Putting $x+y=v$

Different w.e.t x

$1+\dfrac {dy}{dx}=\dfrac {dv}{dx}$

Hence the given equation reduces to

$\dfrac {dv}{dx} + xv =x^3v^3$

$\dfrac 1{v^3}\dfrac {dv}{dx} +\dfrac x{v^2}=x^3$

Putting $\dfrac 1{v^2}=t\hspace {3 cm }\therefore {\dfrac {-2}{v^3}} \dfrac {dv}{dx}=\dfrac {dt}{dx}$

we get

$\dfrac {-1}2\dfrac {dt}{dx} + x\varepsilon =x ^3$

$\therefore \dfrac {dt}{dx}-2xt=-2x^3$

Here $,P=2x,Q=-2 $ & $ n=3$

This is a linear differential equation

$I.F. =e^{\int e\space dx}=e^{\int -2x\space dx}=e^{-x^2}$

$\therefore$ The Solution is

$t.e^{-x^2}=\int e^{-x^2}x(-2x^3)dx + c$

Now put $–x^2=U$

$-2dx=du$

$te^{-x^2}=\int e^U\times - U du + c$

Integration by parts we get

$te^{-x^2}=-[ue^u-\int 1e^U du] $

$te^{-x^2}=-[ue^u- e^U ] + c $

$te^{-x^2}= e^U-ue^u + c $

Re-substituting value of U

$te^{-x^2}= e^{-x^2}+x^2e^{-x^2} + c $

$=e^{-x^2} (x^2+1)+ c$