Laplace transforms and ODEs

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 $2y^{″}+5y^{\prime }+3y=0$ 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 $y=A{e}^{-3x/2}+B{e}^{-x}$. I’ll stop waving now.

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

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 $2y^{″}+2y^{\prime }+3y^{\prime }+3y=0$, so let $z=y^{\prime }+y$” 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 $2y^{″}+5y^{\prime }+3y=0$ is $2s^{2}Y(s)+5sY(s)+3Y(s)-(2s+5)y(0)-2y^{\prime }(0)=0$.

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!