Determine a valid substitution for a differential equation

The problem with $x^3 y^3$ is that its derivative is rather complicated and unnatural to find in the given problem. Using $y/x$ also doesn't work for the same reason.

But if we take $v = y^{-2}$, its derivative is $-2y^{-3} y'$, which does appear naturally: the stated ODE can be written as

$$y' y^{-3} = y^{-2} x^{-2} + x^3$$

at which point the substitution is pretty quick.

Likewise, $v = y^2$ leads to a simpler (but still non-linear) equation. It's less natural because $2y y'$ doesn't appear obviously... but we could write

$$yy' = \frac{y^2}{x^2} + x^3 y^4 \implies \frac 1 2 v' = \frac 1 {x^2} v + x^3 v^2$$ and dealing with $v^2$ could well be simpler than dealing with $x^3$.

Regardless, $v = y^{1 - 3}$ is the standard substitution for a Bernoulli equation like this.


Hint: Write your equation in the form $$-\frac{2y'(x)}{y(x)^3}+\frac{2}{x^2y(x)^2}=-2x^3$$ and Substitute $$v(x)=\frac{1}{y(x)^2}$$

Tags:

Substitution

Ordinary Differential Equations

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