Question

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	$\frac{dy}{dx}+2y=\sin x$	1	$y=(\tan x-1)+C{e}^{-\tan x}$
B	$\frac{dy}{dx}+3y-e^{-2x}$	2	$y=\frac{x^2}{16}(4\log |x|-1)+C{x}^2$
C	$\frac{dy}{dx}+\frac{y}{x}=x^2$	3	$xy=\frac{x^4}{4}+C$
D	$\frac{dy}{dx}+(\sec x)y=\tan x\left(0\le x<\frac{\pi }{2}\right)$	4	$y(\sec x+\tan x)=\sec x+\tan x-x+C$
E	E. ${\cos }^{2}x\frac{dy}{dx}+y=\tan x$ $\left(0\le x<\frac{\pi }{2}\right)$	5	$y=\frac{1}{5}(2\sin x-\cos x)+C{e}^{-2x}$
F	$x\frac{dy}{dx}+2y=x^2\log x$	6	$y=e^{-2x}+C{e}^{-3x}$

• A. A - (6), B - (4), C - (1), D - (3), E - (5), F - (2)
• B. A - (3), B - (1), C - (4), D - (5), E - (2), F - (6)
• C. A - (5), B - (6), C - (3), D - (4), E - (1), F - (2)
• D. A - (6), B - (5), C - (4), D - (3), E - (2), F - (1)

Answer 1

Answer: (C)

A. The given differential equation is $dy/dx+2y=\sin x$. comparing with the form $dy/dx+Py-Q$, we get
$P=2\text{ and }Q=\sin x$
$\therefore IF=e^{\int Pdx}\Rightarrow e^{2\int 1dx}\Rightarrow F=e^{2x}$
Hence, the solution is given by
$y\cdot F=\int (Q\times F)dx+C$
$\Rightarrow y\times e^{2x}=\int \sin x\cdot e^{2x}dx+C....$(i)
Let $I=\int \sin x{e}^{2x}$
$\therefore I=\sin x\int e^{2x}dx-\int \left[\frac{d}{dx}\sin x\int e^{2x}dx\right]dx$
(integrating by parts)
$\Rightarrow I=\sin x\frac{e^{2x}}{2}-\frac{1}{2}\int \cos x{e}^{2x}dx$
$\Rightarrow I=\sin x\frac{e^{2x}}{2}-\frac{1}{2}\cos x\frac{e^{2x}}{2}$
$\Rightarrow +\frac{1}{2}\int \left[\frac{d}{dx}\cos x\int e^{2x}dx\right]dx$
$\Rightarrow I=\sin x\frac{e^{2x}}{2}-\frac{1}{4}\cos x{e}^{2x}+\frac{1}{4}\int (-\sin x)e^{2x}dx$
$\Rightarrow I=\sin x\frac{e^{2x}}{2}-\frac{1}{4}\cos x{e}^{2x}-\frac{1}{4}I$
$\Rightarrow \frac{5I}{4}=\frac{e^{2x}}{4}(2\sin x-\cos x)$
$\Rightarrow I=\frac{e^{2x}}{5}(2\sin x-\cos x)$
Therefore, Eq. (i) becomes
$y\times e^{2x}=\frac{e^{2x}}{5}(2\sin x-\cos x+C)$
$\Rightarrow y=\frac{1}{5}(2\sin x-\cos x)+e^{-2x}C$
B. The given differential equation is $\frac{dy}{dx}+3y=e^{-2x}$. On comparing with the form $\frac{dy}{dx}+Py=Q$, we get $P=3$ and $Q=e^{-2x}$
$\therefore IF=e^{\int Pdx}\Rightarrow e^{\int 3dx}\Rightarrow F=e^{3x}$
The solution of the given differential equation is given by
$y\cdot IF=\int (Q\times I)dx+C$
$\Rightarrow y\times e^{3x}=\int e^{-2x}\cdot e^{3x}dx+C$
$\Rightarrow y{e}^{3x}=\int e^{x}dx+C$
$\Rightarrow y{e}^{3x}=e^x+C$
$\Rightarrow y=e^{-2x}+C{e}^{-3x}$
C. The given differential equation is $\frac{dy}{dx}+\frac{y}{x}=x^2$.
On comparing with the form $\frac{dy}{dx}+Py-Q$, we get $P-\frac{1}{x}$ and $Q=x^2$.
$\therefore F=e^{\int Pdx}=e^{\frac{1}{x}dx}$
$\Rightarrow F=e^{\log |x|}=x\left(\because e^{\log x}=x\right)$
The solution of the given differential equation is given by
$\Rightarrow y\cdot IF=\int (Q\times IF)dx+C$
$\Rightarrow yx=\int x^{3}dx+C$
$\Rightarrow yx=\frac{x^4}{4}+C$
D. The given differential equation is $\frac{dy}{dx}+y\sec x=\tan x$. On comparing with the form $\frac{dy}{dx}+Py=Q$, we get
$P=\sec x\text{ and }Q=\tan x$
$\therefore \mid F=e^{\int Pdx}=e^{\int \sec xdx}$
$\Rightarrow IF=e^{\log |\sec x+\tan x|}=\sec x+\tan x....$(i)
The general solution of the given differential equation is given by
$y\cdot F=\int (Qx\mid F)dx+C$
$\Rightarrow y(\sec x+\tan x)=\int \tan x(\sec x+\tan x)dx+C$
$\Rightarrow y(\sec x+\tan x)=\int \tan x\sec xdx+\int {\tan }^{2}xdx+C$
$\Rightarrow y(\sec x+\tan x)=\sec x+\int {\sec }^{2}xdx-\int 1dx+C$
$\left(\because {\tan }^{2}x={\sec }^{2}x-1\right)$
$\Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C$
E. Given, ${\cos }^{2}x\frac{dy}{dx}+y=\tan x$
On dividing by ${\cos }^{2}x$ both sides, we get
$\frac{dy}{dx}+y{\sec }^{2}x=\tan x{\sec }^{2}x$
On comparing with the form $\frac{dy}{dx}+Py=Q$, we get
$P={\sec }^{2}x\text{ and }Q=\tan x{\sec }^{2}x$
$\therefore IF=e^{\int Pdx}=e^{\int {\sec }^{2}xdx}\Rightarrow IF=e^{\tan x}$
The general solution of the given differential equation is given by
$y\cdot IF=\int (Q\times IF)dx+C$
$\Rightarrow y{e}^{\tan x}=\int e^{\tan x}\cdot \tan x{\sec }^{2}xdx$
Put $\tan x=t\Rightarrow {\sec }^{2}x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{{\sec }^{2}x}$
$\therefore y{e}^{\tan x}=\int e^t\cdot t{\sec }^2\times \frac{dt}{{\sec }^{2}x}$
$\Rightarrow y{e}^{\tan x}=\int e^t\cdot tdt$
$\Rightarrow y{e}^{\tan x}=e^t\cdot t-\int \left[\frac{d}{dt}(t)\cdot \int e^{t}dt\right]dt$
$\Rightarrow y{e}^{\tan x}=e^t\cdot t-\int e^{t}dt$
$\Rightarrow y{e}^{\tan x}=e^{t}t-e^t+C$
$\Rightarrow y{e}^{\tan x}=(\tan x-1)e^{\tan x}+C$
$\Rightarrow y=\tan x-1+C{e}^{-\tan x}$
F. Given, $x\frac{dy}{dx}+2y=x^2\log x$
On dividing by $x$ both sides, we get $\frac{dy}{dx}+2\frac{y}{x}=x\log x$
On comparing with the form $\frac{dy}{dx}+Py=Q$, we get
$P-\frac{2}{x}\text{ and }Q-x\log x$
$\therefore \mid F=e^{\int Pdx}=e^{2\int \frac{1}{x}dx}$
$\Rightarrow \mid F=e^{2\log |x|}=x^2$
T he general solution of the given difterential equation is given by
$y\cdot F=\int (Q\times \mid F)dx+C$
$\Rightarrow x^{2}y=\int x^2\cdot x\log xdx+C$
$\Rightarrow x^{2}y=\int \underset{I}{x^3}\underset{II}{\mathrm{logx}}dx+C$
$\Rightarrow x^{2}y=\log x\cdot \int x^3-\int \left[\frac{d}{dx}\log x\int x^{3}dx\right]dx+C$
$\Rightarrow x^{2}y=\log x\cdot \frac{x^4}{4}-\int \left[\frac{d}{dx}\log x\int x^{3}dx\right]dx+C$
$\Rightarrow x^{2}y=\log x\cdot \frac{x^4}{4}-\int \left[\frac{1}{x}\times \frac{x^4}{4}\right]dx+C$
$\Rightarrow x^{2}y=\log x\cdot \frac{x^4}{4}-\int \frac{x^3}{4}dx+C$
$\Rightarrow x^{2}y=\frac{x^4}{4}\log x-\frac{x^4}{16}+C$
$\Rightarrow y=\frac{x^2}{16}(4\log x-1)+C{x}^{-2}$