NCERT Solutions Class-12-Chapter-9-Differential Equations
Excercise-9.5
Note:
Linear Differential Equation: An equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P \) and \( Q \) are constants or functions of \( x \) only.
Solution Steps:
1. Find the Integrating Factor (I.F.): \( I.F. = e^{\int P dx} \).
2. The general solution is given by: \( y \cdot (I.F.) = \int (Q \cdot I.F.) dx + C \).
Alternatively, for \( \frac{dx}{dy} + P_1 x = Q_1 \), the I.F. is \( e^{\int P_1 dy} \) and solution is \( x \cdot (I.F.) = \int (Q_1 \cdot I.F.) dy + C \).
Q1
Find the general solution: \( \frac{dy}{dx} + 2y = \sin x \)
Solution:
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = 2 \) and \( Q = \sin x \).
Integrating Factor (I.F.) = \( e^{\int P dx} = e^{\int 2 dx} = e^{2x} \).
The general solution is:
\( y \cdot e^{2x} = \int (\sin x \cdot e^{2x}) dx + C \).
Let \( I = \int e^{2x} \sin x \, dx \). Using integration by parts:
\( I = \frac{e^{2x}}{5}(2\sin x - \cos x) \).
So, \( y e^{2x} = \frac{e^{2x}}{5}(2\sin x - \cos x) + C \).
\( y = \frac{1}{5}(2\sin x - \cos x) + C e^{-2x} \).
Q2
Find the general solution: \( \frac{dy}{dx} + 3y = e^{-2x} \)
Solution:
Here, \( P = 3 \) and \( Q = e^{-2x} \).
I.F. = \( e^{\int 3 dx} = e^{3x} \).
Solution: \( y \cdot e^{3x} = \int (e^{-2x} \cdot e^{3x}) dx + C = \int e^x dx + C \).
\( y e^{3x} = e^x + C \Rightarrow y = e^{-2x} + C e^{-3x} \).
Q3
Find the general solution: \( \frac{dy}{dx} + \frac{y}{x} = x^2 \)
Solution:
\( P = \frac{1}{x}, Q = x^2 \).
I.F. = \( e^{\int \frac{1}{x} dx} = e^{\log x} = x \).
Solution: \( y \cdot x = \int (x^2 \cdot x) dx + C = \int x^3 dx + C \).
\( xy = \frac{x^4}{4} + C \Rightarrow y = \frac{x^3}{4} + \frac{C}{x} \).
Q4
Find the general solution: \( \frac{dy}{dx} + (\sec x)y = \tan x \quad (0 \le x < \frac{\pi}{2}) \)
Solution:
\( P = \sec x, Q = \tan x \).
I.F. = \( e^{\int \sec x dx} = e^{\log(\sec x + \tan x)} = \sec x + \tan x \).
Solution: \( y(\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C \).
\( \int (\sec x \tan x + \tan^2 x) dx = \int (\sec x \tan x + \sec^2 x - 1) dx \).
\( = \sec x + \tan x - x + C \).
\( y(\sec x + \tan x) = \sec x + \tan x - x + C \).
Q5
Find the general solution: \( \cos^2 x \frac{dy}{dx} + y = \tan x \quad (0 \le x < \frac{\pi}{2}) \)
Solution:
Divide by \( \cos^2 x \): \( \frac{dy}{dx} + (\sec^2 x)y = \sec^2 x \tan x \).
\( P = \sec^2 x, Q = \sec^2 x \tan x \).
I.F. = \( e^{\int \sec^2 x dx} = e^{\tan x} \).
Solution: \( y e^{\tan x} = \int (\sec^2 x \tan x) e^{\tan x} dx + C \).
Put \( \tan x = t \Rightarrow \sec^2 x dx = dt \). Integral is \( \int t e^t dt = e^t(t-1) \).
\( y e^{\tan x} = e^{\tan x}(\tan x - 1) + C \).
\( y = \tan x - 1 + C e^{-\tan x} \).
Q6
Find the general solution: \( x \frac{dy}{dx} + 2y = x^2 \log x \)
Solution:
Divide by \( x \): \( \frac{dy}{dx} + \frac{2}{x}y = x \log x \).
\( P = \frac{2}{x} \). I.F. = \( e^{2 \log x} = e^{\log x^2} = x^2 \).
Solution: \( y \cdot x^2 = \int (x \log x \cdot x^2) dx + C = \int x^3 \log x \, dx + C \).
Using by parts: \( \int \log x \cdot x^3 dx = \log x \frac{x^4}{4} - \int \frac{1}{x} \frac{x^4}{4} dx \).
\( x^2 y = \frac{x^4}{4} \log x - \frac{x^4}{16} + C \).
Q7
Find the general solution: \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \)
Solution:
Divide by \( x \log x \): \( \frac{dy}{dx} + \frac{1}{x \log x}y = \frac{2}{x^2} \).
I.F. = \( e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x \).
Solution: \( y \log x = \int \frac{2}{x^2} \log x \, dx + C \).
Integrate by parts: \( 2 \int x^{-2} \log x dx = 2 [-\frac{\log x}{x} - \int -\frac{1}{x^2} dx] = 2 [-\frac{\log x}{x} - \frac{1}{x}] \).
\( y \log x = -\frac{2}{x}(1 + \log x) + C \).
Q8
Find the general solution: \( (1+x^2)dy + 2xy dx = \cot x \, dx \quad (x \neq 0) \)
Solution:
\( \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{\cot x}{1+x^2} \).
I.F. = \( e^{\int \frac{2x}{1+x^2} dx} = e^{\log(1+x^2)} = 1+x^2 \).
Solution: \( y(1+x^2) = \int \frac{\cot x}{1+x^2} (1+x^2) dx = \int \cot x \, dx \).
\( y(1+x^2) = \log|\sin x| + C \).
\( y = (1+x^2)^{-1} \log|\sin x| + C(1+x^2)^{-1} \).
Q9
Find the general solution: \( x \frac{dy}{dx} + y - x + xy \cot x = 0 \quad (x \neq 0) \)
Solution:
Rewrite: \( x \frac{dy}{dx} + y(1 + x \cot x) = x \).
Divide by \( x \): \( \frac{dy}{dx} + (\frac{1}{x} + \cot x)y = 1 \).
I.F. = \( e^{\int (\frac{1}{x} + \cot x) dx} = e^{\log x + \log \sin x} = x \sin x \).
Solution: \( y(x \sin x) = \int 1 \cdot (x \sin x) dx + C \).
\( xy \sin x = -x \cos x + \sin x + C \).
\( y = \frac{1}{x} - \cot x + \frac{C}{x \sin x} \).
Q10
Find the general solution: \( (x+y) \frac{dy}{dx} = 1 \)
Solution:
Rewrite as \( \frac{dx}{dy} = x + y \Rightarrow \frac{dx}{dy} - x = y \).
Linear in \( x \). \( P_1 = -1, Q_1 = y \).
I.F. = \( e^{\int -1 dy} = e^{-y} \).
Solution: \( x e^{-y} = \int y e^{-y} dy + C \).
Using integration by parts: \( \int y e^{-y} dy = -y e^{-y} - e^{-y} \).
\( x e^{-y} = -e^{-y}(y+1) + C \Rightarrow x = -y - 1 + C e^y \).
Q11
Find the general solution: \( y dx + (x - y^2) dy = 0 \)
Solution:
\( \frac{dx}{dy} + \frac{1}{y}x = y \).
I.F. = \( e^{\int \frac{1}{y} dy} = y \).
Solution: \( x y = \int y \cdot y \, dy = \int y^2 dy = \frac{y^3}{3} + C \).
\( x = \frac{y^2}{3} + \frac{C}{y} \).
Q12
Find the general solution: \( (x + 3y^2) \frac{dy}{dx} = y \quad (y > 0) \)
Solution:
Rewrite: \( \frac{dx}{dy} = \frac{x + 3y^2}{y} = \frac{x}{y} + 3y \).
\( \frac{dx}{dy} - \frac{1}{y}x = 3y \).
I.F. = \( e^{-\int \frac{1}{y} dy} = e^{-\log y} = \frac{1}{y} \).
Solution: \( x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} dy = \int 3 dy = 3y + C \).
\( x = 3y^2 + Cy \).
Q13
Find particular solution: \( \frac{dy}{dx} + 2y \tan x = \sin x \); \( y=0 \) when \( x=\pi/3 \).
Solution:
I.F. = \( e^{\int 2 \tan x dx} = e^{2 \log \sec x} = \sec^2 x \).
\( y \sec^2 x = \int \sin x \sec^2 x \, dx = \int \sec x \tan x \, dx = \sec x + C \).
At \( x = \pi/3, y = 0 \): \( 0 = \sec(\pi/3) + C = 2 + C \Rightarrow C = -2 \).
\( y \sec^2 x = \sec x - 2 \Rightarrow y = \cos x - 2 \cos^2 x \).
Q14
Find particular solution: \( (1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2} \); \( y=0 \) when \( x=1 \).
Solution:
\( \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{1}{(1+x^2)^2} \).
I.F. = \( 1+x^2 \).
\( y(1+x^2) = \int \frac{1}{(1+x^2)^2}(1+x^2) dx = \int \frac{1}{1+x^2} dx = \tan^{-1} x + C \).
At \( x=1, y=0 \): \( 0 = \frac{\pi}{4} + C \Rightarrow C = -\frac{\pi}{4} \).
\( y(1+x^2) = \tan^{-1} x - \frac{\pi}{4} \).
Q15
Find particular solution: \( \frac{dy}{dx} - 3y \cot x = \sin 2x \); \( y=2 \) when \( x=\pi/2 \).
Solution:
I.F. = \( e^{-3 \int \cot x dx} = e^{-3 \log \sin x} = \text{cosec}^3 x \).
\( y \text{cosec}^3 x = \int \sin 2x \text{cosec}^3 x \, dx = \int 2 \sin x \cos x \frac{1}{\sin^3 x} dx = 2 \int \cot x \text{cosec} x \, dx \).
\( y \text{cosec}^3 x = -2 \text{cosec} x + C \).
At \( x=\pi/2, y=2 \): \( 2(1) = -2(1) + C \Rightarrow C=4 \).
\( y = -2 \sin^2 x + 4 \sin^3 x \).