Class 12-NCERT Solutions-Chapter-09-Differential Equations-Mis

Solution:

(i) Highest order derivative is \( \frac{d^2y}{dx^2} \) (Order 2). Its power is 1.
Order: 2, Degree: 1

(ii) Highest order derivative is \( \frac{dy}{dx} \) (Order 1). The highest power of \( \frac{dy}{dx} \) is 3.
Order: 1, Degree: 3

(iii) Highest order derivative is \( \frac{d^4y}{dx^4} \) (Order 4). The equation involves \( \sin(\text{derivative}) \), so degree is not defined.
Order: 4, Degree: Not defined

Solution:

(i) Differentiating \( xy = ae^x + be^{-x} + x^2 \):
\( xy' + y = ae^x - be^{-x} + 2x \).
Differentiating again:
\( xy'' + y' + y' = ae^x + be^{-x} + 2 \)
\( xy'' + 2y' = (ae^x + be^{-x}) + 2 \)
From original eq, \( ae^x + be^{-x} = xy - x^2 \).
So, \( xy'' + 2y' = xy - x^2 + 2 \Rightarrow xy'' + 2y' - xy + x^2 - 2 = 0 \). Verified.

(ii) \( y' = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x) = y + e^x(-a \sin x + b \cos x) \).
\( y'' = y' + [e^x(-a \sin x + b \cos x) + e^x(-a \cos x - b \sin x)] \)
\( y'' = y' + (y' - y) - y = 2y' - 2y \Rightarrow y'' - 2y' + 2y = 0 \). Verified.

(iii) \( y = x \sin 3x \).
\( y' = \sin 3x + 3x \cos 3x \).
\( y'' = 3 \cos 3x + 3 \cos 3x + 3x(-3 \sin 3x) = 6 \cos 3x - 9x \sin 3x \).
\( y'' + 9y = (6 \cos 3x - 9y) + 9y = 6 \cos 3x \).
\( y'' + 9y - 6 \cos 3x = 0 \). Verified.

Solution:

\( \frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} \). This is a homogeneous equation.

Put \( y = vx \). Simplification leads to:

\( \frac{v^3 - 3v}{1 - 3v^2} - v = x \frac{dv}{dx} \)

Solving the integrals leads to the given general solution.

Solution:

\( \frac{dy}{dx} = - \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} \).

Separating variables:

\( \frac{dy}{\sqrt{1-y^2}} = - \frac{dx}{\sqrt{1-x^2}} \).

Integrating both sides:

\( \sin^{-1} y = - \sin^{-1} x + C \).

Solution: \( \sin^{-1} x + \sin^{-1} y = C \).

Solution:

Separating variables:

\( \frac{dy}{y^2+y+1} = - \frac{dx}{x^2+x+1} \).

Completing the square in denominators: \( (y + 1/2)^2 + (\sqrt{3}/2)^2 \).

Integrating gives terms involving \( \tan^{-1} \). Using the formula \( \tan^{-1} A + \tan^{-1} B = \tan^{-1} \frac{A+B}{1-AB} \), the result can be simplified to the required form.

Solution:

\( \cos x \sin y \, dy = - \sin x \cos y \, dx \).

\( \frac{\sin y}{\cos y} dy = - \frac{\sin x}{\cos x} dx \Rightarrow \tan y \, dy = - \tan x \, dx \).

Integrating:

\( \log |\sec y| = - \log |\sec x| + \log C \).

\( \log |\sec y \cdot \sec x| = \log C \Rightarrow \sec x \sec y = C \Rightarrow \cos x \cos y = \frac{1}{C} = k \).

At \( x=0, y=\pi/4 \): \( \cos 0 \cos(\pi/4) = k \Rightarrow 1 \cdot \frac{1}{\sqrt{2}} = k \).

Equation: \( \cos x \cos y = \frac{1}{\sqrt{2}} \) or \( \sec x \sec y = \sqrt{2} \).

Solution:

\( \frac{dy}{1+y^2} = - \frac{e^x}{1+e^{2x}} dx \).

Integrating both sides:

\( \tan^{-1} y = - \int \frac{e^x}{1+(e^x)^2} dx \).

Put \( e^x = t \Rightarrow e^x dx = dt \). \( \int \frac{dt}{1+t^2} = \tan^{-1} t = \tan^{-1} e^x \).

\( \tan^{-1} y = - \tan^{-1} e^x + C \).

At \( x=0, y=1 \): \( \tan^{-1} 1 = - \tan^{-1} 1 + C \Rightarrow \frac{\pi}{4} = - \frac{\pi}{4} + C \Rightarrow C = \frac{\pi}{2} \).

Solution: \( \tan^{-1} y + \tan^{-1} e^x = \frac{\pi}{2} \).

Solution:

Rearranging: \( \frac{dx}{dy} = \frac{x e^{x/y} + y^2}{y e^{x/y}} = \frac{x}{y} + \frac{y}{e^{x/y}} \).

This suggests substitution \( x = vy \). So, \( \frac{dx}{dy} = v + y \frac{dv}{dy} \).

\( v + y \frac{dv}{dy} = v + \frac{y}{e^v} \Rightarrow y \frac{dv}{dy} = y e^{-v} \).

\( e^v dv = dy \).

Integrating: \( e^v = y + C \).

Substituting back \( v = x/y \):

Solution: \( e^{x/y} = y + C \).

Solution:

\( (x-y)dx + (x-y)dy = dx - dy \).

\( dy(x-y+1) = dx(1-(x-y)) \).

\( \frac{dy}{dx} = \frac{1-(x-y)}{1+(x-y)} \).

Put \( x-y = t \Rightarrow 1 - \frac{dy}{dx} = \frac{dt}{dx} \Rightarrow \frac{dy}{dx} = 1 - \frac{dt}{dx} \).

\( 1 - \frac{dt}{dx} = \frac{1-t}{1+t} \Rightarrow \frac{dt}{dx} = 1 - \frac{1-t}{1+t} = \frac{2t}{1+t} \).

\( \frac{1+t}{2t} dt = dx \Rightarrow \frac{1}{2}(\frac{1}{t} + 1) dt = dx \).

Integrating: \( \frac{1}{2} (\log|t| + t) = x + C \).

\( \log|x-y| + (x-y) = 2x + 2C \).

\( \log|x-y| - x - y = K \).

At \( x=0, y=-1 \): \( \log|1| - 0 - (-1) = K \Rightarrow K=1 \).

Solution: \( \log|x-y| = x + y + 1 \).

Solution:

Rewrite as \( \frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \).

\( \frac{dy}{dx} + \frac{1}{\sqrt{x}}y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \). This is a linear differential equation.

I.F. = \( e^{\int x^{-1/2} dx} = e^{2\sqrt{x}} \).

Solution: \( y \cdot e^{2\sqrt{x}} = \int \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \cdot e^{2\sqrt{x}} dx + C \).

\( y e^{2\sqrt{x}} = \int \frac{1}{\sqrt{x}} dx + C = 2\sqrt{x} + C \).

Solution: \( y = (2\sqrt{x} + C)e^{-2\sqrt{x}} \).

Solution:

Linear Equation. \( P = \cot x \), \( Q = 4x \text{cosec } x \).

I.F. = \( e^{\int \cot x dx} = \sin x \).

Solution: \( y \sin x = \int (4x \text{cosec } x) \sin x dx + C = \int 4x dx + C \).

\( y \sin x = 2x^2 + C \).

At \( x = \pi/2, y=0 \): \( 0 = 2(\pi^2/4) + C \Rightarrow C = -\pi^2/2 \).

Solution: \( y \sin x = 2x^2 - \frac{\pi^2}{2} \).

Solution:

\( \frac{dy}{2e^{-y}-1} = \frac{dx}{x+1} \Rightarrow \frac{e^y dy}{2 - e^y} = \frac{dx}{x+1} \).

Integrate: \( -\log|2-e^y| = \log|x+1| + \log C \).

\( \log|(2-e^y)(x+1)| = -\log C = K \).

\( (2-e^y)(x+1) = A \).

At \( x=0, y=0 \): \( (2-1)(1) = A \Rightarrow A=1 \).

Solution: \( y = \log \left( \frac{2x+1}{x+1} \right) \).

Answer: (C)

Solution:

\( y dx - x dy = 0 \Rightarrow \frac{dx}{x} = \frac{dy}{y} \).

\( \log x = \log y + \log k \Rightarrow x = ky \Rightarrow y = Cx \).

Answer: (C)

Solution:

For a linear differential equation of the form \( \frac{dx}{dy} + P_1 x = Q_1 \), where \( P_1 \) and \( Q_1 \) are constants or functions of \( y \) only:

• The Integrating Factor (I.F.) is \( e^{\int P_1 dy} \).
• The general solution is given by: \[ x \cdot (\text{I.F.}) = \int (Q_1 \cdot \text{I.F.}) dy + C \]

Substituting the I.F., we get:

\[ x e^{\int P_1 dy} = \int (Q_1 e^{\int P_1 dy}) dy + C \]

This matches option (C).

Answer: (C)

Solution:

\( e^x dy + y e^x dx + 2x dx = 0 \).

\( d(y e^x) + 2x dx = 0 \).

Integrating: \( y e^x + x^2 = C \).