Question

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

Answer 1

$\frac{dy}{dx}+xy=x^3{y}^3$
or $\frac{1}{y^3}\frac{dy}{dx}+x.\frac{1}{y^2}=x^3$ ....(1)
Let $1/y^2=v$,
Then $(-2/y^3)\left(dy/dx\right)=\left(dv/dx\right)$
$\Rightarrow -\frac{1}{2}\frac{dv}{dx}+x.v=x^3$
or $\frac{dv}{dx}-2x.v=-2x^3$
This is linear equation in variable $v$.
Here $P=-2x$ and $Q=-2x^3$
$\therefore I.F.=e^{\int Pdx}=e^{\int -2xdx}$
$=e^{-2.\frac{1}{2}{x}^2}=e^{-x^2}$
$\therefore v.(I.F.)=\int [Q.I.F.)]dx+C$
i.e., $v.e^{-x^2}=\int -2x^3.e^{-x^2}dx+C$
$=-\int (-x^2).e^{-x^2}.(-2x)dx+C$
$=-\int t{e}^{t}dt+C$
Let $-x^2=t$ and $-2xdx=dt$
$=-[t.e^t-\int e^{t}dt]+C$
$=(1-t)e^t+C$
$=(1+x^2)e^{-x^2}[\because t=-x^2]$
$\left(1/y^2\right)e^{-x^2}=x^2{e}^{-x^2}+e^{-x^2}+C$,
( Putting $v=1/y^2$)
$y^{-2}=1+x^2+c{e}^{x^2}$