Why is $y'=\frac{x^2}{y^2}$ a Bernoulli equation and not a linear one, but $y'=-\frac{2y}{x}$ is Linear and not Bernoulli? Im going through this book on differential equations:
Differential Equations (Schaum's Outlines) 4th Edition
In chapter 3, Supplementary Problems I have to determine whether an equation is homogeneous and/or linear, and if not linear, whether it is a Bernoulli equation.

Definition of a Linear equation:
$y' + p(x)y=q(x)$
Definition of a Bernoulli equation:
$y' + p(x)y=q(x)y^n$

I understand why:
$y'=\frac{x^2}{y^2}$
is a Bernoulli equation. It cannot be a Linear one because when we rearrange everything we get:
$y' + 0 = x^2 \cdot y^{-2}$
So we get:
$
n=-2 \\
p(x)=0 \\
q(x)=x^2
$
Because we have two $y$'s (although with a - sign in their powers, it doesn't matter), so we cannot formulate a Linear equation.

But here:
$y'=-\frac{2y}{x}$
is not a Bernoulli equation but a Linear one.
I can see why it is a Linear equation, if we rearrange things we get:
$y' + 2\frac{1}{x}\cdot y = 0$
So we get:
$
p(x)=2\frac{1}{x} \\
q(x)=0
$
But why cant it also be a Bernoulli equation with $q(x)=0$?
 A: I am currently taking a Differential Equations class and I think I might be able to help!
$y' = \frac{x^2}{y^2}$ is a Bernoulli's equation since it can be written in the form:
$$
y' + P(x)y = Q(x)y^n \\
\Downarrow \\
y'+ 0 = x^2(y^{-2}) \\
n=-2
$$
**NOTE: In order for an ODE to be considered a Bernoulli's ODE, n can not $= 0$ or $1$.
Since the $y$ is raised to the power $2$ in the ODE, where $2$ is not $1$ or $0$, then this ODE meets the requirements to be a Bernoulli's ODE.
If the $y$ was raised to the power of $1$, I believe it can then be solved as a linear ODE where we get:
$y' - (x^2)(\frac{1}{y})=0$
But it is raised to the power of 2, so it is a Bernoulli's ODE.

$y' = \frac{-2y}{x}$ is a linear ODE because it can be put into the linear ODE general form:
$$
y'+P(x)y = Q(x) \\
\Downarrow \\
y'+(\frac{2}{x})y=0
$$
Since y is raised to the power 1, the ODE does not meet the requirements necessary to be a Bernoulli's ODE as stated above ( y can not be raised to the power of 0 or 1).

Thus...
$y'= \frac{x^2}{y^2}$ is a Bernoulli's ODE and NOT a Linear ODE
&
$y'= \frac{-2y}{x}$ is a Linear ODE and NOT a Bernoulli's ODE
Hopefully, this helps!!
A: The Bernoulli DE is a class of DE with a "trick". The important thing is the trick. And the trick is to substitute $u=y^{1-n}$, resulting in a linear DE in $u$.
This does obviously not work for $n=1$, giving a non-reversible transformation. For $n=0$ it does not make sense, as the term with $y^n$ would not contain $y$.
