Solve ODE by change of variable and separation

Question

I have come across the following question:

Solve by change of variable and separation:

$(x+y+1)^2 \frac{dy}{dx} + (x+y+1)^2 + x^3 = 0$

I have tried $u=(x+y+1)^2$, $u=(x+y+1)$ and also $u=\frac{y}{x}$ (thinking it was a homogeneous DE), but haven't been able to get to the separation stage. Am I on the right lines, or have I completely missed the substitution?

Many Thanks

Answer 1

Your considering $u=x+y+1$ work:

If $x+y+1=0$, then a solution is $y=-x-1$. Else, dividing by $(x+y+1)^2$ gives $$\frac{dy}{dx}=-1-\frac{x^3}{(x+y+1)^2}.$$

Now put $u=x+y+1$, and we have $\frac{du}{dx}=1+\frac{dy}{dx}$, thus $$\frac{du}{dx}-1=-1-\frac{x^3}{u^2}$$ or $$\frac{du}{dx}=-\frac{x^3}{u^2}$$ therefore $$\frac{1}{3}u^3=-\frac{1}{4}x^4+c$$ or $$(x+y+1)^3=-\frac{3}{4}x^4+c.$$

Answer 2

Let $x+y+1=v \implies \frac{dy}{dx}=\frac{dv}{dx}-1$, the ODE becomes $$v^2(\frac{dv}{dx}-1)+v^2+x^3 \implies v^2\frac{dv}{dx}+x^3=0$$ $$\implies \int v^2 dv+ \int x^3 dx \implies \frac{v^3}{3}+\frac{x^4}{4}=C$$ Hence, $4(x+y+1)^3+3x^3=D$$