Question

Find the general solution of given differential equation.

$x\sin x\frac{dy}{dx}+y(x\cos x+\sin x)=\sin x$

• A. $y\left(x\sin x\right)=-\sin x+c.$
• B. $y\left(x\cos x\right)=\sin x+c.$
• C. $y\left(x\cos x\right)=\cos x+c.$
• D. $y\left(x\sin x\right)=-\cos x+c.$

Answer 1

The correct option is D $y\left(x\sin x\right)=-\cos x+c.$

$x\sin x\frac{dy}{dx}+y(x\cos x+\sin x)=\sin x$
$\Rightarrow \frac{dy}{dx}+y(\cot x+\frac{1}{x})=\frac{1}{x}$
which is a linear differential equation with y as dependent variable
Here, $P=\cot x+\frac{1}{x};Q=\frac{1}{x}$
Integrating factor $I.F.=e^{Pdx}$
$=e^{\int (\cot x+\frac{1}{x})dx}$
$=e^{\log \sin x+\log x}$
$=e^{\log \left(x\sin x\right)}$
$\Rightarrow I.F.=x\sin x$
Solution is given by
$y\left(x\sin x\right)=\int x\sin x\frac{1}{x}dx$
$\Rightarrow y\left(x\sin x\right)=-\cos x+C$