Solve the following Bernoulli's DE $y' + xy^3 + \frac{y}{x} = 0$

Question

Here it is the ODE I want to solve:

$$y'+ xy^{3} + \frac{y}{x} = 0$$

I know that this equation is Bernoulli's equation. I can solve it by substitution $u=y^{-2}$.

But a browser differential equation solver uses substitution $u = xy$, so $y=u/x$.

How can I predict this substitution? I will be appreciated if you can help, thank you!

Answer 1

I believe there is not a traditional way to know what substitution to made. But, in some cases, as it is here, we can seek some interesting factors of the differential equation:

Multiply by x to take the x out of the denominator yields to $$xy' + x²y³+ y = 0$$

Note that the second and third term can be written as $x^{a}y^{a+1}$.

To make it most clear, multiply by x the equation to make the exponent equal, we get: $$x²y' + x³y³+ xy = 0$$

Now, at least to me, seems that the equations begs to a substitution. Anyway, this is not a rigorous answer since is based on guess.

Answer 2

Multiplying by $ x $, the equation becomes $$(xy'+y)+x^2y^3=0$$or $$(xy)'+(xy)^3\frac 1x=0$$

From here, we can think about the substitution $\; u=xy \;$ to get $$\frac{-2u'}{u^3}=\frac 2x=(\frac{1}{u^2})'$$