Is this differential equation solvable when I cannot separate the variables?

Question

Suppose $(x+y+1)^2 \frac{dy}{dx}+(x+y+1)^2+x^3=0$. How do I express x in terms of y?

My thoughts:

1. I don't think there is a clean way of separating the variables, especially once I expand the terms.

2. So I try to work through without expanding. I can try dividing the entire equation by $(x+y+1)^2$ and this might give me a cleaner expression, but I still cannot separate the x from the y.

It doesn't seem there is a way to separate the x from the y, so is this still solvable?

Answer 1

Let $u=x+y+1$ and the equation tranforms to \begin{eqnarray*} \frac{du}{dx}= - \frac{x^3}{u^2}. \end{eqnarray*}

Answer 2

This ODE has the form: $M(x,y) + N(x,y)dy/dx = 0$, that is ripe for the application of exact differential equations because $M_y = N_x = 2(1+x+y)$. By this theory, let

$\Psi_x = M(x,y)$ and $\Psi_y = N(x,y)$. Then, our ODE can be written in the form:

$\frac{d}{dx} \left[\Psi(x,y(x))\right]$ =0.

Starting with the first equation (in doesn't matter which one), we have that:

$\Psi_x = M(x,y) = (x+y+1)^2 + x^3$, integrating we get:

$\Psi(x,y) = \frac{x^4}{4}+\frac{1}{3} (x+y+1)^3 + h(y)$.

Now differentiate this with respect to $y$, we get:

$\Psi_y = (1+x+y)^2 + h'(y)$,

but this must be equal to N(x,y):

$\Psi_y = N(x,y) = (1+x+y)^2 + h'(y) = (1+x+y)^2$,

which implies that $h'(y) = 0$, which means that $h(y) = k$, where $k \in \mathbb{R}$ is a constant.

Therefore, our solution is evidently:

$\Psi(x,y) = c$,

or

$x^4/4 + 1/3 (1 + x + y)^3 + k = c$

or absorbing the constant into one constant:

$\frac{x^4}{4} + \frac{1}{3} (1 + x + y)^3 = c$.