Question

The solution of the differential equation $log x \frac{dy}{dx}+\frac{y}{x}=sin 2x$ is

• A. $ y log\begin{vmatrix}x\end{vmatrix}=C-\frac{1}{2}cos x$
• B. $ y log\begin{vmatrix}x\end{vmatrix}=C+\frac{1}{2}cos 2x$
• C. $ y log\begin{vmatrix}x\end{vmatrix}=C-\frac{1}{2}cos 2x$
• D. $ xy log\begin{vmatrix}x\end{vmatrix}=C-\frac{1}{2}cos 2x$
• E. $ y log\begin{vmatrix}x\end{vmatrix}=Cx-\frac{1}{2}cos x$

Answer 1

Answer: (C)

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