Second-order Linear Differential Homogeneous Equations
Consider operator
$$ \gdef\mydef{\gdef}\mydef\s#1{\left(#1\right)}\mydef\m#1{\left[#1\right]}\mydef\b#1{\left\{#1\right\}}\mydef\abs#1{\left\lvert#1\right\rvert}L = D^2 + pD + q \tag*{} $$
where $D$ is differential operator
For the case where $P\s x$ and $Q\s x$ are constant functions:
Let $r_1$ and $r_2$ denote the two roots of the polynomial $\lambda^2 + p\lambda + q$.
Case 1: $r_1 \neq r_2$
It is evident that
$$ \ker\s L = \operatorname{span}\b{e^{r_1x}, e^{r_2x}}. \tag*{} $$
In particular, if $r_1, r_2 \in \mathbb C$, let $r_1 = \alpha + i\beta$ and $r_2 = \alpha - i\beta$.
Through Euler's formula:
$$ \begin{aligned} e^{r_1 x} &= e^{\alpha x}\s{\cos\s{\beta x} + i\sin\s{\beta x}}, \\ e^{r_2 x} &= e^{\alpha x}\s{\cos\s{\beta x} - i\sin\s{\beta x}}. \end{aligned} \tag*{} $$
Therefore, $\ker\s L$ can be rewritten to
$$ \ker\s L = \operatorname{span}\b{e^{\alpha x}\cos\s{\beta x}, e^{\alpha x}\sin\s{\beta x}}. \tag*{} $$
Case 2: $r_1 = r_2$
Let $r = r_1 = r_2$.
We know that $\exp\s{rx} \in \ker\s L$.
Assume $u\s x\exp\s{rx} \in \ker\s L$, where $u\s x$ is not a constant.
Let $y = u\s x\exp\s{rx}$.
$$ \begin{aligned} y' &= e^{rx}\s{u'\s x + ru\s x}, \\ y'' &= e^{rx}\s{u''\s x + 2ru'\s x + r^2}. \end{aligned} \tag*{} $$
Substitute back into $L\m y = 0$:
$$ \begin{aligned} e^{rx}\s{\s{u''\s x + 2ru'\s x + r^2} + p\s{u'\s x + ru\s x} + qu\s x} &= 0 \\ e^{rx}\s{u''\s x + \s{2r + p}u'\s x + \s{r^2 + pr + q}u\s x} &= 0. \end{aligned} \tag*{} $$
Note that $r^2 + pr + q = 0$ since $r$ is a root of $\lambda^2 + p\lambda + q$.
Also that $2r + p = 0$, because of Vieta's formula: $r_1 + r_2 + p = 0$, and $r_1 = r_2$.
Therefore
$$ e^{rx}\s{u''\s x} = 0, \tag*{} $$
so $u''\s x = 0$. Pick $u\s x = x$,
$$ \ker\s L = \operatorname{span}\b{e^{rx}, xe^{rx}}. \tag*{} $$
Second-order Non-homogeneous Differential equations
Consider operator
$$ L = D^2 + P\s x D + Q\s x\tag*{} $$
and two equations
$$ L\m y = 0, \tag{$\dagger$} $$
$$ L\m y = f, \tag{$\ddagger$} $$
If we know the general solution to $\s{\dagger}$, that is, $\ker\s L$, let $\ker\s L = \operatorname{span}\b{y_1\s x, y_2\s x}$.
Let $y_p = v_1y_1 + v_2y_2$ be a particular solution to $\s{\ddagger}$.
Assume that $y_1v_1' + y_2v_2' = 0$.
$$ \begin{aligned} y_p' &= v_1y_1' + v_2y_2' + y_1v_1' + y_2v_2' = v_1y_1' + v_2y_2'. \\ y_p'' &= v_1y_1'' + v_1'y_1' + v_2y_2'' + v_2'y_2'. \end{aligned} \tag*{} $$
Substitute back into $\s{\ddagger}$:
$$ \begin{aligned} v_1y_1'' + v_1'y_1' + v_2y_2'' + v_2'y_2' + P\s x\s{v_1y_1' + v_2y_2'} + Q\s x\s{v_1y_1 + v_2y_2} &= f \\ v_1'y_1' + v_2'y_2' + v_1\s{y_1'' + P\s xy_1' + Q\s x y_1} + v_2\s{y_2'' + P\s xy_2' + Q\s x y_2} &= f \\ v_1'y_1' + v_2'y_2' &= f. \end{aligned} \tag*{} $$
Therefore
$$ \begin{bmatrix} 0 \\ f \end{bmatrix} = \begin{bmatrix} y_1v_1' + y_2v_2' \\ y_1'v_1' + y_2'v_2' \end{bmatrix} = \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \begin{bmatrix} v_1' \\ v_2' \end{bmatrix}. \tag*{} $$
And
$$ \begin{bmatrix} v_1' \\ v_2' \end{bmatrix} = \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix}^{-1} \begin{bmatrix} 0 \\ f \end{bmatrix}. \tag*{} $$
Let
$$ W = \det \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} = y_1y_2' - y_1'y_2. \tag*{} $$
$$ \begin{bmatrix} v_1' \\ v_2' \end{bmatrix} = \frac1W \begin{bmatrix} y_2' & -y_2 \\ -y_1' & y_1 \end{bmatrix} \begin{bmatrix} 0 \\ f \end{bmatrix} = \frac1W \begin{bmatrix} -y_2f \\ y_1f \end{bmatrix}. \tag*{} $$
Hence
$$ \begin{aligned} v_1 &= -\frac1{y_1y_2' - y_1'y_2}\int y_2f, \\ v_2 &= \frac1{y_1y_2' - y_1'y_2}\int y_1f, \end{aligned} \tag*{} $$
and
$$ \b{y : L\m y = f} = y_p + \ker\s L. \tag*{} $$