right)dx+C$ 

$\Rightarrow xy\sin x=I+C\,\,\left( I=\int x\sin xdx \right)$ ……(1) 

Solving the integral $\text{I}$:

$I=\int x\sin xdx$ 

\[\Rightarrow I=x\int \sin xdx-\int \left( \left( x \right)'\int \sin xdx \right)dx\] 

\[\Rightarrow I=x\left( -\cos x \right)+\int \left( \cos x \right)dx\] 

\[\Rightarrow I=-x\cos x+\sin x\] 

Back substitute $\text{I}$ in expression (1):

$xy\sin x=I+C\,$ 

\[\Rightarrow xy\sin x=-x\cos x+\sin x+C\,\] 

\[\Rightarrow y=-\cot x+\frac{1}{x}+\frac{C}{x\sin x}\,\] 

The general solution for given differential equation is \[\text{y=-cotx+}\frac{\text{1}}{\text{x}}\text{+}\frac{\text{C}}{\text{xsinx}}\,\].

10. Find the general solution for the differential equation $\left( \text{x+y} \right)\frac{\text{dy}}{\text{dx}}=1$.

Ans: The given differential equation is:

$\left( x+y \right)\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}=\frac{1}{x+y}$ 

$\Rightarrow \frac{dx}{dy}=x+y$ 

$\Rightarrow \frac{dx}{dy}-x=y$ 

It is differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$, with:

$p=-1$ 

$Q=y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{\int -1dy}}$ 

$\Rightarrow I.F={{e}^{-y}}$

General solution is of the form:

$x\left( I.F \right)=\int \left( Q\times I.F \right)dy+C$ 

$\Rightarrow x\left( {{e}^{-y}} \right)=\int \left( y \right)\left( {{e}^{-y}} \right)dy+C$ 

$\Rightarrow x{{e}^{-y}}=I+C\,\,\left( I=\int y{{e}^{-y}}dy \right)$ ……(1) 

Solving the integral $\text{I}$:

$I=\int y{{e}^{-y}}dy$ 

$\Rightarrow I=y\int {{e}^{-y}}dy-\int \left( \left( y \right)'\int {{e}^{-y}}dy \right)dy$ 

$\Rightarrow I=-y{{e}^{-y}}+\int \left( \left( 1 \right){{e}^{-y}} \right)dy$ 

$\Rightarrow I=-y{{e}^{-y}}+\int {{e}^{-y}}dy$

$\Rightarrow I=-y{{e}^{-y}}-{{e}^{-y}}$  

Back substitute $\text{I}$ in expression (1):

$x{{e}^{-y}}=I+C$ 

\[\Rightarrow x{{e}^{-y}}=-y{{e}^{-y}}-{{e}^{-y}}+C\] 

\[\Rightarrow x=-y-1+C{{e}^{y}}\]

\[\Rightarrow x+y+1=C{{e}^{y}}\]  

The general solution for given differential equation is \[\text{x+y+1=C}{{\text{e}}^{\text{y}}}\].

11. Find the general solution for the differential equation $\text{ydx+}\left( \text{x-}{{\text{y}}^{\text{2}}} \right)\text{dy=0}$.

Ans: The given differential equation is:

$ydx+\left( x-{{y}^{2}} \right)dy=0$

$\Rightarrow \frac{dx}{dy}=\frac{{{y}^{2}}-x}{y}$ 

$\Rightarrow \frac{dx}{dy}+\left( \frac{1}{y} \right)x=y$ 

It is differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$, with:

$p=\frac{1}{y}$ 

$Q=y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{\int \frac{1}{y}dy}}$ 

$\Rightarrow I.F={{e}^{\log y}}$

$\Rightarrow I.F=y$ 

General solution is of the form:

$x\left( I.F \right)=\int \left( Q\times I.F \right)dy+C$ 

$\Rightarrow x\left( y \right)=\int \left( y \right)\left( y \right)dy+C$ 

$\Rightarrow xy=\int {{y}^{2}}dy+C$ 

$\Rightarrow xy=\frac{{{y}^{3}}}{3}+C$ 

$\Rightarrow x=\frac{{{y}^{2}}}{3}+\frac{C}{y}$ 

The general solution for given differential equation is $\text{x=}\frac{{{\text{y}}^{\text{2}}}}{\text{3}}\text{+}\frac{\text{C}}{\text{y}}$.

12. Find the general solution for the differential equation $\left( \text{x+3}{{\text{y}}^{\text{2}}} \right)\frac{\text{dy}}{\text{dx}}\text{=y}\,\,\left( \text{y0} \right)$.

Ans: The given differential equation is:

$\left( x+3{{y}^{2}} \right)\frac{dy}{dx}=y$

$\Rightarrow \frac{dx}{dy}=\frac{x+3{{y}^{2}}}{y}$ 

$\Rightarrow \frac{dx}{dy}-\left( \frac{1}{y} \right)x=3y$ 

It is differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$, with:

$p=-\frac{1}{y}$ 

$Q=3y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{-\int \frac{1}{y}dy}}$ 

$\Rightarrow I.F={{e}^{-\log y}}$

$\Rightarrow I.F={{e}^{\log {{y}^{-1}}}}$ 

$\Rightarrow I.F=\frac{1}{y}$ 

General solution is of the form:

$x\left( I.F \right)=\int \left( Q\times I.F \right)dy+C$ 

$\Rightarrow x\left( \frac{1}{y} \right)=\int \left( 3y \right)\left( \frac{1}{y} \right)dy+C$ 

$\Rightarrow \frac{x}{y}=3\int dy+C$ 

$\Rightarrow \frac{x}{y}=3y+C$ 

$\Rightarrow x=3{{y}^{2}}+Cy$ 

The general solution for given differential equation is $\text{x=3}{{\text{y}}^{\text{2}}}\text{+Cy}$.

13. Find particular solution for $\frac{\text{dy}}{\text{dx}}\text{+2ytanx=sinx}$ satisfying $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+2y\tan x=\sin x$ 

$\frac{dy}{dx}+\left( 2\tan x \right)y=\sin x$ 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=2\tan x$ 

$Q=\sin x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{2\int \tan xdx}}$ 

$\Rightarrow I.F={{e}^{2\log \left| \sec x \right|}}$

\[\Rightarrow I.F={{e}^{\log {{\left( \sec x \right)}^{2}}}}\] 

\[\Rightarrow I.F={{\sec }^{2}}x\] 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y{{\sec }^{2}}x=\int \left( \sin x \right)\left( {{\sec }^{2}}x \right)dx+C$ 

$\Rightarrow y{{\sec }^{2}}x=\int \tan x\sec xdx+C$ 

$\Rightarrow y{{\sec }^{2}}x=\sec x+C$ 

$\Rightarrow y=\cos x+C{{\cos }^{2}}x$ 

Given $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$:

$0=\cos \left( \frac{\pi }{3} \right)+C{{\cos }^{2}}\left( \frac{\pi }{3} \right)$ 

$\Rightarrow 0=\frac{1}{2}+C{{\left( \frac{1}{2} \right)}^{2}}$ 

$\Rightarrow C=-2$ 

Therefore the particular solution will be:

$\Rightarrow y=\cos x-2{{\cos }^{2}}x$ 

The particular solution for given differential equation satisfying the given conditions is $\text{y=cosx-2co}{{\text{s}}^{\text{2}}}\text{x}$.

14. Find particular solution for $\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\frac{\text{dy}}{\text{dx}}\text{+2xy=}\frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}}$ satisfying $\text{y=0}$ when $\text{x=1}$.

Ans: The given differential equation is:

$\left( 1+{{x}^{2}} \right)\frac{dy}{dx}+2xy=\frac{1}{1+{{x}^{2}}}$ 

\[\frac{dy}{dx}+\left( \frac{2x}{1+{{x}^{2}}} \right)y=\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}}\] 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{2x}{1+{{x}^{2}}}$ 

$Q=\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{\int \frac{2x}{1+{{x}^{2}}}dx}}$ 

$\Rightarrow I.F={{e}^{2\log \left| \sec x \right|}}$

$\Rightarrow I.F={{e}^{\log \left( 1+{{x}^{2}} \right)}}$ 

\[\Rightarrow I.F=1+{{x}^{2}}\] 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \left( \frac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)\left( 1+{{x}^{2}} \right)dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \frac{1}{1+{{x}^{2}}}dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)={{\tan }^{-1}}x+C$ 

Given $\text{y=0}$ when $\text{x=1}$:

$0\left( 1+1 \right)={{\tan }^{-1}}\left( 1 \right)+C$ 

$\Rightarrow C+\frac{\pi }{4}=0$ 

$\Rightarrow C=-\frac{\pi }{4}$ 

Therefore the particular solution will be:

$\Rightarrow y\left( 1+{{x}^{2}} \right)={{\tan }^{-1}}x-\frac{\pi }{4}$ 

The particular solution for given differential equation satisfying the given conditions is $\text{y}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\text{x-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$.

15. Find particular solution for $\frac{\text{dy}}{\text{dx}}\text{-3ycotx=sin2x}$ satisfying $\text{y=2}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}-3y\cot x=\sin 2x$ 

\[\frac{dy}{dx}+\left( -3\cot x \right)y=\sin 2x\] 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=-3\cot x$ 

$Q=\sin 2x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-3\int \cot xdx}}$ 

$\Rightarrow I.F={{e}^{-3\log \left| \sin x \right|}}$

$\Rightarrow I.F={{e}^{\log \left( \frac{1}{{{\sin }^{3}}x} \right)}}$ 

\[\Rightarrow I.F=\frac{1}{{{\sin }^{3}}x}\] 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( \frac{1}{{{\sin }^{3}}x} \right)=\int \left( \sin 2x \right)\left( \frac{1}{{{\sin }^{3}}x} \right)dx+C$ 

$\Rightarrow y\left( \frac{1}{{{\sin }^{3}}x} \right)=2\int \left( \sin x\cos x \right)\left( \frac{1}{{{\sin }^{3}}x} \right)dx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=2\int \left( \frac{\cos x}{{{\sin }^{2}}x} \right)dx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=2\int \cot x\text{cosec}xdx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=-2\text{cosec}\,x+C$ 

$\Rightarrow y=-2{{\sin }^{2}}x+C{{\sin }^{3}}x$ 

Given $\text{y=2}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$:

$2=-2{{\sin }^{2}}\left( \frac{\pi }{2} \right)+C{{\sin }^{3}}\left( \frac{\pi }{2} \right)$ 

$\Rightarrow C-2=2$ 

$\Rightarrow C=4$ 

Therefore the particular solution will be:

$\Rightarrow y=-2{{\sin }^{2}}x+4{{\sin }^{3}}x$ 

The particular solution for given differential equation satisfying the given conditions is $\text{y=-2si}{{\text{n}}^{\text{2}}}\text{x+4si}{{\text{n}}^{\text{3}}}\text{x}$.

16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $\left( \text{x,y} \right)$ is equal to the sum of the coordinates of the point.

Ans: According to  the slope of tangent $\frac{\text{dy}}{\text{dx}}$ is equal to sum of the coordinates:

$\frac{dy}{dx}=x+y$ 

The given differential equation is:

$\frac{dy}{dx}=x+y$ 

\[\frac{dy}{dx}+\left( -1 \right)y=x\] 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=-1$ 

$Q=x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-1\int dx}}$ 

$\Rightarrow I.F={{e}^{-x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( {{e}^{-x}} \right)=\int x\left( {{e}^{-x}} \right)dx+C$ 

$\Rightarrow y{{e}^{-x}}=x\int {{e}^{-x}}dx-\int \left( \left( x \right)'\int {{e}^{-x}}dx \right)dx+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}+\int \left( {{e}^{-x}} \right)dx+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}-{{e}^{-x}}+C$ 

$\Rightarrow y=-x-1+C{{e}^{x}}$ 

$\Rightarrow y+x+1=C{{e}^{x}}$ 

Given $\text{y=0}$ when $\text{x=0}$ as it passes through origin:

$0+0+1=C{{e}^{0}}$ 

$\Rightarrow C=1$ 

Therefore the equation of the required curve is $\text{y+x+1=}{{\text{e}}^{\text{x}}}$.

17. Find the equation of a curve passing through the point $\left( \text{0,2} \right)$  given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by $\text{5}$.

Ans: Let the slope of tangent be $\frac{\text{dy}}{\text{dx}}$.

According to :

$x+y=\frac{dy}{dx}+5$ 

The given differential equation is:

$x+y=\frac{dy}{dx}+5$ 

\[\frac{dy}{dx}+\left( -1 \right)y=x-5\] 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=-1$ 

$Q=x-5$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-1\int dx}}$ 

$\Rightarrow I.F={{e}^{-x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( {{e}^{-x}} \right)=\int \left( x-5 \right)\left( {{e}^{-x}} \right)dx+C$ 

$\Rightarrow y\left( {{e}^{-x}} \right)=\int x\left( {{e}^{-x}} \right)dx-5\int {{e}^{-x}}dx+C$ 

$\Rightarrow y{{e}^{-x}}=x\int {{e}^{-x}}dx-\int \left( \left( x \right)'\int {{e}^{-x}}dx \right)dx-5\int {{e}^{-x}}dxC$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}+\int \left( {{e}^{-x}} \right)dx+5{{e}^{-x}}+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}-{{e}^{-x}}+5{{e}^{-x}}+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}+4{{e}^{-x}}+C$ 

$\Rightarrow y=-x+4+C{{e}^{x}}$ 

$\Rightarrow y+x-4=C{{e}^{x}}$ 

Given as it passes through $\left( \text{0,2} \right)$ :

$2+0-4=C{{e}^{0}}$ 

$\Rightarrow C=-2$ 

Therefore the equation of the required curve is:

 $y+x-4=-2{{e}^{x}}$.

18. Find the integrating factor of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}\text{-y=2}{{\text{x}}^{\text{2}}}$.

A. ${{e}^{-x}}$ 

B. ${{e}^{-y}}$ 

C. $\frac{1}{x}$ 

D. $x$ 

Ans: Given differential equation is:

$x\frac{dy}{dx}-y=2{{x}^{2}}$ 

$\Rightarrow \frac{dy}{dx}-\left( \frac{1}{x} \right)y=2{{x}^{2}}$ 

Thus it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$:

$p=-\frac{1}{x}$ 

$I.F={{e}^{\int pdx}}$ 

\[\Rightarrow I.F={{e}^{-\int \frac{1}{x}dx}}\]

$\Rightarrow I.F={{e}^{-\log \left| x \right|}}$  

$\Rightarrow I.F={{e}^{\log {{x}^{-1}}}}$

$\Rightarrow I.F=\frac{1}{x}$

Therefore integrating factor is $\frac{\text{1}}{\text{x}}$. Thus the correct option is (C).

19. Find the integrating factor of the differential equation $\left( \text{1-}{{\text{y}}^{\text{2}}} \right)\frac{\text{dx}}{\text{dy}}\text{+yx=ay}\,\left( \text{-1y1} \right)$.

A. $\frac{1}{{{y}^{2}}-1}$ 

B. $\frac{1}{\sqrt{{{y}^{2}}-1}}$ 

C. $\frac{1}{1-{{y}^{2}}}$ 

D. $\frac{1}{\sqrt{1-{{y}^{2}}}}$ 

Ans: Given differential equation is:

$\left( 1-{{y}^{2}} \right)\frac{dx}{dy}+yx=ay$ 

$\Rightarrow \frac{dx}{dy}+\left( \frac{y}{1-{{y}^{2}}} \right)x=\frac{ay}{\left( 1-{{y}^{2}} \right)}$ 

Thus it is a linear differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$:

$p=\frac{y}{1-{{y}^{2}}}$ 

\[I.F={{e}^{\int pdy}}\] 

\[\Rightarrow I.F={{e}^{\int \frac{y}{1-{{y}^{2}}}dy}}\]

\[\Rightarrow I.F={{e}^{\frac{1}{-2}\int \frac{-2y}{1-{{y}^{2}}}dy}}\]  

\[\Rightarrow I.F={{e}^{\frac{1}{-2}\log \left( 1-{{y}^{2}} \right)}}\]

\[\Rightarrow I.F={{e}^{\log {{\left( 1-{{y}^{2}} \right)}^{-\frac{1}{2}}}}}\]

\[\Rightarrow I.F=\frac{1}{\sqrt{1-{{y}^{2}}}}\] 

Therefore integrating factor is $\frac{\text{1}}{\sqrt{\text{1-}{{\text{y}}^{\text{2}}}}}$. Thus the correct option is (D).

Miscellaneous Exercise 

1. For each of the differential equations given below, indicate its order and degree (if defined).

(i)$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+5x}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}\text{-6y=logx}$ 

Ans: The given differential equation is:

 $\frac{{{d}^{2}}y}{d{{x}^{2}}}+5x{{\left( \frac{dy}{dx} \right)}^{2}}-6y-\log x=0$ 

The highest order derivative in the equation is of the term $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order of the equation is $\text{2}$ and its highest power is $\text{1}$. Therefore its degree is $\text{1}$.

(ii)${{\left( \frac{\text{dy}}{\text{dx}} \right)}^{3}}-4{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{2}}\text{+7y=sin}\,\text{x}$ 

Ans: The given differential equation is:

 ${{\left( \frac{dy}{dx} \right)}^{3}}-4{{\left( \frac{dy}{dx} \right)}^{2}}+7y-\sin x=0$ 

The highest order derivative in the equation is of the term ${{\left( \frac{\text{dy}}{\text{dx}} \right)}^{3}}$, thus the order of the equation is $1$ and its highest power is $3$. Therefore its degree is $3$.

(iii)$\frac{{{\text{d}}^{\text{4}}}\text{y}}{\text{d}{{\text{x}}^{\text{4}}}}\text{-sin}\left( \frac{{{\text{d}}^{\text{3}}}\text{y}}{\text{d}{{\text{x}}^{\text{3}}}} \right)\text{=0}$ 

Ans: The given differential equation is:

 $\frac{{{d}^{4}}y}{d{{x}^{4}}}-\sin \left( \frac{{{d}^{3}}y}{d{{x}^{3}}} \right)=0$ 

The highest order derivative in the equation is of the term $\frac{{{\text{d}}^{4}}\text{y}}{\text{d}{{\text{x}}^{4}}}$, thus the order of the equation is $4$.

As the differential equation is not polynomial in its derivative, therefore its degree is not defined.

2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

1. $\text{xy=a}{{\text{e}}^{\text{x}}}\text{+b}{{\text{e}}^{\text{-x}}}\text{+}{{\text{x}}^{\text{2}}}\,\,\,\text{:x}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+2}\frac{\text{dy}}{\text{dx}}\text{-xy+}{{\text{x}}^{\text{2}}}\text{-2=0}$ 

Ans: The given function is:
$xy=a{{e}^{x}}+b{{e}^{-x}}+{{x}^{2}}$ 

Take derivative on both side:

$\Rightarrow y+x\frac{dy}{dx}=a{{e}^{x}}-b{{e}^{-x}}+2x$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}+x\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}=a{{e}^{x}}+b{{e}^{-x}}+2$ 

$\Rightarrow x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}=a{{e}^{x}}+b{{e}^{-x}}+2$ ……(1)

The given differential equation is:
$x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}-xy+{{x}^{2}}-2=0$ 

Solving LHS:

Substitute $\text{x}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+2}\frac{\text{dy}}{\text{dx}}$ from the result (1) and $\text{xy}$:

$\Rightarrow \left( x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx} \right)-xy+{{x}^{2}}-2$ 

$\Rightarrow \left( a{{e}^{x}}+b{{e}^{-x}}+2 \right)-\left( a{{e}^{x}}+b{{e}^{-x}}+{{x}^{2}} \right)+{{x}^{2}}-2$ 

$\Rightarrow 2-{{x}^{2}}+{{x}^{2}}-2$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

2. $\text{y=}{{\text{e}}^{\text{x}}}\left( \text{acosx+bsinx} \right)\,\,\text{:}\,\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-2}\frac{\text{dy}}{\text{dx}}\text{+2y=0}$ 

Ans: The given function is:
$y={{e}^{x}}\left( a\cos x+b\sin x \right)$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\left( a\cos x+b\sin x \right)+{{e}^{x}}\left( -a\sin x+b\cos x \right)$ 

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)$ 

Take derivative on both side:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)+{{e}^{x}}\left( -\left( a+b \right)\sin x+\left( b-a \right)\cos x \right)$

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( \left( a+b+b-a \right)\cos x+\left( b-a-a-b \right)\sin x \right)$ 

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( 2b\cos x-2a\sin x \right)$  

The given differential equation is:
$\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}+2y=0$ 

Solving LHS:

$\Rightarrow {{e}^{x}}\left( 2b\cos x-2a\sin x \right)-2{{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)+2y$ 

$\Rightarrow {{e}^{x}}\left( \left( 2b-2a-2b \right)\cos x+\left( -2a-2b+2a \right)\sin x \right)-2y$ 

$\Rightarrow {{e}^{x}}\left( -2a\cos x-2b\sin x \right)-2y$ 

$\Rightarrow -2{{e}^{x}}\left( a\cos x+b\sin x \right)-2y$

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

3. $\text{y=x}\,\text{sin3x}\,\,\text{:}\,\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+9y-6cos3x=0}$ 

Ans: The given function is:
$y=x\sin 3x$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}=\sin 3x+3x\cos 3x$ 

Take derivative on both side:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=3\cos 3x+3\left( \cos 3x+x\left( -3\sin 3x \right) \right)$

\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=3\cos 3x+3\cos 3x-9x\sin 3x\] 

\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=6\cos 3x-9x\sin 3x\]  

The given differential equation is:
$\frac{{{d}^{2}}y}{d{{x}^{2}}}+9y-6\cos 3x=0$ 

Solving LHS:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}+9y-6\cos 3x$ 

$\Rightarrow \left( 6\cos 3x-9x\sin 3x \right)+9\left( x\sin 3x \right)-6\cos 3x$ 

$\Rightarrow 6\cos 3x-9x\sin 3x+9x\sin 3x-6\cos 3x$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

4. ${{\text{x}}^{\text{2}}}\text{=2}{{\text{y}}^{\text{2}}}\,\text{log y}\,\,\text{:}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)\frac{\text{dy}}{\text{dx}}\text{-xy=0}$ 

Ans: The given function is:
${{x}^{2}}=2{{y}^{2}}\log y$ 

Take derivative on both side:

\[\Rightarrow 2x=2\left( 2y\log y+{{y}^{2}}\left( \frac{1}{y} \right) \right)\frac{dy}{dx}\] 

\[\Rightarrow \frac{dy}{dx}=\frac{x}{\left( 2y\log y+y \right)}\]

Multiply numerator and denominator by $\text{y}$:

\[\Rightarrow \frac{dy}{dx}=\frac{xy}{\left( 2{{y}^{2}}\log y+{{y}^{2}} \right)}\] 

\[\Rightarrow \frac{dy}{dx}=\frac{xy}{\left( {{x}^{2}}+{{y}^{2}} \right)}\]  

The given differential equation is:
$\left( {{x}^{2}}+{{y}^{2}} \right)\frac{dy}{dx}-xy=0$ 

Solving LHS:

$\Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)\frac{dy}{dx}-xy$ 

$\Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)\left( \frac{xy}{{{x}^{2}}+{{y}^{2}}} \right)-xy$ 

$\Rightarrow xy-xy$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

3. Prove that ${{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=c}{{\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)}^{\text{2}}}$  is the general solution of differential equation $\left( {{\text{x}}^{\text{3}}}\text{-3x}{{\text{y}}^{\text{2}}} \right)\text{dx=}\left( {{\text{y}}^{\text{3}}}\text{-3}{{\text{x}}^{\text{2}}}\text{y} \right)\text{dy}$ , where $\text{c}$ is a parameter.

Ans: Given differential equation:

$\left( {{x}^{3}}-3x{{y}^{2}} \right)dx=\left( {{y}^{3}}-3{{x}^{2}}y \right)dy$ 

$\Rightarrow \frac{dy}{dx}=\frac{{{x}^{3}}-3x{{y}^{2}}}{{{y}^{3}}-3{{x}^{2}}y}$ 

As it can be seen that this is an homogenous equation. Substitute $\text{y=vx}$:

$\Rightarrow \frac{d\left( vx \right)}{dx}=\frac{{{x}^{3}}-3x{{\left( vx \right)}^{2}}}{{{\left( vx \right)}^{3}}-3{{x}^{2}}\left( vx \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{{{x}^{3}}\left( 1-3{{v}^{2}} \right)}{{{x}^{3}}\left( {{v}^{3}}-3v \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{1-3{{v}^{2}}}{{{v}^{3}}-3v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-3{{v}^{2}}}{{{v}^{3}}-3v}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-3{{v}^{2}}-{{v}^{4}}+3{{v}^{2}}}{{{v}^{3}}-3v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-{{v}^{4}}}{{{v}^{3}}-3v}$ 

Separate the differentials:

$\frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\log x+\log C$ 

\[\Rightarrow I=\log x+\log C\,\,\left( I=\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv \right)\] ……(1)

Solving integral $\text{I}$:

\[\Rightarrow I=\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv\] 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{{{v}^{3}}-3v}{\left( 1-{{v}^{2}} \right)\left( 1+{{v}^{2}} \right)}\] 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{{{v}^{3}}-3v}{\left( 1-v \right)\left( 1+v \right)\left( 1+{{v}^{2}} \right)}\] 

Using partial fraction:

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{A}{1-v}+\frac{B}{1+v}+\frac{Cv+D}{1+{{v}^{2}}}\] 

Solving for $\text{A,B,C}\,\text{and}\,\text{D}$:

$A=-\frac{1}{2}$ 

$B=\frac{1}{2}$ 

$C=-2$ 

$D=0$ 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{-\frac{1}{2}}{1-v}+\frac{\frac{1}{2}}{1+v}+\frac{-2v+0}{1+{{v}^{2}}}\] 

$I=-\frac{1}{2}\int \frac{1}{1-v}dv+\frac{1}{2}\int \frac{1}{1+v}dv-\int \frac{2v}{1+{{v}^{2}}}dv$

$I=-\frac{1}{2}\left( -\log \left( 1-v \right) \right)+\frac{1}{2}\left( \log \left( 1+v \right) \right)-\log \left( 1+{{v}^{2}} \right)$ 

\[I=\frac{1}{2}\left( \log \left( 1-{{v}^{2}} \right) \right)-\frac{2}{2}\log \left( 1+{{v}^{2}} \right)\] 

\[I=\frac{1}{2}\left( \log \frac{\left( 1-{{v}^{2}} \right)}{{{\left( 1+{{v}^{2}} \right)}^{2}}} \right)\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( 1-\frac{{{y}^{2}}}{{{x}^{2}}} \right)}{{{\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right)}^{2}}} \right)\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{{{x}^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)\]

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\frac{1}{2}\log {{x}^{2}}\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\log x\] 

Back substitute $\text{I}$ in expression (1):

\[\Rightarrow I=\log x+\log C\,\] 

\[\Rightarrow \frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\log x=\log x+\log C\]

\[\Rightarrow \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}=2\log C\] 

\[\Rightarrow \frac{{{x}^{2}}-{{y}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}={{C}^{2}}\] 

\[\Rightarrow {{x}^{2}}-{{y}^{2}}=c{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}\,\,\left( c={{C}^{2}} \right)\] 

Thus for given differential equation, its general solution is \[{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=c}{{\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)}^{\text{2}}}\].

4. Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\sqrt{\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1-}{{\text{x}}^{\text{2}}}}}\text{=0}$.

Ans: The given differential equation is:

 $\frac{dy}{dx}+\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}=0$ 

$\frac{dy}{dx}=-\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}$ 

$\Rightarrow \frac{dy}{\sqrt{1-{{y}^{2}}}}=-\frac{dx}{\sqrt{1-{{x}^{2}}}}$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{\sqrt{1-{{y}^{2}}}}=-\int \frac{dx}{\sqrt{1-{{x}^{2}}}}$ 

$\Rightarrow {{\sin }^{-1}}y=-{{\sin }^{-1}}x+C$

$\Rightarrow {{\sin }^{-1}}y+{{\sin }^{-1}}x=C$ 

Thus the general solution of given differential equation is $\text{si}{{\text{n}}^{\text{-1}}}\text{y+si}{{\text{n}}^{\text{-1}}}\text{x=C}$.

5. Show that the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\frac{{{\text{y}}^{\text{2}}}\text{+y+1}}{{{\text{x}}^{\text{2}}}\text{+x+1}}\text{=0}$ is given by $\left( \text{x+y+1} \right)\text{=A}\left( \text{1-x-y-2xy} \right)$  where $\text{A}$ is a parameter. 

Ans: The given differential equation is:

$\frac{dy}{dx}+\frac{{{y}^{2}}+y+1}{{{x}^{2}}+x+1}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{{{y}^{2}}+y+1}{{{x}^{2}}+x+1}$

$\Rightarrow \frac{dy}{dx}=-\frac{{{y}^{2}}+2\left( \frac{1}{2} \right)y+\frac{1}{4}-\frac{1}{4}+1}{{{x}^{2}}+2\left( \frac{1}{2} \right)x+\frac{1}{4}-\frac{1}{4}+1}$ 

$\Rightarrow \frac{dy}{dx}=-\frac{{{\left( y+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}{{{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}$   

$\Rightarrow \frac{dy}{{{\left( y+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}=-\frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{{{\left( y+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}=-\int \frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}$

$\Rightarrow \frac{1}{\left( \frac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left[ \frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right]=-\frac{1}{\left( \frac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left[ \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right]+C$ 

$\Rightarrow \frac{2}{\sqrt{3}}\left( {{\tan }^{-1}}\left[ \frac{2y+1}{\sqrt{3}} \right]+{{\tan }^{-1}}\left[ \frac{2x+1}{\sqrt{3}} \right] \right)=C$ 

$\Rightarrow {{\tan }^{-1}}\left[ \frac{2y+1}{\sqrt{3}} \right]+{{\tan }^{-1}}\left[ \frac{2x+1}{\sqrt{3}} \right]=\frac{\sqrt{3}}{2}C$ 

Thus the general solution for given differential equation is $\text{ta}{{\text{n}}^{\text{-1}}}\left[ \frac{\text{2y+1}}{\sqrt{\text{3}}} \right]\text{+ta}{{\text{n}}^{\text{-1}}}\left[ \frac{\text{2x+1}}{\sqrt{\text{3}}} \right]\text{=}\frac{\sqrt{\text{3}}}{\text{2}}\text{C}$.

6. Find the equation of the curve passing through the point $\left( 0,\frac{\pi }{4} \right)$ whose differential equation is $\text{sin}\,\text{x}\,\text{cos}\,\text{ydx+cos}\,\text{x}\,\text{sin}\,\text{ydy=0}$.

Ans: Given differential equation is:

$\sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy=0$

$\Rightarrow \sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy=0$

Divide both side by $\text{cos}\,\text{x}\,\text{cos}\,\text{y}$:

$\Rightarrow \frac{\sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy}{\cos x\cos y}=0$ 

$\Rightarrow \tan xdx+\tan ydy=0$ 

$\Rightarrow \tan ydy=-\tan xdx$ 

Integrate both side:

$\Rightarrow \int \tan ydy=-\int \tan xdx$ 

$\Rightarrow \log \left( \sec y \right)=-\log \left( \sec x \right)+C$ 

$\Rightarrow \log \left( \sec y \right)+\log \left( \sec x \right)=C$ 

$\Rightarrow \log \left( \sec x\sec y \right)=C$ 

$\Rightarrow \sec x\sec y=k\,\,\left( k={{e}^{C}} \right)$ 

As curve passes through $\left( \text{0,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$:

$\sec 0\sec \left( \frac{\pi }{4} \right)=k$ 

$\Rightarrow k=\sqrt{2}$ 

$\Rightarrow \sec x\sec y=\sqrt{2}$ 

Thus the equation of required curve is $\text{sec}\,\text{x}\,\text{sec}\,\text{y=}\sqrt{\text{2}}$.

7. Find the particular solution of the differential equation $\left( \text{1+}{{\text{e}}^{\text{2x}}} \right)\text{dy+}\left( \text{1+}{{\text{y}}^{\text{2}}} \right){{\text{e}}^{\text{x}}}\text{dx=0}$ given that $\text{y=1}$ when $\text{x=0}$.

Ans: The given differential equation is:

$\left( 1+{{e}^{2x}} \right)dy+\left( 1+{{y}^{2}} \right){{e}^{x}}dx=0$ 

Divide both side $\left( \text{1+}{{\text{e}}^{\text{2x}}} \right)\left( \text{1+}{{\text{y}}^{\text{2}}} \right)$:

\[\frac{dy}{\left( 1+{{y}^{2}} \right)}+\frac{{{e}^{x}}}{\left( 1+{{e}^{2x}} \right)}dx=0\] 

\[\int \frac{dy}{\left( 1+{{y}^{2}} \right)}=-\int \frac{{{e}^{x}}}{\left( 1+{{e}^{2x}} \right)}dx\] 

\[{{\tan }^{-1}}y=-\int \frac{{{e}^{x}}}{\left( 1+{{\left( {{e}^{x}} \right)}^{2}} \right)}dx\]

Substitute $t={{e}^{x}}$:

$dt={{e}^{x}}dx$ 

\[\Rightarrow {{\tan }^{-1}}y=-\int \frac{1}{\left( 1+{{t}^{2}} \right)}dt\] 

\[\Rightarrow {{\tan }^{-1}}y=-{{\tan }^{-1}}t+C\] 

\[\Rightarrow {{\tan }^{-1}}y=-{{\tan }^{-1}}{{e}^{x}}+C\] 

$\Rightarrow {{\tan }^{-1}}y+{{\tan }^{-1}}{{e}^{x}}=C$ 

As $\text{y=1}$ when $\text{x=0}$:

${{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( {{e}^{0}} \right)=C$ 

\[\Rightarrow \frac{\pi }{4}+\frac{\pi }{4}=C\] 

\[\Rightarrow C=\frac{\pi }{2}\] 

$\Rightarrow {{\tan }^{-1}}y+{{\tan }^{-1}}{{e}^{x}}=\frac{\pi }{2}$ 

Thus the required particular solution is $\text{ta}{{\text{n}}^{\text{-1}}}\text{y+ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{x}}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

8. Solve the differential equation $\text{y}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{dx=}\left( \text{x}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{+}{{\text{y}}^{\text{2}}} \right)\text{dy}\,\,\left( \text{y}\ne \text{0} \right)$.

Ans: The given differential equation is:

$y{{e}^{\frac{x}{y}}}dx=\left( x{{e}^{\frac{x}{y}}}+{{y}^{2}} \right)dy$ 

$y{{e}^{\frac{x}{y}}}\frac{dx}{dy}=x{{e}^{\frac{x}{y}}}+{{y}^{2}}$ 

$\Rightarrow y{{e}^{\frac{x}{y}}}\frac{dx}{dy}-x{{e}^{\frac{x}{y}}}={{y}^{2}}$ 

$\Rightarrow {{e}^{\frac{x}{y}}}\frac{\left[ y\frac{dx}{dy}-x \right]}{{{y}^{2}}}=1$ 

Substitute $\text{z=}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}$:

$z={{e}^{\frac{x}{y}}}$ 

$\frac{d}{dy}z=\frac{d}{dy}{{e}^{\frac{x}{y}}}$ 

$\Rightarrow \frac{dz}{dy}=\frac{d}{dy}\left( {{e}^{\frac{x}{y}}} \right)$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\frac{d}{dy}\left( \frac{x}{y} \right)$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\left[ \left( \frac{1}{y} \right)\frac{dx}{dy}-\frac{x}{{{y}^{2}}} \right]$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\left[ \frac{y\frac{dx}{dy}-x}{{{y}^{2}}} \right]$ 

$\Rightarrow \frac{dz}{dy}=1$ 

\[\Rightarrow dz=dy\] 

\[\Rightarrow \int dz=\int dy\]

\[\Rightarrow z=y+C\] 

\[\Rightarrow {{e}^{\frac{x}{y}}}=y+C\] 

Thus the required general solution is \[{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{=y+C}\].

9. Find a particular solution of the differential equation $\left( \text{x-y} \right)\left( \text{dx+dy} \right)\text{=dx-dy}$ given that $\text{y=-1}$ when $\text{x=0}$. Hint (put $\text{x-y=t}$).

Ans: Given differential equation is:

$\left( x-y \right)\left( dx+dy \right)=dx-dy$ 

$\Rightarrow \left( x-y \right)dx-dx=\left( y-x \right)dy-dy$

$\Rightarrow \left( x-y+1 \right)dy=\left( 1-x+y \right)dx$ 

$\Rightarrow \frac{dy}{dx}=\frac{1-x+y}{x-y+1}$ 

Put $\text{x-y=t}$:

$x-y=t$ 

$\Rightarrow 1-\frac{dy}{dx}=\frac{dt}{dx}$ 

$\Rightarrow 1-\frac{dt}{dx}=\frac{dy}{dx}$ 

$\Rightarrow 1-\frac{dt}{dx}=\frac{1-t}{1+t}$ 

$\Rightarrow \frac{dt}{dx}=1-\frac{1-t}{1+t}$

$\Rightarrow \frac{dt}{dx}=\frac{1+t-1+t}{1+t}$ 

$\Rightarrow \frac{dt}{dx}=\frac{2t}{1+t}$ 

$\Rightarrow \frac{1+t}{t}dt=2dx$ 

Integrate both side:

$\Rightarrow \int \frac{1+t}{t}dt=2\int dx$

$\Rightarrow \int \frac{1}{t}dt+\int dt=2x+C$ 

$\Rightarrow \log \left| t \right|+t=2x+C$ 

$\Rightarrow \log \left| x-y \right|+x-y=2x+C$ 

$\Rightarrow \log \left| x-y \right|-y=x+C$ 

As $\text{y=-1}$ when $\text{x=0}$:

$\Rightarrow \log \left| 0-\left( -1 \right) \right|-\left( -1 \right)=0+C$ 

$\Rightarrow \log 1+1=C$ 

$\Rightarrow C=1$ 

Thus the required particular solution is:

$\log \left| x-y \right|-y=x+1$.

10. Solve the differential equation $\left[ \frac{{{\text{e}}^{\text{-2}\sqrt{\text{x}}}}}{\sqrt{\text{x}}}\text{-}\frac{\text{y}}{\sqrt{\text{x}}} \right]\frac{\text{dx}}{\text{dy}}\text{=1}\,\left( \text{x}\ne \text{0} \right)$.

Ans: Given differential equation is:
$\left[ \frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right]\frac{dx}{dy}=1$ 

$\Rightarrow \frac{dy}{dx}=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}$ 

$\Rightarrow \frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}$ 

It is linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$:

$p=\frac{1}{\sqrt{x}}$ 

$Q=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}$ 

Calculating integrating factor:

$I.F={{e}^{\int pdx}}$

$I.F={{e}^{\int \frac{1}{\sqrt{x}}dx}}$ 

$I.F={{e}^{2\sqrt{x}}}$  

The general solution is given by:

$y\times I.F=\int \left( Q\times I.F \right)dx+C$ 

$y\times \left( {{e}^{2\sqrt{x}}} \right)=\int \left( \frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}\times {{e}^{2\sqrt{x}}} \right)dx+C$ 

$\Rightarrow y{{e}^{2\sqrt{x}}}=\int \left( \frac{{{e}^{-2\sqrt{x}+2\sqrt{x}}}}{\sqrt{x}} \right)dx+C$ 

$\Rightarrow y{{e}^{2\sqrt{x}}}=\int \frac{1}{\sqrt{x}}dx+C$ 

$\Rightarrow y{{e}^{2\sqrt{x}}}=2\sqrt{x}+C$ 

Thus the general solution for the given differential equation is

$y{{e}^{2\sqrt{x}}}=2\sqrt{x}+C$.

11. Find a particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+ycotx=4x}\,\text{cosec}\,\text{x}\left( \text{x}\ne \text{0} \right)$ given that $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

Ans:
The given differential equation is:

 $\frac{dy}{dx}+y\cot x=4x\,\text{cosec}x$ 

It is linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$:

$p=\cot x$ 

$Q=4x\text{cosec}\,x$ 

Calculating integrating factor:

$I.F={{e}^{\int pdx}}$

$I.F={{e}^{\int \cot xdx}}$ 

$I.F={{e}^{\log \left| \sin x \right|}}$ 

$I.F=\sin x$ 

The general solution is given by:

$y\times I.F=\int \left( Q\times I.F \right)dx+C$

$\Rightarrow y\times \sin x=\int \left( 4x\cos \text{ec}x \right)\sin xdx+C$ 

$\Rightarrow y\sin x=4\int xdx+C$ 

$\Rightarrow y\sin x=4\left( \frac{{{x}^{2}}}{2} \right)+C$ 

$\Rightarrow y\sin x=2{{x}^{2}}+C$ 

As $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$:

$0\times \sin \left( \frac{\pi }{2} \right)=2{{\left( \frac{\pi }{2} \right)}^{2}}+C$ 

$C=-2\left( \frac{{{\pi }^{2}}}{4} \right)$ 

$C=-\frac{{{\pi }^{2}}}{2}$ 

Thus the required particular solution is:

$y\sin x=2{{x}^{2}}-\frac{{{\pi }^{2}}}{2}$

12. Find a particular solution of the differential equation $\left( \text{x+1} \right)\frac{\text{dy}}{\text{dx}}\text{=2}{{\text{e}}^{\text{-y}}}\text{-1}$ given that $\text{y=0}$ when $\text{x=0}$.

Ans:
The given differential equation is:

 $\left( x+1 \right)\frac{dy}{dx}=2{{e}^{-y}}-1$

$\Rightarrow \frac{dy}{2{{e}^{-y}}-1}=\frac{dx}{x+1}$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\int \frac{dx}{x+1}$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \left( x+1 \right)+\log C$ ……(1) 

Evaluating LHS integral:

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\int \frac{{{e}^{y}}dy}{2-{{e}^{y}}}$ 

Put $\text{t=2-}{{\text{e}}^{\text{y}}}$:

$t=2-{{e}^{y}}$

$dt=-{{e}^{y}}dy$  

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=-\int \frac{dt}{t}$

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=-\log \left( t \right)$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \frac{1}{t}$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \frac{1}{2-{{e}^{y}}}$ 

Back substituting in expression (1):

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \left( x+1 \right)+\log C$ 

$\Rightarrow \log \left( \frac{1}{2-{{e}^{y}}} \right)=\log C\left( x+1 \right)$ 

$\Rightarrow 2-{{e}^{y}}=\frac{1}{C\left( x+1 \right)}$ 

As $\text{y=0}$ when $\text{x=0}$:

$\Rightarrow 2-{{e}^{0}}=\frac{1}{C\left( 0+1 \right)}$ 

$\Rightarrow 2-1=\frac{1}{C}$ 

$\Rightarrow C=1$ 

Thus the required particular solution is:

$\Rightarrow 2-{{e}^{y}}=\frac{1}{\left( x+1 \right)}$

$\Rightarrow {{e}^{y}}=2-\frac{1}{\left( x+1 \right)}$ 

$\Rightarrow {{e}^{y}}=\frac{2x+2-1}{\left( x+1 \right)}$ 

$\Rightarrow {{e}^{y}}=\frac{2x+1}{x+1}$ 

$\Rightarrow y=\log \left( \frac{2x+1}{x+1} \right)$  

Thus for given conditions the particular solution is $\text{y=log}\left( \frac{\text{2x+1}}{\text{x+1}} \right)$ .

13. The general solution of the differential equation $\frac{\text{ydx-xdy}}{\text{y}}\text{=0}$.

A. $xy=C$ 

B. $x=C{{y}^{2}}$ 

C. $y=Cx$ 

D. $y=C{{x}^{2}}$ 

Ans: Given differential equation:

$\frac{ydx-xdy}{y}=0$

Divide both side by $\text{x}$ :

$\Rightarrow \frac{ydx-xdy}{xy}=0$

$\Rightarrow \frac{dx}{x}-\frac{dy}{y}=0$ 

Integrate both side:

$\Rightarrow \int \frac{dx}{x}-\int \frac{dy}{y}=0$ 

$\Rightarrow \log \left| x \right|-\log \left| y \right|=\log k$ 

$\Rightarrow \log \left| \frac{x}{y} \right|=\log k$ 

$\Rightarrow \frac{x}{y}=k$ 

$\Rightarrow y=Cx\,\,\,\left( C=\frac{1}{k} \right)$

Thus the correct option is (C)

14. Find the general solution of a differential equation of the type $\frac{\text{dx}}{\text{dy}}\text{+}{{\text{P}}_{\text{1}}}\text{x=}{{\text{Q}}_{\text{1}}}$.

A. \[y{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C\]   

B. \[y{{e}^{\int {{P}_{1}}dx}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dx}} \right)dy+C\] 

C. \[x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C\]   

D. \[x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dx}} \right)dy+C\]   

Ans: The given differential equation is:

$\frac{dx}{dy}+{{P}_{1}}x={{Q}_{1}}$ 

It is a linear differential equation and its general solution is:

$x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C$ 

With integrating factor $I.F={{e}^{\int {{P}_{1}}dy}}$.

Thus the correct option is (C).

15. Find the general solution of the differential equation ${{\text{e}}^{\text{x}}}\text{dy+}\left( \text{y}{{\text{e}}^{\text{x}}}\text{+2x} \right)\text{dx=0}$.

A. $\text{x}{{\text{e}}^{\text{y}}}\text{+}{{\text{x}}^{\text{2}}}\text{=C}$ 

B. $\text{x}{{\text{e}}^{\text{y}}}\text{+}{{\text{y}}^{\text{2}}}\text{=C}$ 

C. $\text{y}{{\text{e}}^{\text{x}}}\text{+}{{\text{x}}^{\text{2}}}\text{=C}$ 

D. $\text{y}{{\text{e}}^{\text{y}}}\text{+}{{\text{x}}^{\text{2}}}\text{=C}$ 

Ans: The given differential equation is:
${{e}^{x}}dy+\left( y{{e}^{x}}+2x \right)dx=0$ 

$\Rightarrow {{e}^{x}}\frac{dy}{dx}+y{{e}^{x}}=-2x$ 

$\Rightarrow \frac{dy}{dx}+y=-2x{{e}^{-x}}$   

The given differential equation is of the form:

$\frac{dy}{dx}+Py=Q$ 

$\Rightarrow P=1$ 

$\Rightarrow Q=-2x{{e}^{-x}}$ 

Calculating integrating factor:
$I.F={{e}^{\int Pdx}}$

$\Rightarrow I.F={{e}^{\int dx}}$ 

$\Rightarrow I.F={{e}^{x}}$  

It is a linear differential equation and its general solution is:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$

 $y\left( {{e}^{x}} \right)=\int \left( -2x{{e}^{-x}}\times {{e}^{x}} \right)dy+C$ 

$\Rightarrow y{{e}^{x}}=-2\int xdx+C$ 

$\Rightarrow y{{e}^{x}}=-2\left( \frac{{{x}^{2}}}{2} \right)+C$ 

$\Rightarrow y{{e}^{x}}+{{x}^{2}}=C$ 

Thus the correct answer is option (C).

NCERT Solutions for Class 12 Maths Chapter 9 Important Points

• Differential Equation: A differential equation has an independent variable, a dependent variable, derivatives of the dependent variable with respect to the independent variable, and a constant.

• Ordinary Differential Equation: An ordinary differential equation is one that involves derivatives of the dependent variable with respect to only one independent variable.

• Order of a Differential Equation: The order of a differential equation is defined as the highest order derivative of the dependent variable with respect to the independent variable.

• Degree of a Differential Equation: The degree of a differential equation is the highest exponent of the highest order derivative if the exponent of each derivative is a non-negative integer and the unknown variable in the differential equation is a non-negative integer.

• General solution: The general solution of a differential equation is one that contains as many arbitrary constants as the order of the differential equation, i.e., if the solution of a differential equation of order n has n arbitrary constants, it is the general solution.

• Particular Solution: The particular solution is a solution obtained by giving particular values to arbitrary constants in the general solution of a differential equation.

Overview of Deleted Syllabus for CBSE Class 12 Maths Differential Equations

Class 12 Maths Chapter 9: Exercises Breakdown

Conclusion

The Solutions for Maths Chapter 9 class 12 differential equations NCERT solutions, provided by Vedantu, are designed to help students understand the ability to analyse and solve problems involving rates of change and relationships between variables - a valuable skill in various scientific disciplines. This chapter provides tools to analyse and solve equations, which have numerous applications in various scientific fields.

From previous year's question papers, around 8 questions are typically asked from this chapter. Understanding and practising these solutions will help students score well on their exams.

Other Study Material for CBSE Class 12 Maths Chapter 9

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

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