Question

Solve the following differential equation:
$\frac{dy}{dx}+xy=x^3{y}^3$

Answer 1

$\frac{dy}{dx}+xy=x^3{y}^3$

$\Rightarrow \frac{1}{y^3}\frac{dy}{dx}+\frac{x}{y^2}=x^3$

put $\frac{1}{y^2}=t$

$\Rightarrow \frac{-2}{y^3}\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{1}{y^3}\frac{dy}{dx}=-\frac{1}{2}\frac{dt}{dx}$

$\Rightarrow -\frac{1}{2}\frac{dt}{dx}+tx=x^3$

$\Rightarrow \frac{dt}{dx}+(-2t).x=-2x^3$

$\Rightarrow \frac{dt}{dx}+(-2x).t=-2x^3$

$I.F=e^{-x^2}$

$\Rightarrow \frac{d}{dx}(e^{-x^2}.t)=-2x^3{e}^{-x^2}$

$\Rightarrow t{e}^{-x^2}=-2\int x^3{e}^{-x^2}dx$

$t{e}^{-x^2}=-2\int x.x^2{e}^{-x^2}dx$

put $-x^2=-z$

$-2xdx=d-z$

$\Rightarrow t{e}^{-x^2}=\int --z.e^{z}d-z$

$\Rightarrow \frac{1}{y^2}{e}^{-x^2}=-[-z{e}^z-e^z]+c$

$\Rightarrow \frac{1}{y^2}{e}^{-x^2}=e^z(1--z)+c$

$\Rightarrow \frac{1}{y^2}{e}^{-x^2}=e^{-x^2}(1+x^2)+c$