Let's check if the ODE is exact. To do that, we want to show that if
[tex]\underbrace{(x+y)^2}_M\,\mathrm dx+\underbrace{(2xy+x^2-2)}_N\,\mathrm dy=0[/tex]
then [tex]M_y=N_x[/tex]. We have
[tex]M_y=2(x+y)[/tex]
[tex]N_x=2y+2x=2(x+y)[/tex]
so the equation is indeed exact. We're looking for a solution of the form [tex]\Psi(x,y)=C[/tex]. Computing the total differential yields the original ODE,
[tex]\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0[/tex]
[tex]\implies\begin{cases}\Psi_x=(x+y)^2\\\Psi_y=2xy+x^2-2\end{cases}[/tex]
Integrate both sides of the first PDE with respect to [tex]x[/tex]; then
[tex]\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\dfrac{(x+y)^3}3+f(y)[/tex]
where [tex]f(y)[/tex] is a function of [tex]y[/tex] alone. Differentiate this with respect to [tex]y[/tex] so that
[tex]\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)[/tex]
[tex]\implies2xy+x^2-2=x^2+2xy+y^2+f'(y)[/tex]
[tex]f'(y)=-2-y^2\implies f(y)=-2y-\dfrac{y^3}3+C[/tex]
So the solution to this ODE is
[tex]\Psi(x,y)=\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3+C=C[/tex]
i.e.
[tex]\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C[/tex]
