Determine a valid substitution for a differential equation I am trying to determine whether any of the $v(x)$ substitutions that I'm given are are possible to make the equation first order linear in terms of $v$.
$$y' = \frac{y}{x^2} + x^3y^3$$
The given possible substitutions are
$$v(x)=x^3y^3
\\v(x)=y^2
\\v(x)=y^{-2}
\\v(x)=y/x$$
I don't know how to infer the answer by looking at it. So I went with a brute force approach.
Solving for y in the first three yields $y = \frac{v^{1/3}}{x}$, $y=\sqrt{v}$, and $y=v^{-1/2}$ respectively. I went with the logic that this would not satisfy the linearity of the problem since $v$ in each case is nonlinear.
Is the last one also false because it would make the equation
$$vx = y
\\v'x + v = y'
\\v'x + v = \frac{v}{x} + x^7v^3$$
Thus keeping the equation in a nonlinear form.
This is just my thinking on the matter. I want to make sure it is correct.
 A: 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.
A: Hint: Write your equation in the form $$-\frac{2y'(x)}{y(x)^3}+\frac{2}{x^2y(x)^2}=-2x^3$$ and Substitute $$v(x)=\frac{1}{y(x)^2}$$
A: More generally, consider the substitution $v = x^p y^q$ (with $q \ne 0$).  We have $$v' = p x^{p-1} y^q + q x^p y^{q-1} y'$$
If this satisfies a first-order linear equation
$$ v' = a(x) v + b(x) $$
that becomes 
$$ p x^{p-1} y^q + q x^p y^{q-1} y'= a(x) x^p y^q + b(x) $$
or
$$ y' = \frac{a(x)x - p}{q x} y  + \frac{b(x)}{q x^p} y^{1-q} $$ 
This can only match with your original equation in the case $q=-2$.
