Question

The solution of the differential equation $logx\frac{dy}{dx}+\frac{y}{x}=sin2x$ is

• A. $ylog\left|\begin{array}{c}x\end{array}\right|=C-\frac{1}{2}cosx$
• B. $ylog\left|\begin{array}{c}x\end{array}\right|=C+\frac{1}{2}cos2x$
• C. $ylog\left|\begin{array}{c}x\end{array}\right|=C-\frac{1}{2}cos2x$
• D. $xylog\left|\begin{array}{c}x\end{array}\right|=C-\frac{1}{2}cos2x$
• E. $ylog\left|\begin{array}{c}x\end{array}\right|=Cx-\frac{1}{2}cosx$

Answer 1

Answer: (C)

Given differential equation is
$\log x\frac{dy}{dx}+\frac{y}{x}=\sin 2x\dots (i)$
$\Rightarrow \frac{dy}{dx}+\frac{y}{x\log x}=\frac{\sin 2x}{\log x}$
$IF=e^{\int \frac{1}{x\log x}dx}=e^{\log (\log x)}$
$=\log |x|$
$\therefore$ Complete solution is
$y\cdot (\log |x|)=\int \log x\cdot \frac{\sin 2x}{\log x}dx+C$
$=\int \sin 2xdx+C$
$=-\frac{1}{2}\cos 2x+C$
$\Rightarrow y\log |x|=C-\frac{1}{2}\cos 2x$