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NCERT Solutions

Class 12

Maths

Chapter 9 Differential Equations Exercise 9.1

NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations Exercise 9.1 - 2025-26

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In NCERT Solutions Class 12 Maths Chapter 9 Exercise 9 1, you’ll dive into the basics of differential equations. You’ll learn how to find the order and degree of different equations and understand the types of questions that pop up in board exams. These step-by-step NCERT Solutions make tricky problems much easier, so you can build your confidence as you study.

If you’re feeling stuck, Vedantu’s easy-to-follow explanations and downloadable PDF will be super helpful. They are aligned with the latest CBSE syllabus and perfect for your exam revision. For a quick look at what’s important this year, don’t forget to check the Class 12 Maths syllabus.

Understanding differential equations is important for higher studies and covers MCQs, short answers, and how to form equations from solutions. The best part? This chapter carries 7 marks in your CBSE exam, making it a great way to boost your score.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.1

```html 1. Determine order and degree (if defined) of differential equation $\dfrac{{{d^4}y}}{{d{x^4}}} + \sin \left( {y'''} \right) = 0$.
The order of this equation is four. The degree can’t be defined since there’s a sine function of a derivative in the equation.

Highest derivative present: $y''''$ (which is $\dfrac{{d^4}y}{d{x^4}}$).
Order = 4, since fourth derivative is the highest present.
Degree is not defined, because the equation has the trigonometric function $\sin(y''')$ (not a polynomial in derivatives).

2. Determine order and degree (if defined) of differential equation $y' + 5y = 0$.
Order is one and degree is also one.

Highest derivative present is $y'$ (or $\dfrac{dy}{dx}$).
Order = 1 (first derivative).
It is a polynomial in $y'$ with maximum power 1, so degree = 1.

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

Highest order derivative: $s''$ (or $\dfrac{{d^2}s}{{dt^2}}$).
Order = 2 (second derivative).
It’s a polynomial in derivatives, with maximum power of $s''$ as 1, so degree = 1.

4. Determine order and degree (if defined) of differential equation ${\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^2} + \cos \left( {\dfrac{{dy}}{{dx}}} \right) = 0$.
Order is two, but degree is not defined due to the cosine function.

Highest order derivative: $y''$ (or $\dfrac{{d^2}y}{{dx^2}}$).
Order = 2 (second derivative is highest).
Degree not defined, as the equation is not a polynomial in derivatives (because of $\cos(y')$).

5. Determine order and degree (if defined) of differential equation $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\cos 3x+\sin 3x$.
Order is two and degree is one.

Highest derivative: $y''$ (or $\dfrac{d^2y}{dx^2}$).
Order = 2.
The equation is polynomial in $y''$ (first power), so degree = 1.

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

Highest order derivative: $y'''$ (or $\dfrac{d^3y}{dx^3}$).
Order = 3.
This is a polynomial in derivatives, with $y'''$ raised to power 2, so degree = 2.

7. Determine order and degree (if defined) of differential equation $y''' + 2y'' + y' = 0$.
Order is three and degree is one.

Highest order derivative: $y'''$ (third derivative).
Order = 3.
It’s a polynomial in derivatives, with highest power of $y'''$ as 1, so degree = 1.

8. Determine order and degree (if defined) of differential equation $y' + y = e'$.
Order is one and degree is one.

Highest derivative: $y'$ (first derivative).
Order = 1.
Equation is polynomial in $y'$, highest power is 1, so degree = 1.

9. Determine order and degree (if defined) of differential equation $y'' + (y')^2 + 2y = 0$.
Order is two and degree is one.

Highest order derivative: $y''$ (second derivative).
Order = 2.
Polynomial in $y''$ with highest power 1, so degree = 1.

10. Determine order and degree (if defined) of differential equation $y'' + 2y' + \sin y = 0$.
Order is two and degree is one.

Highest order derivative: $y''$ (second derivative).
Order = 2.
Degree is 1 because it's a polynomial in $y''$ (and $y'$).

11. Determine degree (if defined) of differential equation ${\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \sin \left( {\dfrac{{dy}}{{dx}}} \right) + 1 = 0$ is (A) 3 (B) 2 (C) 1 (D) Not defined
The degree is not defined because the equation contains the term $\sin\left(\dfrac{dy}{dx}\right)$, which is not a polynomial in derivatives.

12. Determine order of the differential equation $2{x^2}\dfrac{{{d^2}y}}{{d{x^2}}} + 3\dfrac{{dy}}{{dx}} + y = 0$ (A) 2 (B) 1 (C) 0 (D) Not defined
The order of the equation is 2 since the highest order derivative present is second order, that is, $y''$.

``` --- ### Main Ideas from Differential Equations Class 12 Exercise 9.1 - Learn how to find the order and degree of different differential equations in exercise 9.1 class 12 maths. - Understand which equations have an undefined degree because of functions like sine or cosine. - Practice identifying the highest order derivative and the degree in both simple and complex equations. - These are common types of questions asked in board exams on ex 9.1 class 12 maths NCERT solutions. - Knowing these basics helps you solve and identify different forms of differential equations confidently.

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FAQs on NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations Exercise 9.1 - 2025-26

1. Where can I find reliable, step-by-step NCERT Solutions for Class 12 Maths Chapter 9, Differential Equations?

You can find comprehensive and expert-verified NCERT Solutions for Class 12 Maths Chapter 9 on Vedantu. These solutions are crafted to provide clear, step-by-step explanations for every problem in the textbook, strictly following the CBSE 2025-26 syllabus and marking scheme to help students understand the correct methodology for solving differential equations.

2. How do you determine the order and degree of a differential equation as per the NCERT guidelines?

To determine the order and degree of a differential equation, follow these two steps:

Order: The order is the highest order of the derivative present in the equation. For example, in the equation d²y/dx² + y = 0, the highest order derivative is d²y/dx², so the order is 2.
Degree: The degree is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. In (d²y/dx²)³ + dy/dx = 0, the degree is 3. If the equation contains terms like sin(dy/dx), the degree is not defined.

3. What is the correct method to solve a differential equation using variable separation?

The variable separable method is used when the differential equation can be expressed in the form f(x)dx = g(y)dy. The step-by-step method is as follows:

Rearrange the equation to separate all terms involving 'x' and 'dx' on one side, and all terms involving 'y' and 'dy' on the other.
Integrate both sides of the equation independently.
Add a single constant of integration, 'C', to one side of the equation to get the general solution.

4. How are homogeneous differential equations solved in the Class 12 NCERT solutions?

A differential equation is homogeneous if it can be written as dy/dx = F(y/x). The standard procedure to solve it is:

Substitute y = vx. This implies that dy/dx = v + x(dv/dx).
Substitute these into the original equation, which will transform it into a new equation with variables 'v' and 'x'.
This new equation will be solvable using the variable separation method.
Solve for 'v' and then substitute back v = y/x to get the final solution in terms of 'x' and 'y'.

5. What is the step-by-step process for solving a linear differential equation of the form dy/dx + Py = Q?

To solve a linear differential equation, follow these precise steps:

First, identify the functions P and Q from the equation, ensuring it matches the standard form dy/dx + P(x)y = Q(x).
Calculate the Integrating Factor (I.F.) using the formula: I.F. = e^{∫P dx}.
The general solution is then found using the standard result: y × (I.F.) = ∫(Q × I.F.) dx + C, where C is the constant of integration.

6. Why is the degree of a differential equation 'not defined' if it contains a term like sin(dy/dx)?

The concept of 'degree' applies only to differential equations that are polynomials in their derivatives (like y'', y', etc.). A function like sin(dy/dx) is a transcendental function. When expanded using a Maclaurin series, it becomes an infinite series of powers of dy/dx. Since a polynomial must have a finite number of terms and a finite highest power, the equation is not a polynomial in its derivatives, and thus its degree cannot be defined.

7. What is the fundamental difference between a general solution and a particular solution of a differential equation?

The key difference lies in the presence of arbitrary constants:

A general solution contains one or more arbitrary constants (like 'C'). It represents a family of curves that all satisfy the differential equation. The number of constants equals the order of the equation.
A particular solution is derived from the general solution by using given initial conditions (e.g., y=2 when x=1) to find specific values for the arbitrary constants. It represents a single, unique curve from the family of solutions.

8. How can a student quickly identify if a differential equation is variable separable, homogeneous, or linear?

Here is a quick checklist to identify the type of first-order, first-degree differential equation:

Variable Separable: Check if you can algebraically move all 'x' terms to one side with 'dx' and all 'y' terms to the other side with 'dy'.
Homogeneous: Check if the equation can be expressed as a function of y/x or if every term in the equation has the exact same degree. For example, in y² + x² = xy(dy/dx), all terms have a degree of 2.
Linear: Check if the equation can be perfectly structured into the form dy/dx + P(x)y = Q(x) or dx/dy + P(y)x = Q(y). If it fits this template, it is linear.

Always check in this order, as some equations might fit multiple criteria, but one method may be simpler.

9. How do you form a differential equation from a given general solution containing arbitrary constants?

To form a differential equation from a solution, the goal is to eliminate the arbitrary constants. The correct procedure is:

Count the number of independent arbitrary constants in the given equation. Let's say there are 'n' constants.
Differentiate the equation successively 'n' times. This will give you 'n' new equations in addition to the original one.
Use these (n+1) equations to algebraically eliminate the 'n' constants.
The resulting equation, which is free of arbitrary constants, is the required differential equation. Its order will be 'n'.

NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations Exercise 9.1 - 2025-26

Differential Equations Questions and Answers - Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.1

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FAQs on NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations Exercise 9.1 - 2025-26