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## Karnataka 2nd PUC Maths Question Bank Chapter 9 Differential Equations Ex 9.2

### 2nd PUC Maths Differential Equations NCERT Text Book Questions and Answers Ex 9.2

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

Question 1.
y = ex + 1 : y’’- y’ = 0

Given, y’’- y’= ex – ex =0
∴ Hence given function is solution of
corresponding differential equation.

Question 2.
y5= x2 + 2x + C : y’ – 2x – 2 = 0
Given y = x2 + 2x + C
⇒ $$y^{\prime}=\frac{d}{d x}$$ to + 2x + C) = 2x + 2
Given, y’ = 2x -2 = 2x + 2 – 2x – 2 = 0
LHS = RHS
Hence y = x2 + 2x + C, is solution of given differential equation.

Question 3.
y = cos x + C : y’+ sin x = 0
Given y = cos x + C
⇒ $$y^{\prime}=\frac{d}{d x}$$ (cos x + C) = – sin x
Given, y’ + sinx = 0sinx + sinx = 0
LHS = RHS
Hence y = cos x + C, is solution of given differential equation.

Question 4.
$$y=\sqrt{1+x^{2}} \quad: y^{\prime}=\frac{x y}{1+x^{2}}$$

(1) and (2), Hence $$y=\sqrt{\left(1+x^{2}\right)}$$is solution of given differential equation.

Question 5.
y = Ax : xy’ = y (x ≠ 0)
Given, y = Ax ⇒ y’=A
Given, x y’ = y ⇒ x x A = Ax => Ax = Ax
Hence y = Ax, is solution of given differential equation.

Question 6.
$${ y }={ x }\sin { x } \quad :{ x }{ y }^{ \prime }={ y }+{ x }\sqrt { { x }^{ 2 }-{ y }^{ 2 } } \\ (x\neq 0{ \quad and }\quad x\quad >\quad y{ \quad or }\quad x<-y)$$

Question 7.
$$x y=\log y+C \quad: y^{\prime}=\frac{y^{2}}{1-x y}(x y \neq 1)$$
Given xy = log y + C
Differentiating w.r.t x, we get

Hence xy = log y + C, is a solution of given differential equation.

Question 8.
y – cos y = x : (y sin y + cos y + x) y’ = y
Given y-cos y = x ⇒ x + cos y = y …(1)
Differentiating w.r.t x, we get
⇒ y’+sin yy’= 1 ⇒ y'(1+sin y)= 1 ….(2)
LHS = (y sin y + cos y + x) y’ ………… (3)
putting (1) in (3) = (y sin y + y) y’
= {y (siny +1)} y’
= { y'(1+ siny)} y putting (2) in (3)
= 1 x y = y = RHS
LHS = RHS
Hence y – cos y = x, is solution of given differential equation.

Question 9.
x + y = tan1 y : y2 y’ + y2 + 1 = 0
x + y = tan-1 y (given)                       .
differentiating w.r.t x
$$1+\mathrm{y}^{\prime}=\frac{1}{\left(1+\mathrm{y}^{2}\right)}\left(\mathrm{y}^{\prime}\right) \Rightarrow\left(1+\mathrm{y}^{\prime}\right)\left(1+\mathrm{y}^{2}\right)=\mathrm{y}^{\prime}$$
⇒ 1+ y’ + y2 + y2 y’ = y’
⇒ 1 + y2 + y2 y’ = 0
⇒ y2 y’ + y2 + 1 = 0 (given)
Hence x + y = tan’1 y, is solution of given differential equation.

Question 10.
\begin{aligned}y=& \sqrt{a^{2}-x^{2}} & x \in(-a, a): x+y \frac{d y}{d x}=0(y \neq 0)\end{aligned}

Hence $$y=\sqrt{a^{2}-x^{2}}$$, is solution of given differential equation.

Question 11.
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4