Question

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

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

Answer 1

Answer: (C)

$y\log |x|=C-\frac{1}{2}\cos x$

Given differential equation is

$\log x\frac{dy}{dx}+\frac{y}{x}=\sin 2x$ .... (i)

$\Rightarrow \frac{dy}{dx}+\frac{y}{x\log x}=\frac{\sin 2x}{\log x}$

$IF=e^{\int \frac{1}{x\log x}dx}=e^{\log \left(\log x\right)}$

$=\log |x|$

Therefore, complete solution is

$y\cdot \left(\log |x|\right)=\int \log x\cdot \frac{\sin 2x}{\log x}dx+C$

$=\int \sin 2xdx+C$

$=-\frac{1}{2}\cos 2x+C$

$\Rightarrow y\log |x|=C-\frac{1}{2}\cos 2x$