Say, following on from that thing about Laplace transforms, it reminded me that I typically wave my hands at the ODE characteristic equation trick.

You know the one. You have something like $2y’’ + 5y’ + 3y = 0$ and you magically turn it into $2\lambda^2 + 5\lambda + 3=0$, factorise it as $(2\lambda + 3)(\lambda + 1)=0$ and deduce that your complementary function is $y=Ae^{-3x/2 } + Be^{-x}$. I’ll stop waving now.

However, I will throw some initial conditions at it: let’s say that when $x=0$, $y=0$ and $y’ = 1$.

That means $A + B = 0$ and $-\frac{3}{2}A-B = 1$. That leads to $A=-2$ and $B = 2$.

Usually I justify the handwaves either in an Ansatzy sort of way (“let’s assume solutions of the form…”) or via something like “we can write it as $2y’’ + 2y’ + 3y’ + 3y=0$, so let $z = y’ + y$” and it all integrates out nicely. Both of those are fine. But the Laplace transform rang a bell and I started salivating.

The Laplace transform of $2y’’ + 5y’ + 3y=0$ is $2s^2 Y(s) + 5s Y(s) + 3Y(s) - (2s+5)y(0) - 2y’(0)=0$.

I’m going to throw the initial conditions in here, and tidy up because it’s getting messy:

$Y(s)(2s^2 + 5s + 3) = 2 Rearranging gives$Y(s) = \frac{2}{(2s+3)(s+1)}$. This partial-fractions nicely as$Y(s) = \frac{2}{s+1} - \frac{4}{2s+3}$. In turn, this inverts to$2e^{-x} - 2e^{-3x/2}\$, just as before. There’s no simultaneous equation to solve, but there is a partial fraction, so it’s swings and roundabouts – but this method should also be able to handle a non-zero right-hand-side without needing to come up with a template solution. I like it!