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

Question:2 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = x^2 + 2x + C\qquad:\ y' -2x - 2 =0$

Answer:

Given,

$y = x^2 + 2x + C$

Now, differentiating both sides w.r.t. x,

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

Substituting the values of y’ in the given differential equations,

$y' -2x - 2 =2x + 2 - 2x - 2 = 0= RHS$ .

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

Question:3. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = \cos x + C\qquad :\ y' + \sin x = 0$

Answer:

Given,

$y = \cos x + C$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(cosx + C) = -sinx$

Substituting the values of y’ in the given differential equations,

$y' - \sin x = -sinx -sinx = -2sinx \neq RHS$ .

Therefore, the given function is not the solution of the corresponding differential equation.

Question:4. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = \sqrt{1 + x^2}\qquad :\ y' = \frac{xy}{1 + x^2}$

Answer:

Given,

$y = \sqrt{1 + x^2}$

Now, differentiating both sides w.r.t. x,

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

Substituting the values of y in RHS,

$\frac{x\sqrt{1+x^2}}{1 + x^2} = \frac{x}{\sqrt{1+x^2}} = LHS$ .

Therefore, the given function is a solution of the corresponding differential equation.

Question:5 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = Ax\qquad :\ xy' = y\;(x\neq 0)$

Answer:

Given,

$y = Ax$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(Ax) = A$

Substituting the values of y' in LHS,

$xy' = x(A) = Ax = y = RHS$ .

Therefore, the given function is a solution of the corresponding differential equation.

Question:6. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = x\sin x\qquad :\ xy' = y + x\sqrt{x^2 - y^2}\ (x\neq 0\ \textup{and} \ x > y\ or \ x < -y)$

Answer:

Given,

$y = x\sin x$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(xsinx) = sinx + xcosx$

Substituting the values of y' in LHS,

$xy' = x(sinx + xcosx)$

Substituting the values of y in RHS.

$\\xsinx + x\sqrt{x^2 - x^2sin^2x} = xsinx + x^2\sqrt{1-sinx^2} = x(sinx+xcosx) = LHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question:7 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$xy = \log y + C\qquad :\ y' = \frac{y^2}{1 - xy}\ (xy\neq 1)$

Answer:

Given,

$xy = \log y + C$

Now, differentiating both sides w.r.t. x,

$\\ y + x\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(logy) = \frac{1}{y}\frac{\mathrm{d}y }{\mathrm{d} x}$ \\ \\

$\implies y^2 + xyy' = y' $

$\implies y^2 = y'(1-xy) $

$\implies y' = \frac{y^2}{1-xy}$

Substituting the values of y' in LHS,

$y' = \frac{y^2}{1-xy} = RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question:8 In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y - cos y = x \qquad :(\ y\sin y + \cos y + x) y' = y$

Answer:

Given,

$y - cos y = x$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} +siny\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x) = 1$

$\implies$ y' + siny.y' = 1

$\implies$ y'(1 + siny) = 1

$\implies y' = \frac{1}{1+siny}$

Substituting the values of y and y' in LHS,

$(\ (x+cosy)\sin y + \cos y + x) (\frac{1}{1+siny})$

$= [x(1+siny) + cosy(1+siny)]\frac{1}{1+siny}$

= (x + cosy) = y = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:9 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$x + y = \tan^{-1}y\qquad :\ y^2y' + y^2 + 1 = 0$

Answer:

Given,

$x + y = \tan^{-1}y$

Now, differentiating both sides w.r.t. x,

$\\ 1 + \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{1 + y^2}\frac{\mathrm{d} y}{\mathrm{d} x}$

$\implies1+y^2 = y'(1-(1+y^2)) = -y^2y'$

$\implies y' = -\frac{1+y^2}{y^2}$

Substituting the values of y' in LHS,

$y^2(-\frac{1+y^2}{y^2}) + y^2 + 1 = -1- y^2+ y^2 +1 = 0 = RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question:10 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = \sqrt{a^2 - x^2}\ x\in (-a,a)\qquad : \ x + y \frac{dy}{dx} = 0\ (y\neq 0)$

Answer:

Given,

$y = \sqrt{a^2 - x^2}$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{a^2 - x^2}) = \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}$

Substituting the values of y and y' in LHS,

$x + y \frac{dy}{dx} = x + (\sqrt{a^2 - x^2})(\frac{-x}{\sqrt{a^2 - x^2}}) = 0 = RHS$

Therefore, the given function is a solution of the corresponding 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:

(D) 4

The number of constants in the general solution of a differential equation of order n is equal to its order.

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:

(D) 0

In a particular solution of a differential equation, there is no arbitrary constan