Question

Find the general solution of the differential equation $y{e}^{y}dx=(y^3+2x{e}^y)dy$

Answer 1

$y{e}^{y}dx=(y^3+2x{e}^y)dy$

$\frac{dx}{dy}=\frac{y^3+2x{e}^y}{y{e}^y}$ by dividing the above equation by $y{e}^y$

$\Rightarrow \frac{dx}{dy}-\frac{2}{y}x=y^2{e}^{-y}$ by seperating the variables

$\frac{dx}{dy}-\frac{2}{y}x=y^2{e}^{-y}$ is of the form $\frac{dx}{dy}+Px=Q$

where $P=\frac{2}{y}$ and $Q=y^2{e}^{-y}$

Integrating Factor$=$ I.F. = $e^{\int P\left(x\right).dx}$

Now,

$x\times$ I.F$=\int Q\times I.Fdy+c$

$\frac{x}{y^2}=\int y^2{e}^{-y}\times \frac{1}{y^2}dy+c$

$=-e^{-y}+c$ where $\int e^{-x}dx=-e^{-x}+c$

$\Rightarrow \frac{x}{y^2}=-e^{-y}+c$

$\Rightarrow x=-y^2{e}^{-y}+c{y}^2$