Question

Solution of the differential equation $(x{e}^{\frac{y}{x}}-y\sin \frac{y}{x})dx+x\sin \frac{y}{x}dy=0;x>0$ is:
(where $c$ is integration constant and $\log$ is given with base ${}^{\prime }{e}^{\prime }$)

• A. $\log x=c+\frac{1}{2}{e}^{\frac{y}{x}}\cdot (\sin \frac{y}{x}+4\cos \frac{y}{x})$
• B. $2\log x=c+\frac{3}{2}{e}^{\frac{-y}{x}}\cdot (\sin \frac{y}{x}-4\cos \frac{y}{x})$
• C. $2\log x=c+\frac{3}{2}{e}^{\frac{y}{x}}\cdot (\sin \frac{y}{x}+4\cos \frac{y}{x})$
• D. $\log x=c+\frac{1}{2}{e}^{\frac{-y}{x}}\cdot (\sin \frac{y}{x}+\cos \frac{y}{x})$

Answer 1

The correct option is D $\log x=c+\frac{1}{2}{e}^{\frac{-y}{x}}\cdot (\sin \frac{y}{x}+\cos \frac{y}{x})$

$(e^{\frac{y}{x}}-\frac{y}{x}\sin \frac{y}{x})+\sin \frac{y}{x}\frac{dy}{dx}=0$
Put $y=vx$ such that $v+x\frac{dv}{dx}=\frac{dy}{dx}$

$\therefore (e^v-v\sin v)+\sin v(v+x\frac{dv}{dx})=0$
$\Rightarrow e^v+x\sin v\frac{dv}{dx}=0$
$\Rightarrow \int \frac{dx}{x}+\int e^{-v}\sin vdv=0$
Integrating by parts, we get
$\log x-\frac{1}{2}{e}^{-v}(\sin v+\cos v)=c$

$\Rightarrow \log x=c+\frac{1}{2}{e}^{\frac{-y}{x}}\cdot (\sin \frac{y}{x}+\cos \frac{y}{x})$