Question

Solve the differential equation $ydx+(x-y{e}^y)dy=0$

Answer 1

Given differential equation is $ydx+(x-y{e}^y)dy=0$

$⟹ydx=-(x-y{e}^y)dy$

$⟹\frac{dx}{dy}=-\frac{(x-y{e}^y)}{y}$

$⟹\frac{dx}{dy}+\frac{x}{y}=e^y$ which is in the form of $\frac{dx}{dy}+P\left(y\right)x=Q\left(y\right)$

$P\left(y\right)=\frac{1}{y},Q\left(y\right)=e^y$

$\text{I.F}=e^{\int -P\left(y\right)}=e^{\int -\frac{1}{y}dy}=e^{-\log y}=\frac{1}{y}$

Then solution of the differential equation is

$x\text{I.F}=\int Q\left(y\right)\text{I.F}dy$

$=\int \frac{e^y}{y}dy$

$=E_i\left(y\right)+C$

$⟹x=\frac{E_i\left(y\right)}{e^y}+C_1$