First-order non-linear ordinary differential equation I know how to solve the Bernoulli differential equation.
How to solve the following first-order non-linear ordinary differential equation.
$y' + p(x)y + q(x) = y^2 r(x)$.
 A: Substitutions or transformations can often be used to simplify such equations.
We have
$$y' + p(x)y + q(x) = y^2 r(x) $$
For example we can do :
$$y = \frac{z'}{z\cdot r(x)}$$
And then solve for $z$ instead. Hmm, maybe there is some sign error, but the idea is fruitful.
A: I think there is a sign mistake in your question (Even if there isn't, this answer will lead you to the concept). I will solve with the equation
$$y'=r(x)y^2+p(x)y+q(x)....(1)$$
Now this is a Riccati's equation so there is a specific method to solve this (Just some substitutions)
Now consider $$y=y_1+\frac{1}{v}....(2)$$ where $y_1$ is a particular solution for this equation and $v$ is nothing, its just a function of $x$ i.e. $v(x)$
Also from now on in this solution, I wont write $r(x)$ or $p(x)$, I will just simple write them as $r$ and $p$.


Now differentiating the  2 equation above we get-
$$y'=y_1'-\frac{1}{v^2}v'....(3)$$
Now at this point you may be confused what's happening but this is nothing more than a substitution.
So by substituting values in equation 1 we get-
$$ry^2+py+q=ry_1^2+py_1+q-\frac{1}{v^2}v'....(4)$$
Now replace $y$ by equation 2.
$$r\left(y_1+\frac{1}{v}\right)^2+p\left(y_1+\frac{1}{v}\right)+q=ry_1^2+py_1+q-\frac{1}{v^2}v'....(5)$$
Now after simplifying the 5th equation you will end up with something like-
$$v'+v(2ry_1+p)=-r....(6)$$
Now this is a standard linear differential equation and you can simply assume $2ry_1+p=Q$ and proceed with the solution.

Also for further knowledge I will suggest some good articles and videos for you-

*

*Given by Alex

*A video

*Another video
