NCERT Solutions for Class 12 Mathematics Chapter 9 Differential Equations

NCERT Solutions for Class 12 Mathematics Chapter 9 - Differential Equations

Welcome to the Chapter 9 - Differential Equations, Class 12 Mathematics NCERT Solutions page. Here, we provide detailed question answers for Chapter 9 - Differential Equations. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Differential Equations and excel in their exams. By going through these Differential Equations question answers, you can strengthen your foundation and improve your performance in Class 12 Mathematics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1

Q 1:

Determine order and degree(if defined) of differential equation \begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0\end{align}

\begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0 \end{align}

\begin{align} \Rightarrow y^{m\;'}+\;\sin(y^m)\;=0 \end{align}

The highest order derivative present in the differential equation is y^{m '}. Therefore, its order is four.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Q 10:

Determine order and degree(if defined) of differential equation yⁿ + 2y^' + siny = 0

yⁿ + 2y^' + siny = 0

The highest order derivative present in the differential equation is yⁿ. Therefore, its order is two.

This is a polynomial equation in yⁿand y' and the highest power raised to yⁿis one. Hence, its degree is one.

Q 11:

The degree of the differential equation

\begin{align}\left(\frac{d^2y}{dx^2}\right)^3\;+ \left(\frac{dy}{dx}\right)^2+\;sin\left(\frac{dy}{dx}\right)\;+ 1=\;0\end{align}

is (A) 3 (B) 2 (C) 1 (D) not defined

\begin{align}\left(\frac{d^2y}{dx^2}\right)^3\;+ \left(\frac{dy}{dx}\right)^2+\;sin\left(\frac{dy}{dx}\right)\;+ 1=\;0\end{align}

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.

Hence, the correct answer is D.

Q 12:

The order of the differential equation

\begin{align}2x^2\frac{d^2y}{dx^2}\;- \;3\frac{dy}{dx}\;+ y=\;0\end{align}

is (A) 2 (B) 1 (C) 0 (D) not defined

\begin{align}2x^2\frac{d^2y}{dx^2}\;- \;3\frac{dy}{dx}\;+ y=\;0\end{align}

The highest order derivative present in the given differential equation is

\begin{align}\frac{d^2y}{dx^2}\end{align}

Therefore, its order is two.

Hence, the correct answer is A.

Q 2:

Determine order and degree(if defined) of differential equation y' + 5y = 0

The given differential equation is:

y' + 5y = 0

The highest order derivative present in the differential equation isy'. Therefore, its order is one.

It is a polynomial equation in y'. The highest power raised to y' is 1. Hence, its degree is one.

Q 3:

Determine order and degree(if defined) of differential equation \begin{align}\left(\frac{ds}{dt}\right)^4\;+\;3s\frac{d^2s}{dt^2}\;=\;0\end{align}

\begin{align}\left(\frac{ds}{dt}\right)^4\;+\;3s\frac{d^2s}{dt^2}\;=\;0\end{align}

The highest order derivative present in the given differential equation is\begin{align}\frac{d^2s}{dt^2}.\end{align}

Therefore, its order is two. It is a polynomial equation in

\begin{align}\frac{d^2s}{dt^2} and \frac{ds}{dt}.\end{align}

The power raised to is 1. \begin{align} \frac{d^2s}{dt^2} \end{align}

Q 4:

Determine order and degree(if defined) of differential equation

\begin{align}\left(\frac{d^2y}{dx^2}\right)^2\;+\;cos\left(\frac{dy}{dx}\right)\;=\;0\end{align}

The highest order derivative present in the given differential equation is \begin{align}\frac{d^2y}{dx^2}.\end{align}

Therefore, its order is 2. The given differential equation is not a polynomial equation in its derivatives.

Hence, its degree is not defined.

Q 5:

Determine order and degree(if defined) of differential equation \begin{align}\frac{d^2y}{dx^2}=\cos3x + sin3x\end{align}

\begin{align}\frac{d^2y}{dx^2}=\cos3x + sin3x\end{align}

\begin{align}\Rightarrow\frac{d^2y}{dx^2} - \cos3x - sin3x = 0\end{align}

The highest order derivative present in the differential equation is\begin{align}\frac{d^2y}{dx^2}.\end{align}

Therefore, its order is two.It is a polynomial equation in \begin{align}\frac{d^2y}{dx^2}\end{align}

and the power raised to is 1.

\begin{align}\frac{d^2y}{dx^2}\end{align}

Hence, its degree is one.

Q 6:

Determine order and degree(if defined) of differential equation (y^m)² + (yⁿ)³ + (y')⁴ + y⁵ =0

(y^m)² + (yⁿ)³ + (y')⁴ + y⁵ =0

The highest order derivative present in the differential equation isy^m. Therefore, its order is three.

The given differential equation is a polynomial equation in y^{m ,}y^{n , y'.}

The highest power raised to y^mis 2. Hence, its degree is 2.

Q 7:

Determine order and degree(if defined) of differential equation y^m + 2yⁿ + y' =0

The highest order derivative present in the differential equation is y^m. Therefore, its order is three.

It is a polynomial equation in y^m,yⁿand y' . The highest power raised to y^m is 1. Hence, its degree is 1.

Q 8:

Determine order and degree(if defined) of differential y^' + y =e^x

y^' + y =e^x

y^' + y - e^x=0

The highest order derivative present in the differential equation is y^'. Therefore, its order is one.

The given differential equation is a polynomial equation in y^'and the highest power raised to y^'is one. Hence, its degree is one.

Q 9:

Determine order and degree(if defined) of differential equation yⁿ + (y')² + 2y =0

yⁿ + (y')² + 2y =0

The highest order derivative present in the differential equation is yⁿ. Therefore, its order is two.

The given differential equation is a polynomial equation in yⁿand y'and the highest power raised to yⁿis one.

Hence, its degree is one.

Exercise 2

Q 1:

y = e^x +1 : yⁿ -y^' = 0

y = e^x +1

Differentiating both sides of this equation with respect to x, we get:

\begin{align}\frac{dy}{dx}=\frac{d}{dx}(e^x + 1)\end{align}

=> y^' = e^x ...(1)

Now, differentiating equation (1) with respect to x, we get:

\begin{align}\frac{d}{dx}(y^{'})=\frac{d}{dx}(e^x)\end{align}

=> y^'' = e^x

Substituting the values of y^' and y^'' in the given differential equation, we get the L.H.S. as:

y^'' - y^' = e^x - e^x = 0 = R.H.S.

Thus, the given function is the solution of the corresponding differential equation.

Q 2:

y = x² + 2x + C : y^' - 2x - 2 = 0

y = x² + 2x + C

Differentiating both sides of this equation with respect to x, we get:

\begin{align}y^{'}=\frac{d}{dx}(x^2 + 2x + C)\end{align}

=> y^' = 2x + 2

Substituting the value of y^' in the given differential equation, we get:

L.H.S. = y^' - 2x - 2 = 2x + 2 - 2x - 2 = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Q 3:

y = cosx + C : y^' + sinx = 0

y = cosx + C

Differentiating both sides of this equation with respect to x, we get:

\begin{align}y^{'}=\frac{d}{dx}(cosx + C)\end{align}

=> y^' = - sinx

Substituting the value of y^' in the given differential equation, we get:

L.H.S. = y^' + sinx = - sinx + sinx = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Q 4:

\begin{align} y= \sqrt{1+x^2} : y^{'}=\frac{xy}{1+x^2}\end{align}

\begin{align} y= \sqrt{1+x^2}\end{align}

Differentiating both sides of the equation with respect to x, we get:

\begin{align} y^{'}=\frac{d}{dx}\left(\sqrt{1+x^2} \right)\end{align}

\begin{align} y^{'}=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}\left(1+x^2\right)\end{align}

\begin{align} y^{'}=\frac{2x}{2\sqrt{1+x^2}}\end{align}

\begin{align} y^{'}=\frac{x}{\sqrt{1+x^2}}\end{align}

\begin{align}\Rightarrow y^{'}=\frac{x}{\sqrt{1+x^2}}\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\end{align}

\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{\sqrt{1+x^2}}\end{align}

\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{y}\end{align}

\begin{align}\Rightarrow y^{'}=\frac{xy}{1+x^2}\end{align}

∴ L.H.S. = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Q 5:

y = Ax : xy^' = y (x ≠ 0)

y = Ax

Differentiating both sides with respect to x, we get:

\begin{align}y^{'}=\frac{d}{dx}(Ax)\end{align}

⇒ y ^' = A

Substituting the value of y^' in the given differential equation, we get:

L.H.S. = xy^'= xA = Ax = y = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Q 6:

\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}

y= x.sinx

Differentiating both sides of this equation with respect to x, we get:

\begin{align} y^{'} =\frac{d}{dx}\left(x.sinx\right)\end{align}

\begin{align}\Rightarrow y^{'} =sinx. \frac{d}{dx}\left(x\right)+ x. \frac{d}{dx}\left(sinx\right)\end{align}

\begin{align} \Rightarrow y^{'} =sinx + x.cosx\end{align}

Differentiating both sides of this equation with respect to x, we get:

L.H.S. =xy^'= x(sinx + xcosx)

\begin{align} =x.sinx + x^2.cosx\end{align}

\begin{align} =y + x^2.\sqrt{1-sin^2x}\end{align}

\begin{align} =y + x^2.\sqrt{1-\left(\frac{y}{x}\right)^2}\end{align}

\begin{align} =y + x^2.\sqrt{\frac{x^2-y^2}{x^2}}\end{align}

\begin{align} =y + x.\sqrt{x^2-y^2}\end{align}

R.H.S.

Frequently Asked Questions about Differential Equations - Class 12 Mathematics

- 1. How many questions are covered in Differential Equations solutions?
- All questions from Differential Equations are covered with detailed step-by-step solutions including exercise questions, additional questions, and examples.
- 2. Are the solutions for Differential Equations helpful for exam preparation?
- Yes, the solutions provide comprehensive explanations that help students understand concepts clearly and prepare effectively for both board and competitive exams.
- 3. Can I find solutions to all exercises in Differential Equations?
- Yes, we provide solutions to all exercises, examples, and additional questions from Differential Equations with detailed explanations.
- 4. How do these solutions help in understanding Differential Equations concepts?
- Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
- 5. Are there any tips for studying Differential Equations effectively?
- Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

Exam Preparation Tips for Differential Equations

The Differential Equations is an important chapter of 12 Mathematics. This chapter’s important topics like Differential Equations are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

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