In this case we are going to consider differential equations in the form,

y′ + p ( x ) y = q ( x ) y n

Here p(x) and q(x) are continuous functions in the interval we're working on and n is a real number.  Differential equations in this form are termed as Bernoulli Equations.

First notice that if n = 0 or n = 1 so the equation is linear and we already identify how to resolve it in these cases. Thus, in this case we're going to be considering solutions for values of n other than these two.

In order to resolve these we'll first divide the differential equation via yⁿ to find,

y^-n y' + p(x) y^1-n = q (x)

We are now uses the substitution v = y^1-n to convert this in a differential equation in terms of v.  When we'll see this will cause a differential equation which we can resolve.

We are going to have to be careful along with this though as it comes to dealing along with the derivative, y′.  We require determining just what y′ is in terms of our substitution. It is simple to do than it might at first look to be. All which we require to do is differentiate both sides of our substitution regarding x. Note here that both v and y are functions of x and so we'll require using the chain rule on the right side.  If you keep in mind your Calculus I you'll recall it is just implicit differentiation.  Thus, taking the derivative provides us:

n' = (1 - n) y^-n y'

Then, plugging it and also our substitution in the differential equation provides:

1/(1- n) n' + p(x) n = q(x)

It is a linear differential equation which we can solve for v and once we get this in hand we can also find the solution to the original differential equation through plugging v back in our substitution and solving for y.