A calculus of the absurd

13.4.2 Homogenous second-order ODEs

These are a bit icky, because they’re not solvable in general. Fortunately a lot of them are solvable.

Homogenous (i.e. everything is a function of only one variable) second-order, linear differential equations can be solved without too much difficulty. We can reduce an equation in the form

\begin{equation} \label {2nd order homogenous linear differential equation} a\frac {d^2y}{dx^2} + b\frac {dy}{dx} + cy = 0 \end{equation}

to a quadratic by setting $y=e^{\lambda x} \implies \frac {dy}{dx} = \lambda e ^{\lambda x} \implies \frac {d^2y}{dx^2} = \lambda ^2 e^{\lambda x}$.

From here,

\begin{align*} & a \lambda ^2 e^{\lambda x} + b \lambda e^{\lambda x} + c e^{\lambda x} = 0 \\ & a \lambda ^2 + b \lambda + c = 0 \text { Which is fine as $e^x > 0$} \\ & \lambda = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a} \end{align*}

Here there are a number of possibilities for the value of the discriminant ⁹²⁹² If you’ve no idea what this is, review the earlier section on quadratics..

• $b^2 - 4ac > 0$, which is the "straightforward" case
• $b^2 - 4ac = 0$, in which case there’s only one value of $\lambda $
• $b^2 - 4ac < 0$, in which case we can use complex numbers and trigonometry.

In the case where $\Delta > 0$ ⁹³⁹³ note that $\Delta $ means "the disciminant" we have two possible solutions to the differential equation,

\begin{equation} e^{\frac {-b \pm \sqrt {b^2-4ac}}{2a} x} \label {quadratic e} \end{equation}

In the case where $\Delta = 0$ we have only

\begin{equation*} y(x) = e^{-\frac {b}{2a}x} \end{equation*}

And in the case where $\Delta < 0$ we have the same case as in Equation 13.10, except that there’s a complex part to the root⁹⁴⁹⁴ And, as Euler might point out, $e^{i\theta } = \cos (\theta ) + i\sin (\theta )$