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

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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 = e^x + 1 : y’’- y’ = 0
Answer:

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

Question 2.
y⁵= x² + 2x + C : y’ – 2x – 2 = 0
Answer:
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
Answer:
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.
\(y=\sqrt{1+x^{2}} \quad: y^{\prime}=\frac{x y}{1+x^{2}}\)
Answer:

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

Question 7.
\(x y=\log y+C \quad: y^{\prime}=\frac{y^{2}}{1-x y}(x y \neq 1)\)
Answer:
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
Answer:
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
Answer:
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.
\(\begin{aligned}y=& \sqrt{a^{2}-x^{2}} & x \in(-a, a): x+y \frac{d y}{d x}=0(y \neq 0)\end{aligned}\)
Answer:

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
Answer:
The number of arbitrary constants in general solution of a differential equation of n^th order is n. Hence answer is (D) 4.

Question 12.
The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B)2
(C)1
(D) 0
Answer:
In particular solution of differential equation, there is no arbitrary constant. Hence answer is (D) 0.