Second-order Linear Differential Homogeneous Equations

Consider operator

$$ L = D^2 + pD + q \tag*{} $$

where $D$ is differential operator

For the case where $P(x)$ and $Q(x)$ 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

$$ \ker(L) = \operatorname{span}\{e^{r_1x}, e^{r_2x}\}. \tag*{} $$

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:

$$ \begin{aligned} e^{r_1 x} &= e^{\alpha x}\bigl(\cos(\beta x) + i\sin(\beta x)\bigr), \\ e^{r_2 x} &= e^{\alpha x}\bigl(\cos(\beta x) - i\sin(\beta x)\bigr). \end{aligned} \tag*{} $$

Therefore, $\ker(L)$ can be rewritten to

$$ \ker(L) = \operatorname{span}\{e^{\alpha x}\cos(\beta x), e^{\alpha x}\sin(\beta x)\}. \tag*{} $$

Case 2: $r_1 = r_2$

Let $r = r_1 = r_2$.

We know that $\exp(rx) \in \ker(L)$.

Assume $u(x)\exp(rx) \in \ker(L)$, where $u(x)$ is not a constant.

Let $y = u(x)\exp(rx)$.

$$ \begin{aligned} y' &= e^{rx}\bigl(u'(x) + ru(x)\bigr), \\ y'' &= e^{rx}\bigl(u''(x) + 2ru'(x) + r^2\bigr). \end{aligned} \tag*{} $$

Substitute back into $L[y] = 0$:

$$ \begin{aligned} e^{rx}\Bigl(\bigl(u''(x) + 2ru'(x) + r^2\bigr) + p\bigl(u'(x) + ru(x)\bigr) + qu(x)\Bigr) &= 0 \\ e^{rx}\bigl(u''(x) + (2r + p)u'(x) + (r^2 + pr + q)u(x)\bigr) &= 0. \end{aligned} \tag*{} $$

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

$$ e^{rx}\bigl(u''(x)\bigr) = 0, \tag*{} $$

so $u''(x) = 0$. Pick $u(x) = x$,

$$ \ker(L) = \operatorname{span}\{e^{rx}, xe^{rx}\}. \tag*{} $$

Second-order Non-homogeneous Differential equations

Consider operator

$$ L = D^2 + P(x) D + Q(x)\tag*{} $$

and two equations

$$ L[y] = 0, \tag{$\dagger$} $$

$$ L[y] = f, \tag{$\ddagger$} $$

If we know the general solution to $(\dagger)$, that is, $\ker(L)$, let $\ker(L) = \operatorname{span}\{y_1(x), y_2(x)\}$.

Let $y_p = v_1y_1 + v_2y_2$ be a particular solution to $(\ddagger)$.

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 $(\ddagger)$:

$$ \begin{aligned} v_1y_1'' + v_1'y_1' + v_2y_2'' + v_2'y_2' + P(x)(v_1y_1' + v_2y_2') + Q(x)(v_1y_1 + v_2y_2) &= f \\ v_1'y_1' + v_2'y_2' + v_1\bigl(y_1'' + P(x)y_1' + Q(x) y_1\bigr) + v_2\bigl(y_2'' + P(x)y_2' + Q(x) y_2\bigr) &= f \\ v_1'y_1' + v_2'y_2' &= f. \end{aligned} \tag*{} $$

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

$$ \{y : L[y] = f\} = y_p + \ker(L). \tag*{} $$