Question

Find particular solution of differential equation.
$x.e^{y/x}-y+x\frac{dy}{dx}=0$ under the initial condition $y\left(e\right)=0$

Answer 1

$x{e}^{\frac{y}{x}}-y+x\frac{dy}{dx}=0$
$\therefore x\frac{dy}{dx}=y-x{e}^{\frac{y}{x}}$
$\therefore \frac{dy}{dx}=\frac{y-x{e}^{\frac{y}{x}}}{x}$
$\therefore \frac{dy}{dx}=\frac{y}{x}-e^{\frac{y}{x}}...\left(1\right)$
Which is a homogeneous differential equation
Now take $\frac{y}{x}=v$
We have $\frac{y}{x}=v$
Differentiate w.r.to $x$
$\therefore \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore$ from equation (1) becomes as follows
$\therefore v+x\frac{dv}{dx}=v-e^v$
$\therefore$ $y=vx$
$\therefore x\frac{dv}{dx}=-e^v$
$\therefore \frac{-dv}{e^v}=\frac{dx}{x}$
integrating both sides
$\therefore -\int e^{v}dv=\int \frac{dx}{x}$
$\therefore e^{-v}=\log \left|x\right|+c$
$\therefore e^{-\frac{y}{x}}=\log x+c...\left(2\right)$
Now $y\left(e\right)=0$
We take $x=e;y=0$
$\therefore e^0=\log e+c$
$\therefore 1=1+c$
$\therefore c=0$
Particular solution is
$\therefore e^{-\frac{y}{x}}=\log \left|x\right|$