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Q.
Match the terms of column I with the terms of column II and choose the correct option from the codes given below.
Column I (Differential equation)
Column II(solutions)
A
$\frac{d y}{d x}+2 y=\sin x$
1
$y=(\tan x-1)+C e^{-\tan x}$
B
$\frac{d y}{d x}+3 y-e^{-2 x}$
2
$y=\frac{x^2}{16}(4 \log |x|-1)+C x^2$
C
$\frac{d y}{d x}+\frac{y}{x}=x^2$
3
$x y=\frac{x^4}{4}+C$
D
$\frac{d y}{d x}+(\sec x) y=\tan x \left(0 \leq x < \frac{\pi}{2}\right)$
4
$y(\sec x+\tan x)=\sec x+\tan x-x+C$
E
E. $\cos ^2 x \frac{d y}{d x}+y=\tan x$
$\left(0 \leq x < \frac{\pi}{2}\right)$
5
$ y=\frac{1}{5}(2 \sin x-\cos x) +C e^{-2 x}$
F
$x \frac{d y}{d x}+2 y=x^2 \log x$
6
$y=e^{-2 x}+C e^{-3 x}$
Differential Equations
Solution:
A. The given differential equation is $d y / d x+2 y=\sin x$. comparing with the form $d y / d x+P y-Q$, we get
$ P =2 \text { and } Q=\sin x$
$\therefore I F=e^{\int P d x} \Rightarrow e^{2 \int 1 d x} \Rightarrow F =e^{2 x}$
Hence, the solution is given by
$ y \cdot F =\int(Q \times F ) d x+C $
$\Rightarrow y \times e^{2 x} =\int \sin x \cdot e^{2 x} d x+C ....$(i)
Let $I =\int \sin x e^{2 x} $
$\therefore I =\sin x \int e^{2 x} d x-\int\left[\frac{d}{d x} \sin x \int e^{2 x} d x\right] d x$
(integrating by parts)
$\Rightarrow I =\sin x \frac{e^{2 x}}{2}-\frac{1}{2} \int \cos x e^{2 x} d x$
$ \Rightarrow I=\sin x \frac{e^{2 x}}{2}-\frac{1}{2} \cos x \frac{e^{2 x}}{2} $
$\Rightarrow +\frac{1}{2} \int\left[\frac{d}{d x} \cos x \int e^{2 x} d x\right] d x$
$ \Rightarrow I=\sin x \frac{e^{2 x}}{2}-\frac{1}{4} \cos x e^{2 x}+\frac{1}{4} \int(-\sin x) e^{2 x} d x$
$ \Rightarrow I=\sin x \frac{e^{2 x}}{2}-\frac{1}{4} \cos x e^{2 x}-\frac{1}{4} I $
$ \Rightarrow \frac{5 I}{4}=\frac{e^{2 x}}{4}(2 \sin x-\cos x)$
$\Rightarrow I=\frac{e^{2 x}}{5}(2 \sin x-\cos x)$
Therefore, Eq. (i) becomes
$y \times e^{2 x} =\frac{e^{2 x}}{5}(2 \sin x-\cos x+C)$
$\Rightarrow y =\frac{1}{5}(2 \sin x-\cos x)+e^{-2 x} C$
B. The given differential equation is $\frac{d y}{d x}+3 y=e^{-2 x}$. On comparing with the form $\frac{d y}{d x}+P y=Q$, we get $P=3$ and $Q=e^{-2 x}$
$\therefore IF =e^{\int P d x} \Rightarrow e^{\int 3 d x} \Rightarrow F =e^{3 x}$
The solution of the given differential equation is given by
$y \cdot I F =\int(Q \times I ) d x+C $
$\Rightarrow y \times e^{3 x} =\int e^{-2 x} \cdot e^{3 x} d x+C$
$\Rightarrow y e^{3 x} =\int e^x d x+C $
$\Rightarrow y e^{3 x} =e^x+C$
$\Rightarrow y =e^{-2 x}+C e^{-3 x}$
C. The given differential equation is $\frac{d y}{d x}+\frac{y}{x}=x^2$.
On comparing with the form $\frac{d y}{d x}+P y-Q$, we get $P-\frac{1}{x}$ and $Q=x^2$.
$\therefore F =e^{\int P d x}=e^{\frac{1}{x} d x} $
$\Rightarrow F =e^{\log |x|}=x \left(\because e^{\log x}=x\right)$
The solution of the given differential equation is given by
$\Rightarrow y \cdot IF =\int(Q \times IF ) d x+C $
$\Rightarrow y x =\int x^3 d x+C$
$\Rightarrow y x =\frac{x^4}{4}+C$
D. The given differential equation is $\frac{d y}{d x}+y \sec x=\tan x$. On comparing with the form $\frac{d y}{d x}+P y=Q$, we get
$P=\sec x \text { and } Q=\tan x $
$\therefore \mid F =e^{\int P d x}=e^{\int \sec x d x} $
$\Rightarrow IF =e^{\log |\sec x+\tan x|}=\sec x+\tan x....$(i)
The general solution of the given differential equation is given by
$y \cdot F =\int(Q x \mid F) d x+C$
$\Rightarrow y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+C $
$\Rightarrow y(\sec x+\tan x)=\int \tan x \sec x d x+\int \tan ^2 x d x+C$
$\Rightarrow y(\sec x+\tan x)=\sec x+\int \sec ^2 x d x-\int 1 d x+C$
$\left(\because \tan ^2 x=\sec ^2 x-1\right) $
$\Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C$
E. Given, $\cos ^2 x \frac{d y}{d x}+y=\tan x$
On dividing by $\cos ^2 x$ both sides, we get
$\frac{d y}{d x}+y \sec ^2 x=\tan x \sec ^2 x$
On comparing with the form $\frac{d y}{d x}+P y=Q$, we get
$P =\sec ^2 x \text { and } Q=\tan x \sec ^2 x $
$\therefore IF =e^{\int P d x}=e^{\int \sec ^2 x d x} \Rightarrow IF =e^{\tan x}$
The general solution of the given differential equation is given by
$y \cdot IF =\int(Q \times IF ) d x+C$
$ \Rightarrow y e^{\tan x}=\int e^{\tan x} \cdot \tan x \sec ^2 x d x $
Put $ \tan x=t \Rightarrow \sec ^2 x=\frac{d t}{d x} \Rightarrow d x=\frac{d t}{\sec ^2 x} $
$ \therefore y e^{\tan x}=\int e^t \cdot t \sec ^2 \times \frac{d t}{\sec ^2 x} $
$ \Rightarrow y e^{\tan x}=\int e^t \cdot t d t$
$\Rightarrow y e^{\tan x} =e^t \cdot t-\int\left[\frac{d}{d t}(t) \cdot \int e^t d t\right] d t $
$\Rightarrow y e^{\tan x} =e^t \cdot t-\int e^t d t $
$\Rightarrow y e^{\tan x} =e^t t-e^t+C $
$\Rightarrow y e^{\tan x} =(\tan x-1) e^{\tan x}+C$
$\Rightarrow y =\tan x-1+C e^{-\tan x}$
F. Given, $x \frac{d y}{d x}+2 y=x^2 \log x$
On dividing by $x$ both sides, we get $\frac{d y}{d x}+2 \frac{y}{x}=x \log x$
On comparing with the form $\frac{d y}{d x}+P y=Q$, we get
$P-\frac{2}{x} \text { and } Q-x \log x $
$\therefore \mid F=e^{\int P d x}=e^{2 \int \frac{1}{x} d x}$
$\Rightarrow \mid F=e^{2 \log |x|}=x^2$
T he general solution of the given difterential equation is given by
$y \cdot F =\int(Q \times \mid F ) d x+C$
$\Rightarrow x^2 y=\int x^2 \cdot x \log x d x+C$
$\Rightarrow x^2 y=\int \underset{I}{x^3 } \underset{II}{\log x} d x+C$
$\Rightarrow x^2 y=\log x \cdot \int x^3-\int\left[\frac{d}{d x} \log x \int x^3 d x\right] d x+C$
$\Rightarrow x^2 y=\log x \cdot \frac{x^4}{4}-\int\left[\frac{d}{d x} \log x \int x^3 d x\right] d x+C$
$\Rightarrow x^2 y=\log x \cdot \frac{x^4}{4}-\int\left[\frac{1}{x} \times \frac{x^4}{4}\right] d x+C$
$\Rightarrow x^2 y=\log x \cdot \frac{x^4}{4}-\int \frac{x^3}{4} d x+C$
$\Rightarrow x^2 y=\frac{x^4}{4} \log x-\frac{x^4}{16}+C$
$\Rightarrow y=\frac{x^2}{16}(4 \log x-1)+C x^{-2}$