Exercise 9.2 Class 12 Maths Differential Equations

NCERT Solutions Class 12 Maths Chapter 9 Ex 9.2 | LearnCBSEHub

💡 Verification Method

To verify if a function is a solution to a differential equation:

Differentiate the given function $y$ as many times as the order of the differential equation.
Substitute values of $y, y’, y”$, etc., into the given differential equation.
If L.H.S = R.H.S, then it is a solution.

Note: General Solution has arbitrary constants. Particular Solution has fixed values.

Question 1

$y = e^x + 1 \quad : \quad y” – y’ = 0$

Step 1: Differentiate

$y = e^x + 1$
$y’ = \frac{d}{dx}(e^x + 1) = e^x$
$y” = \frac{d}{dx}(e^x) = e^x$

Step 2: Substitute

LHS $= y” – y’ = e^x – e^x = 0 = $ RHS.

Verified: It is a solution.

Question 2

$y = x^2 + 2x + C \quad : \quad y’ – 2x – 2 = 0$

Step 1: Differentiate

$y’ = \frac{d}{dx}(x^2 + 2x + C) = 2x + 2$

Step 2: Substitute

LHS $= y’ – 2x – 2 = (2x + 2) – 2x – 2 = 0 = $ RHS.

Verified: It is a solution.

Question 3

$y = \cos x + C \quad : \quad y’ + \sin x = 0$

Step 1: Differentiate

$y’ = \frac{d}{dx}(\cos x + C) = -\sin x$

Step 2: Substitute

LHS $= y’ + \sin x = -\sin x + \sin x = 0 = $ RHS.

Verified: It is a solution.

Question 4

$y = \sqrt{1+x^2} \quad : \quad y’ = \frac{xy}{1+x^2}$

Step 1: Differentiate

$y’ = \frac{d}{dx}(\sqrt{1+x^2}) = \frac{1}{2\sqrt{1+x^2}} \cdot (2x) = \frac{x}{\sqrt{1+x^2}}$

Step 2: Verify RHS

RHS $= \frac{xy}{1+x^2} = \frac{x(\sqrt{1+x^2})}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$
LHS = RHS.

Verified: It is a solution.

Question 5

$y = Ax \quad : \quad xy’ = y \quad (x \neq 0)$

Step 1: Differentiate

$y’ = \frac{d}{dx}(Ax) = A$
Since $y = Ax \implies A = y/x$.

Step 2: Substitute

LHS $= xy’ = x(A) = x(y/x) = y = $ RHS.

Verified: It is a solution.

Question 6

$y = x \sin x \quad : \quad xy’ = y + x\sqrt{x^2-y^2}$

Step 1: Differentiate

$y’ = x \cos x + \sin x$

Step 2: LHS and RHS

LHS $= x(x \cos x + \sin x) = x^2 \cos x + x \sin x$
RHS $= y + x\sqrt{x^2 – y^2} = x \sin x + x\sqrt{x^2 – x^2\sin^2 x}$
$= x \sin x + x\sqrt{x^2(1-\sin^2 x)} = x \sin x + x(x \cos x)$
$= x \sin x + x^2 \cos x$.
LHS = RHS.

Verified: It is a solution.

Question 7

$xy = \log y + C \quad : \quad y’ = \frac{y^2}{1-xy}$

Step 1: Differentiate Implicitly

$x y’ + y(1) = \frac{1}{y} y’$
$y’ (x – \frac{1}{y}) = -y$
$y’ (\frac{xy-1}{y}) = -y \implies y’ = \frac{-y^2}{xy-1} = \frac{y^2}{1-xy}$

Verified: Matches the Diff. Eq.

Question 8

$y – \cos y = x \quad : \quad (y \sin y + \cos y + x)y’ = y$

Step 1: Differentiate

$y’ – (-\sin y)y’ = 1 \implies y'(1+\sin y) = 1 \implies y’ = \frac{1}{1+\sin y}$

Step 2: Check LHS

LHS $= ((y \sin y + \cos y) + x) \cdot \frac{1}{1+\sin y}$
Substitute $x = y – \cos y$:
$= (y \sin y + \cos y + y – \cos y) \cdot \frac{1}{1+\sin y}$
$= (y \sin y + y) \cdot \frac{1}{1+\sin y} = \frac{y(\sin y + 1)}{1+\sin y} = y = $ RHS.

Verified: It is a solution.

Question 9

$x + y = \tan^{-1}y \quad : \quad y^2 y’ + y^2 + 1 = 0$

Step 1: Differentiate

$1 + y’ = \frac{1}{1+y^2} y’ \implies 1 = y'(\frac{1}{1+y^2} – 1)$
$1 = y'(\frac{1 – 1 – y^2}{1+y^2}) \implies 1 = y'(\frac{-y^2}{1+y^2})$
$1+y^2 = -y^2 y’ \implies y^2 y’ + y^2 + 1 = 0$.

Verified: Matches the Diff. Eq.

Question 10

$y = \sqrt{a^2-x^2} \quad : \quad x + y\frac{dy}{dx} = 0$

Step 1: Differentiate

$y’ = \frac{1}{2\sqrt{a^2-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{a^2-x^2}} = \frac{-x}{y}$

Step 2: Rearrange

$y’ = \frac{-x}{y} \implies y y’ = -x \implies x + y y’ = 0$.

Verified: Matches the Diff. Eq.

Questions 11 — 12

Multiple Choice Questions.

11. Number of arbitrary constants in the general solution of a differential equation of fourth order are:

The number of arbitrary constants in the general solution is equal to the order of the differential equation.

Correct Option: (D) 4

12. Number of arbitrary constants in the particular solution of a differential equation of third order are:

A particular solution is obtained by giving specific values to the arbitrary constants. Therefore, it contains zero arbitrary constants.

Correct Option: (D) 0