This course jumps right in, good luck!
A general linear (no $y^2$ or $y y'$ terms) second order differential equation can be written as
$$ \boxed{p_0 (x) y''(x)+p_1(x)y'(x)+q(x)y(x)=f(x)} $$
where $p_0,p_1,q,f$ are known, they may be complex and $y(x)$ is defined within $a\le x \le b$
We can define the differential operator $\mathcal L$ such that we write the previous expression as
$$ \mathcal L y(x)= f(x) $$
$\mathcal L=p_0(x) \frac{\partial^2}{\partial x^2}+p_1(x)\frac{\partial }{\partial x} +q(x)$
🗒️ Note: These functions can be described as vectors with an inner product in a Hilbert space
$$ \braket{y_1|y_2}=\int^b_a y_1^* (x)y_2(x)\,\text dx $$
1️⃣ Homogenous case
when $f(x)=0$ the ODE becomes $\mathcal L y(x)= 0$ and since it is linear:
if $y_1$ and $y_2$ are solutions to the ODE then so is $Ay_1+By_2$ for any $A,B$ constant
2️⃣ Singular point classification
Dividing our general ODE by $p_0(x)$ we get
$$ y''(x)+P(x)y'(x)+Q(x)y(x)=f(x) $$
📖 Definition: singularity of the point
Ordinary point: If $P(x_0)$ and $Q(x_0)$ are finite at $x_0$ then the point is ordinary
Solutions are analytic and can be Taylor expanded
if $P(x_0)$ and $Q(x_0)$ are not finite at $x_0$ we have 2 options
Regular singular point: if $(x-x_0)P(x)$ and $(x-x_0)^2Q(x)$ are finite at $x_0$
Solutions may be analytic or at worst will have a pole or a branch point
Irregular singular point: if $(x-x_0)P(x)$ and $(x-x_0)^2Q(x)$ are not finite at $x_0$
Solutions cannot be expressed in a simple series expansion