Question

Match the following with the integrating factor of the D.E., A, B, C, D

List I	List II
A) $\frac{dy}{dx}-y\cot x=\csc x$	1. $x\sin x$
B) $x\log x\frac{dy}{dx}+y=2\log x$	2. $x{e}^x$
C) $x\sin x\frac{dy}{dx}+y(x\cos x+\sin x)=\sin x$	3. $\cos ecx$
D) $x\frac{dy}{dx}+y(1+x)=1$	4. $-\cos ecx$
	5. $logx$

• A. 3, 5, 1, 4
• B. 3, 5, 1, 2
• C. 3, 1, 5, 4
• D. 2, 3, 4, 5

Answer 1

The correct option is B 3, 5, 1, 2

$I.F=e^{\int p\left(x\right)dx}$

$A)$ $p\left(x\right)=-\cot x$

$\Rightarrow \int p\left(x\right)dx=\int -\cot xdx=\log \csc x$

$\Rightarrow I.F=e^{\log \csc x}=\csc x$

$B)$ $\frac{dy}{dx}+\frac{y}{x\log x}$

$\int p\left(x\right)dx=\int \frac{1}{x\log x}dx$

put $\log x=t$

$\frac{1}{x}dx=dt$

$\Rightarrow \int \frac{1}{t}dt=\log t$

$\int p\left(x\right)dx=\log \log x$

$\Rightarrow I.F=e^{\log \log x}=\log x$

$C)$ $\frac{dy}{dx}+y\left[\frac{x\cos x+\sin x}{x\sin x}\right]=\frac{1}{x}$

$\int p\left(x\right)dx=\int \cot xdx+\int \frac{1}{x}dx$

$=\log \sin x+\log x=\log x\sin x$

$\Rightarrow I.F=e^{\int p\left(x\right)dx}=e^{\log x\sin x}=x\sin x$

$D)$ $\frac{dy}{dx}+y\left(\frac{1+x}{x}\right)=\frac{1}{x}$

$\int p\left(x\right)dx=\int \frac{1}{x}dx+\int 1dx$

$=\log x+x$

$\Rightarrow I.F=e^{\log x+x}=x.e^x$