[Math] 2nd order linear homogeneous differential equations

Form

Second-order differential equation

$$ y'' + p(x) y' + q(x) y = f(x) $$

Homogeneous

$$ y'' + p(x) y' + q(x) y = 0 $$

General Solution

$$ y(x) = c_1 y_1(x) + c_2 y_2(x) $$

Solving Homogeneous

$$
\begin{split}

y'' + p(x) y' + q(x) y & = 0 \\
\\
\text{let } y & = e^{rx} \\
y' &= r e^{rx} \\
y'' &= r^2 e^{rx} \\
\\
( r^2 e^{rx} ) + p(x) ( r e^{rx} ) + q(x) ( e^{rx} ) &= 0 \\
e^{r x} ( r^2 + p(x) r + q(x) ) &= 0 \\
\\
\text{case } p(x) \text{ and } q(x) \text{ are constant} \\
e^{r x} (r^2 + ar + b) &= 0 \\
r &= \frac{1}{2} \left ( -a \pm \sqrt{a^2 - 4b} \right ) \\
\\
\text{if } a^2 - 4 b > 0 \\
y(x) &= c_1 e^{r_1 x} + c_2 e^{r_2 x} \\
\\
\text{if } a^2 - 4 b = 0 \\
y(x) &= (c_1 + c_2 x) e^{-a x/2} \\
\\
\text{if } a^2 - 4 b < 0 \\
r \text{ has complex roots } \alpha \pm i \beta \\
y(x) &= c_1 e^{\alpha x} \cos(\beta x) + c_2 e^{\alpha x} \sin(\beta x)

\end{split}
$$

Examples

Ex. 1

$$
\begin{split}

y'' - y' - 6 y &= 0 \\
r^2 - r - 6 = (r-3)(r+2) &= 0 \\
y(x) &= c_1 e^{3 x} + c_2 e^{-2 x}

\end{split}
$$

Ex. 2

$$
\begin{split}

y'' + 8 y' + 16y &= 0 \\
r^2 + 8r + 16 = (r+4)^2 &= 0 \\
y(x) &= (c_1+c_2 x)e^{-4 x}

\end{split}
$$

Ex. 3

$$
\begin{split}

y'' + 2 y' + 3y &= 0 \\
r^2 + 2 r + 3 &= 0 \\
r &= -1 \pm \sqrt{2}i \\
y(x) &= c_1 e^{-x} \cos(\sqrt{2}x) + c_2 e^{-x} \sin(\sqrt{2}x)

\end{split}
$$

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