Question

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

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

Answer 1

The correct option is D $e^{-y/x}+\ln x=1$

Given differential equation can be written as, $\frac{dy}{dx}=\frac{y}{x}+e^{y/x}$

Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{x}$

So above diff. equation becomes

$v+x\frac{dv}{dx}=v+e^v$

$\Rightarrow e^{-v}dv=\frac{dx}{x}$

$\Rightarrow -e^{-v}=\ln x+c$, .........On integrating the above one

$\Rightarrow -e^{-y/x}=\ln x+c$

$\Rightarrow -1=0+c\Rightarrow c=-1$ use $y\left(1\right)=0$

Hence solution is, $e^{-y/x}+\ln x=1$