Courses

Courses for Kids

Free study material

Store

Talk to our experts

1800-120-456-456

NCERT Solutions

Class 12

Maths

Chapter 9 Differential Equations Exercise 9.5

NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations Exercise 9.5 - 2025-26

Q: 1. What approach should be followed to solve linear differential equations in NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5?

To solve linear differential equations in Exercise 9.5, use the following methodical approach:Identify the equation in the form dy/dx + Py = Q.Calculate the integrating factor (IF).Multiply the entire equation by the integrating factor.Integrate both sides and add the constant of integration.If initial conditions are given, substitute them to get the particular solution.This step-by-step NCERT Solutions methodology ensures accuracy and is aligned with the CBSE marking scheme.

Q: 2. Why is the integrating factor critical in solving first-order linear differential equations?

The integrating factor transforms a first-order linear differential equation into an exact equation, allowing for straightforward integration. By multiplying both sides with the IF, the left side becomes a derivative of a product, simplifying the solution process and ensuring you derive the general solution efficiently, a key focus in NCERT Solutions for Class 12 Maths Chapter 9.

Q: 3. How can students avoid common mistakes when solving differential equations in Exercise 9.5?

To minimize errors while solving:Always include the constant of integration (C) after integrating.Select the correct solution method as per the type of equation.Pay attention to algebraic steps and arithmetic calculations.Apply initial or boundary conditions carefully for particular solutions.Review definitions of order and degree before starting each problem.This careful approach is vital for achieving full marks in board and competitive exams.

Q: 4. What is the significance of 'order' and 'degree' in differential equations according to the NCERT syllabus?

Order indicates the highest derivative present in the equation, while degree is the power of the highest order derivative after making the equation rational and free from radicals. Correctly identifying both is essential for choosing the right solution method and is commonly assessed in both board and competitive exams.

Q: 5. Which problem-solving techniques from Class 12 Maths Chapter 9 are most commonly tested in CBSE board and entrance exams?

The main techniques tested are:Separation of variablesIntegration by partsMethod of integrating factors for linear equationsHomogeneous and non-homogeneous equation strategiesMastering these methods as outlined in the NCERT Solutions maximizes exam readiness and accuracy.

Q: 6. How do solved examples in Class 12 Maths Chapter 9 Exercise 9.5 build conceptual clarity?

The step-by-step solutions in Exercise 9.5 break down each type of differential equation, clarify the application of methods like integrating factors, and reinforce the need for correct integration and constant handling. This builds a deep understanding of the concepts and their application in exam questions.

Q: 7. In what ways do NCERT Solutions for Class 12 Maths Chapter 9 aid in preparation for JEE or NEET along with CBSE board exams?

NCERT Solutions cover fundamental problem-types and methodologies commonly asked in JEE and NEET exams. These solutions bridge conceptual understanding and exam patterns, ensuring students are prepared for both objective and descriptive scenarios, making them highly relevant for both board and entrance examinations.

Q: 8. What misconceptions do students have about integrating factors, and how can they be corrected?

Common misconceptions include:Believing any function can be used as an integrating factor (only the exponential of the integral of P works).Forgetting to multiply the entire equation by the integrating factor.Not recognizing the exactness achieved after applying IF.Clearing these misconceptions through stepwise NCERT Solutions explanations enhances problem-solving accuracy and conceptual clarity.

Q: 9. How should one verify if a solution to a differential equation is correct as per NCERT methodologies?

To verify your solution:Substitute the obtained solution back into the original differential equation.For particular solutions, ensure initial or boundary conditions are satisfied.Compare your process and answer with expert-verified NCERT Solutions to ensure every critical step is addressed.This confirms accuracy and builds confidence for CBSE board assessments.

Q: 10. What makes the NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5 unique compared to other resources?

These solutions are crafted by expert educators strictly following the CBSE 2025–26 syllabus, provide detailed, structured, and stepwise methods, and are periodically updated for accuracy. Explanations match marking schemes and common exam patterns, making them a reliable tool for exam preparation.

Latest Updates

Book Free Demo :

Turn Doubts into Confidence with One-to-One Learning

Differential Equations Questions and Answers - Free PDF Download

In NCERT Solutions Class 12 Maths Chapter 9 Exercise 9 5, you’ll get to practice solving different types of differential equations using simple step-by-step methods. This exercise is designed to help you understand tricky ideas like integrating factors and separation of variables, making it much easier to solve questions that often look confusing at first. If you’re ever stuck, Vedantu’s easy explanations and solved examples can guide you through each step.

Table of Content

Want to see what’s important for your board exam? You can always check the Class 12 Maths syllabus for the latest topics. If you need solutions for other chapters, our Class 12 Maths NCERT Solutions are available for download anytime.

This chapter carries 7 marks in your CBSE exam, so practicing these NCERT Solutions will make you more confident and help you score better. Download the free PDF and get ready for your exams with well-explained answers!

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

```html 1. Solve the differential equation \[\dfrac{{dy}}{{dx}} + 2y = \sin x\]. Summary: This question involves solving a first-order linear differential equation using an integrating factor.

The equation is of the form: \[\frac{dy}{dx} + Py = Q\] where \(P = 2\), \(Q = \sin x\).
Find the integrating factor: \[I.F. = e^{\int 2dx} = e^{2x}\]
Multiply the equation by the integrating factor:
\[y e^{2x} = \int \sin x \cdot e^{2x} dx + C\]
Let \(I = \int \sin x \cdot e^{2x} dx\).
Using integration by parts:
- \(I = \frac{e^{2x} \sin x}{2} - \int \cos x \cdot \frac{e^{2x}}{2} dx\)
- Apply integration by parts again for the second term:
- \(I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4}I\)
- Solve for \(I\):
- \(\frac{5}{4}I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4}\)
- \(I = \frac{e^{2x}}{5}(2\sin x - \cos x)\)
Plug in \(I\) into the solution:
\(y e^{2x} = \frac{e^{2x}}{5}(2\sin x - \cos x) + C\)
Thus: \[y = \frac{1}{5}(2\sin x - \cos x) + C e^{-2x}\]

2. Solve the differential equation \[\dfrac{{dy}}{{dx}} + 3y = {e^{ - 2x}}\]. Summary: This problem involves solving a linear ODE using an integrating factor.

Given form: \(\frac{dy}{dx} + 3y = e^{-2x}\)
Integrating factor: \(I.F. = e^{\int 3 dx} = e^{3x}\)
Multiply both sides by integrating factor:
\(y e^{3x} = \int e^{x} dx + C\)
\(y e^{3x} = e^{x} + C\)
Final answer: \[y = e^{-2x} + C e^{-3x} e^{3x}\] (simplifies to \(y = e^{-2x} + C e^{-3x} e^{3x}\))

(Since this is just over 100 words, answer is kept concise and step-by-step.)

3. Solve the differential equation \[\dfrac{{dy}}{{dx}} + \dfrac{y}{x} = {x^2}\]. This is a linear ODE which can be solved by integrating factor.

Equation: \(\frac{dy}{dx} + \frac{y}{x} = x^2\)
Integrating factor: \(I.F. = e^{\int \frac{1}{x} dx} = e^{\log x} = x\)
Solution: \(y x = \int x^3 dx + C\)
\(y x = \frac{x^4}{4} + C\)

4. Solve the differential equation \[\dfrac{{dy}}{{dx}} + \left( {\sec x} \right)y = \tan x \quad (0 \leq x \leq \frac{\pi}{2})\] Summary: Here, a linear differential equation is solved using an integrating factor.

The form is \(\frac{dy}{dx} + Py = Q\), with \(P = \sec x\) and \(Q = \tan x\).
Find integrating factor: \(I.F. = e^{\int \sec x dx} = e^{\log (\sec x + \tan x)} = \sec x + \tan x\)
Multiply the original equation by \(I.F.\):
\(y (\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C\)
Break into integrals: \(\int \sec x \tan x dx + \int \tan^2 x dx\)
The solution becomes: \(y (\sec x + \tan x) = \sec x + \tan x - x + C\)

5. Solve the differential equation \[{\cos ^2}x \frac{dy}{dx} + y = \tan x \quad (0 \leq x \leq \frac{\pi}{2})\] Summary: The equation is linear and is solved using substitution and integrating factor.

Rearrange: \(\frac{dy}{dx} + \sec^2 x \cdot y = \sec^2 x \tan x\)
Integrating factor: \(e^{\int \sec^2 x dx} = e^{\tan x}\)
Multiply by IF: \(y e^{\tan x} = \int \tan x \sec^2 x e^{\tan x} dx + C\)
Let \(\tan x = t\), then \(\sec^2 x dx = dt\), so \(y e^{t} = \int t e^{t} dt + C\).
By parts: \(t e^{t} - e^{t} + C\)
Thus: \(y = \tan x - 1 + C e^{-\tan x}\)

6. Solve the differential equation \[x \frac{dy}{dx} + 2y = x^2 \log x\] Summary: We use integrating factor and then integration by parts for solution.

Rearrange: \(\frac{dy}{dx} + \frac{2}{x} y = x \log x\)
Integrating factor: \(e^{\int \frac{2}{x} dx} = e^{\log x^2} = x^2\)
Multiply by IF: \(y x^2 = \int x^3 \log x dx + C\)
Solve \(\int x^3 \log x dx\) by parts:
= \(\log x \cdot \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx\)
= \(\frac{x^4 \log x}{4} - \frac{x^4}{16}\)
Thus: \(y = \frac{x^2 \log x}{4} - \frac{x^2}{16} + C x^{-2}\)
Or, \(y = \frac{x^2}{16}(4\log x - 1) + C x^{-2}\)

7. Solve the differential equation \[x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x\] Summary: This equation is linear and solved by finding the integrating factor and then integrating by parts.

Rearrange: \(\frac{dy}{dx} + \frac{y}{\log x} = \frac{2}{x^2}\)
Integrating factor: \(I.F. = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x\)
After multiplying by IF: \(y \log x = \int \frac{2}{x^2}\log x dx + C\)
By parts: \(= 2[\log x \cdot -\frac{1}{x} + \int \frac{1}{x} \cdot \frac{1}{x} dx]\)
Simplify: \(= -\frac{2}{x}(1 + \log x) + C\)
Final answer: \(y \log x = -\frac{2}{x}(1 + \log x) + C\)

8. Solve the differential equation \[(1 + x^2)dy + 2x y dx = \cot x dx\] Summary: The given equation can be rearranged to standard linear form and solved using an integrating factor.

Rearrange: \(\frac{dy}{dx} + \frac{2x y}{1 + x^2} = \frac{\cot x}{1 + x^2}\)
Integrating factor: \(e^{\int \frac{2x}{1 + x^2} dx} = e^{\log(1 + x^2)} = 1 + x^2\)
Multiply both sides: \(y (1 + x^2) = \int \cot x dx + C\)
Solution: \(y (1 + x^2) = \log(\sin x) + C\)

9. Solve the differential equation \[x \frac{dy}{dx} + y - x + x y \cot x = 0\] Summary: The solution involves rearranging to linear form, finding the integrating factor, and integrating by parts.

Rearrange: \(\frac{dy}{dx} + (\frac{1}{x} + \cot x) y = 1\)
Integrating factor: \(e^{\int (\frac{1}{x} + \cot x) dx} = e^{\log x + \log \sin x} = x \sin x\)
Multiply both sides: \(y x \sin x = \int x \sin x dx + C\)
By parts: \(= x \int \sin x dx - \int \frac{d}{dx}(x) \int \sin x dx dx\)
= \(x(-\cos x) - \int -\cos x dx\)
= \(-x \cos x + \sin x + C\)
\(y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}\)

10. Solve the differential equation \[(x+y)\frac{dy}{dx} = 1\] Summary: This problem is solved by separating variables and using integrating factor method.

Rewrite as: \(\frac{dx}{dy} - x = y\)
Form: \(\frac{dx}{dy} + P x = Q\), with \(P = -1\), \(Q = y\)
Integrating factor: \(e^{\int -1 dy} = e^{-y}\)
\(x e^{-y} = \int y e^{-y} dy + C\)
By parts: \(= -y e^{-y} - e^{-y} + C\)
Final: \(x = -y - 1 + C e^{y}\), or \(x + y + 1 = C e^y\)

11. Solve the differential equation \[y dx + (x - y^2) dy = 0\]

Rewrite as: \(\frac{dx}{dy} + \frac{x}{y} = y\)
Integrating factor: \(e^{\int \frac{1}{y} dy} = e^{\log y} = y\)
Solution: \(x y = \int y^2 dy + C\)
\(x y = \frac{y^3}{3} + C\)

12. Solve the differential equation \[(x + 3y^2) \frac{dy}{dx} = y\), where \(y > 0\)

Rewrite as: \(\frac{dx}{dy} - \frac{x}{y} = 3y\)
Integrating factor: \(e^{\int -\frac{1}{y} dy} = e^{\log \frac{1}{y}} = \frac{1}{y}\)
Solution: \(x \cdot \frac{1}{y} = \int 3 dy + C\)
\(\frac{x}{y} = 3y + C\)
\(x = 3y^2 + C y\)

13. Solve the differential equation \[\frac{dy}{dx} + 2y \tan x = \sin x;\; y=0\ \text{when }x=\frac{\pi}{3}\] Summary: This problem requires finding a particular solution using the integrating factor and initial condition.

Standard form: \(\frac{dy}{dx} + 2y \tan x = \sin x\)
Integrating factor: \(e^{\int 2\tan x dx} = e^{\log \sec^2 x} = \sec^2 x\)
\(y \sec^2 x = \int \sec x \tan x dx + C\)
\(y \sec^2 x = \sec x + C\)
Plug in \(y=0\) when \(x=\frac{\pi}{3}\): \(0 = 2 + C \implies C = -2\)
Final: \(y = \cos x - 2 \sec^2 x\)

14. Solve the differential equation \[(1+x^2)\frac{dy}{dx} + 2x y = \frac{1}{(1+x^2)}; \quad y=0\ \text{when }x=1\] Summary: Here, integrating factor and initial condition are used to find the particular solution.

Rearrange: \(\frac{dy}{dx} + \frac{2x y}{1 + x^2} = \frac{1}{(1 + x^2)^2}\)
Integrating factor: \(e^{\int \frac{2x}{1 + x^2} dx} = e^{\log(1 + x^2)} = 1 + x^2\)
Multiply through: \(y (1 + x^2) = \int \frac{1}{1 + x^2} dx + C = \tan^{-1} x + C\)
Apply condition \(y = 0\) at \(x = 1\):
\(0 = \frac{\pi}{4} + C \implies C = -\frac{\pi}{4}\)
Final: \(y (1 + x^2) = \tan^{-1} x - \frac{\pi}{4}\)

15. Solve the differential equation \[\frac{dy}{dx} - 3y\cot x = \sin 2x;\; y=2\ \text{when }x=\frac{\pi}{2}\] Summary: The answer involves an integrating factor, integration by substitution, and applying initial conditions.

Equation: \(\frac{dy}{dx} - 3y\cot x = \sin 2x\)
Integrating factor: \(e^{\int -3\cot x dx} = e^{-3\log \sin x} = \frac{1}{\sin^3 x}\)
\(y \csc^3 x = \int \sin 2x \csc^3 x dx + C\)
\(\int \sin 2x \csc^3 x dx = 2 \int \cot x \csc x dx = -2 \csc x\)
So, \(y \csc^3 x = -2\csc x + C\)
Simplify: \(y = -2\sin^2 x + C\sin^3 x\)
Apply condition \(y=2\) when \(x = \frac{\pi}{2}\): \(2 = -2(1) + C(1)\) ⇒ \(C = 4\)
Final: \(y = 4\sin^3 x - 2\sin^2 x\)

16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point \((x,y)\) is equal to the sum of the coordinates of the point. Summary: The question is a linear ODE involving curve passing through the origin.

Given: \(\frac{dy}{dx} - y = x\)
Integrating factor: \(e^{\int -1 dx} = e^{-x}\)
General Solution: \(y e^{-x} = \int x e^{-x} dx + C\)
By parts: \(= -x e^{-x} - e^{-x} + C\)
\(y = -x - 1 + C e^{x}\)
Passes through (0,0): \(C = 1\)
So, \(x + y + 1 = e^{x}\)

17. Find the equation of a curve passing through the point \((0,2)\) 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 5. Summary: This is another differential equation with initial condition.

Given: \(\frac{dy}{dx} + 5 = x + y \Rightarrow \frac{dy}{dx} - y = x - 5\)
Integrating factor: \(e^{-x}\)
Solution: \(y e^{-x} = \int (x - 5)e^{-x} dx + C\)
By parts: \(= (5 - x)e^{-x} - e^{-x} + C\)
\(y = 4 - x + C e^{x}\)
Point (0,2): \(2 + 0 - 4 = C\) thus \(C = -2\)
Final: \(y + x - 4 = -2 e^{x}\)

18. The Integrating Factor of the differential equation \[x\frac{dy}{dx} - y = 2x^2\] is

On rearranging: \(\frac{dy}{dx} - \frac{y}{x} = 2x\)
Integrating factor: \(e^{ - \int \frac{1}{x} dx } = e^{ - \log x } = \frac{1}{x}\)
The correct option is (C).

19. The Integrating Factor of the differential equation \[(1 - y^2) \frac{dx}{dy} + yx = ay\ ( -1 < y < 1 )\] is

Rewrite: \(\frac{dx}{dy} + \frac{y}{1 - y^2}x = \frac{a y}{1 - y^2}\)
Integrating factor: \(e^{\int \frac{y}{1 - y^2} dy} = e^{\log \left( \frac{1}{\sqrt{1 - y^2}} \right)} = \frac{1}{\sqrt{1 - y^2}}\)
The correct option is (D).

``` --- **Chapter Highlights: Differential Equations Class 12 Exercise 9.5**

exercise 9.5 class 12 maths covers solving different types of first-order differential equations step by step.
Learn how to use integrating factors to solve linear equations found in 9.5 class 12 maths.
Practice problems often ask you to apply methods like integration by parts and separation of variables.
ex 9.5 class 12 maths ncert solutions help strengthen concepts for both board and entrance exams.
Special focus is given to finding particular solutions when initial conditions are provided.

```

FAQs on NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations Exercise 9.5 - 2025-26

1. What approach should be followed to solve linear differential equations in NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5?

To solve linear differential equations in Exercise 9.5, use the following methodical approach:

Identify the equation in the form dy/dx + Py = Q.
Calculate the integrating factor (IF).
Multiply the entire equation by the integrating factor.
Integrate both sides and add the constant of integration.
If initial conditions are given, substitute them to get the particular solution.

This step-by-step NCERT Solutions methodology ensures accuracy and is aligned with the CBSE marking scheme.

2. Why is the integrating factor critical in solving first-order linear differential equations?

The integrating factor transforms a first-order linear differential equation into an exact equation, allowing for straightforward integration. By multiplying both sides with the IF, the left side becomes a derivative of a product, simplifying the solution process and ensuring you derive the general solution efficiently, a key focus in NCERT Solutions for Class 12 Maths Chapter 9.

3. How can students avoid common mistakes when solving differential equations in Exercise 9.5?

To minimize errors while solving:

Always include the constant of integration (C) after integrating.
Select the correct solution method as per the type of equation.
Pay attention to algebraic steps and arithmetic calculations.
Apply initial or boundary conditions carefully for particular solutions.
Review definitions of order and degree before starting each problem.

This careful approach is vital for achieving full marks in board and competitive exams.

4. What is the significance of 'order' and 'degree' in differential equations according to the NCERT syllabus?

Order indicates the highest derivative present in the equation, while degree is the power of the highest order derivative after making the equation rational and free from radicals. Correctly identifying both is essential for choosing the right solution method and is commonly assessed in both board and competitive exams.

5. Which problem-solving techniques from Class 12 Maths Chapter 9 are most commonly tested in CBSE board and entrance exams?

The main techniques tested are:

Separation of variables
Integration by parts
Method of integrating factors for linear equations
Homogeneous and non-homogeneous equation strategies

Mastering these methods as outlined in the NCERT Solutions maximizes exam readiness and accuracy.

6. How do solved examples in Class 12 Maths Chapter 9 Exercise 9.5 build conceptual clarity?

The step-by-step solutions in Exercise 9.5 break down each type of differential equation, clarify the application of methods like integrating factors, and reinforce the need for correct integration and constant handling. This builds a deep understanding of the concepts and their application in exam questions.

7. In what ways do NCERT Solutions for Class 12 Maths Chapter 9 aid in preparation for JEE or NEET along with CBSE board exams?

NCERT Solutions cover fundamental problem-types and methodologies commonly asked in JEE and NEET exams. These solutions bridge conceptual understanding and exam patterns, ensuring students are prepared for both objective and descriptive scenarios, making them highly relevant for both board and entrance examinations.

8. What misconceptions do students have about integrating factors, and how can they be corrected?

Common misconceptions include:

Believing any function can be used as an integrating factor (only the exponential of the integral of P works).
Forgetting to multiply the entire equation by the integrating factor.
Not recognizing the exactness achieved after applying IF.

Clearing these misconceptions through stepwise NCERT Solutions explanations enhances problem-solving accuracy and conceptual clarity.

9. How should one verify if a solution to a differential equation is correct as per NCERT methodologies?

To verify your solution:

Substitute the obtained solution back into the original differential equation.
For particular solutions, ensure initial or boundary conditions are satisfied.
Compare your process and answer with expert-verified NCERT Solutions to ensure every critical step is addressed.

This confirms accuracy and builds confidence for CBSE board assessments.

10. What makes the NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5 unique compared to other resources?

These solutions are crafted by expert educators strictly following the CBSE 2025–26 syllabus, provide detailed, structured, and stepwise methods, and are periodically updated for accuracy. Explanations match marking schemes and common exam patterns, making them a reliable tool for exam preparation.

11. How does the type of differential equation affect the choice of solution method in NCERT Solutions?

The classification—whether the equation is linear, homogeneous, separable, or exact—determines the preferred solution method. Linear equations require integrating factor methods, while separable equations are solved by separating and integrating variables. Choosing the correct method is crucial for full marks in CBSE board and entrance exams.

12. Why is it recommended to practice every type of problem from Exercise 9.5 in Class 12 Maths?

Practicing every type of problem ensures:

Exposure to all exam-pattern variations.
Confidence in handling both general and particular solutions.
Reduction of mistakes in the actual CBSE and entrance exams.

This comprehensive practice strengthens both conceptual and procedural mastery.

13. How do initial or boundary conditions change the solution strategy for a differential equation?

When an initial or boundary condition is provided, the general solution’s arbitrary constant is substituted and solved for, yielding the particular solution. This aligns the solution to specific scenarios as required in both board and competitive exams, a process emphasized in NCERT Solutions.

14. What is the general form of a first-order, first-degree differential equation according to the NCERT Class 12 syllabus?

The general form is dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x. Recognizing this form helps in applying correct techniques such as integrating factors, as outlined in the chapter.

15. How can understanding the method of integrating factors help in solving real-world problems modeled by differential equations?

The method of integrating factors provides a systematic process for finding exact solutions to first-order linear equations, which frequently model real-life phenomena in physics, engineering, and biology. This NCERT Solutions methodology translates mathematical theory into practical applications, a critical exam and life skill.