Question

Solve the following differential equation
$y{e}^{y}dx=(y^3+2x{e}^y)dy$, given that $x=0,y=1.$

Answer 1

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

$\Rightarrow y{e}^y=\frac{dx}{dy}=y^3+2x{e}^y$

$\Rightarrow \frac{dx}{dy}=\frac{2}{y}.x+y^2{e}^{-}y$

$\Rightarrow \frac{dx}{dy}-\frac{2}{y}.x=y^2{e}^{-}y$ ..... (i)

It is a linear differential equation in $x$.
$I.F.=e^f-\frac{2}{y}.dy$
$=e^{-2\log |y|}$
$=e^{\log |y{|}^{-2}}$
$=|y^{-2}|=\frac{1}{y^2}$

Therefore, general solution of (i) is
$x.\frac{1}{y^2}=\int y^2{e}^{-y}.\frac{1}{y^2}dy+c$
$x.\frac{1}{y^2}=\int e^{-y}dy+c$
$\Rightarrow x.\frac{1}{y^2}=\frac{e^{-y}}{-1}+c$ ...... (ii)

Given that x = 0, where y = 1
$\Rightarrow 0=-e^{-1}+c$
$\Rightarrow c=e^{-1}$

Substituting the value of $c$ in (ii), the required particular solution of the given equation is
$\frac{x}{y^2}=-e^{-y}+e^{-1}$
i.e., $x=y^2(-e^{-y}+e^{-1})$