# Class 12 Maths NCERT Solutions for Chapter 9 Differential Equations Exercise 9.2

### Differential Equations Exercise 9.2 Solutions

**1. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equationy = e ^{x} + 1 : y″ – y′ = 0**

**Solution**

y = e^{x} + 1

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

⇒ y' = e^{x} **...(1) **

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

⇒ 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.

**2. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equationy = x ^{2} + 2x + C : y′ – 2x – 2 = 0 **

**Solution**

y = x^{2} + 2x + C

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

⇒ 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.

**3. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equationy = cos x + C : y′ + sin x = 0**

**Solution**

y = cos x + C

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

⇒ y' = - sin x

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

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

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

**4. Verify that the given functions (explicit of implicit) is a solution of the corresponding differential equation y = √(1 + x ^{2}) : y' = xy/(1+ x^{2})**

**Solution**

y = √(1 + x^{2} )

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

∴ L.H.S. = R.H.S.

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

**5. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equationy = Ax : xy′ = y (x ≠ 0)**

**Solution**

y = Ax

Differentiating both sides with respect to x, we get :

y' = d/dx (Ax)

⇒ y' = A

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

L.H.S. = xy' = x.A = Ax = y = R.H.S

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

**6. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation y = x sin x : xy′ = y + x√(x ^{2} - y^{2} ) (x ≠ 0 and x > y, or x < -y) **

**Solution**

y = x sin x

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

⇒ y' = sin x + x cosx

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

L.H.S = xy' = x(sin x + x cos x)

= x sin x + x^{2} cos x

= y + x^{2} .√(1 - sin^{2} x)

= R.H.S

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

**7. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation xy = log y + C : y' = y ^{2}/(1 - xy)(xy ≠ 1) **

**Solution**

xy = log y + C

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

∴ L.H.S = R.H.S

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

**8. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y – cos y = x : (y sin y + cos y + x) y′ = y**

**Solution**

y - cos y = x **...(1) **

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

⇒ y' + sin y . y' = 1

⇒ y'(1 + sin y( = 1

⇒ y' = 1/(1 + sin y)

Substituting the value of y' in equation (1), we get :

L.H.S = (y sin y + cos y + x)y'

(y sin y + cos y + y - cos y) × ([1/(1 + siny)]

= y(1 + sin y). 1/(1 + sin y)

= y

= R.H.S.

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

**9. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equationx + y = tan ^{–1}y : y^{2} y′ + y^{2} + 1 = 0**

**Solution**

x + y = tan^{-1} y

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

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

= -1 - y^{2} + y^{2} + 1

= 0

= R.H.S.

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

**10. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equationy = √(a ^{2} - x^{2} x) ∈ (-a, a) : x + y (dy/dx) = 0 (y ≠ 0) **

**Solution**

y = √(a^{2} - x^{2} )

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

Substituting the value of dy/dx in the given differential equation, we get :

= x - x

= 0

= R.H.S

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

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

**Solution**

We know that the number of constants in the general solution of a differential equation of order *n* is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential equation is four.

Hence, the correct answer is D.

**12.The numbers of arbitrary constants in the particular solution of a differential equation of third order are:(A) 3(B) 2(C) 1(D) 0**

**Solution**

In a particular solution of a differential equation, there are no arbitrary constants.

Hence, the correct answer is D.