Question

The solution of $x\frac{dx}{dy}=y+x{e}^{y/x}$ with $y(1)=0$ is

• A. $e^{y/x}+\log x=1$
• B. $e^{-y/x}=\log x$
• C. $e^{-y/x}+2\log x=1$
• D. $e^{-y/x}+\log x=1$

Answer 1

Answer: (A)

$x\frac{dy}{dx}=y+x{e}^{yx}$
$\Rightarrow \frac{dy}{dx}-\frac{y}{x}=e^{yx}$
Put $y=vx$
$\Rightarrow \frac{dy}{dx}=\upsilon +x\frac{d\upsilon }{dx}$
$\therefore v+x\frac{dv}{dx}-\upsilon =e^{\upsilon }$
$\Rightarrow x\frac{dv}{dx}=e^{\upsilon }$
$\Rightarrow \int e^{-\upsilon }=\int \frac{dx}{x}+c$
$\Rightarrow -e^{-\upsilon }=logx+c$
$\Rightarrow e^{-yx}=logx+c$
When $x=1$, $y=0$
$\therefore -e^{-0}=log1+c\Rightarrow -1=0+c\Rightarrow c=-1$
$\therefore x=1$, $y=0$
$\therefore -e^{-yx}=logx-1$
$\Rightarrow logx+e^{-yx}=1$