These notes are set so that you get to prove the main results
by solving smaller problems that when put together give the big result.
The answers to the problems are in the videos.
You will get the most out of these notes if you do (or try) the problems
before looking at the videos.
Separable first order differential equations are ones of the form
$$
y'(x) = f(x) g(y(x)).
$$
These will also be written in the form
$$
y' = f(x) f(y)
$$
and the Leibniz notation
$$
\frac{dy}{dx} = f(x) g(y).
$$
Assume that we are on a set where \(g(y)\ne 0\) and that \(f\) and \(g\) are
continuous the function \( f(x) \) and \( \dfrac{1}{g(y)} \) have
anti-derivatives. That is there are functions \(F(x)\) and \(G(y)\)
with
$$
F'(x) = f(x)\qquad \text{and}\qquad G'(y) = \dfrac{1}{g(y)}
$$
Let \(y(x)\) be a function such that
$$
G(y(x)) = F(x) + c
$$
for some constant \(c\). Then \(y(x)\) is a solution to the
differential equation \(y' = f(x) g(y)\). To see this note that
by the chain rule
$$
\frac{d}{dx} G(y(x)) = G'(y(x)) y'(x) = \frac{1}{g(y(x))} y'(x)
$$
and therefore if we take \( \dfrac{d}{dx} \) of the equation
\( G(y(x)) = F(x)\)
$$
\frac{d}{dx} G(y(x)) = \frac{1}{g(y(x))} y'(x) = \frac{d}{dx} (F(x) + c) = F'(x) = f(x).
$$
That is
$$
\frac{1}{g(y(x))} y'(x) = f(x)
$$
and multiplying by \(g(y(x)\) gives \(y'(x) = f(x) g(y(x))\) as required.
Rather than remembering this we can derive the solution as follows.
Write the equation in Leibniz notation
$$
\frac{dy}{dx} = f(x) g(y).
$$
Then formally we can cross multiply to get
$$
\frac{dy}{g(y)} = f(x)\,dx.
$$
Then just formally integrate
$$
\int \frac{dy}{g(y)} =\int f(x)\,dx.
$$
Then the left side of this is an anti-derivative, \(G(y)\) of \( \dfrac{1}{g(y)} \)
and the right side is an anti-derivative, \(F(x)\) of \(f(x)\) so, after including
a constant of integration, we have the solution of the equation.
Problem:
Here is a first example. Find both the general solution and the particular
solution with \(y(1) = 5\) to the equation
$$
y'= (6x^2+ 2x) e^{3y}
$$
Solution:
Problem:
Here is an example where you need to do some algebra before it becomes
clear the equation is separable. Find the general solution
to
$$
\frac{du}{dt} = 1 + t + u^2 + tu^2.
$$
Hint: Factor.