Solving Tricky Laplace Transforms with S-Shifting

fact

If we know the transform of $f(t) \implies F(s)$, then we can use a fancy little trick to get the transform of $e^{at}f(t)$. $$ \Laplace\{e^{at}f(t)\} = F(s-a) $$

This might not look hugely useful, how many times have you had to solve a differential equation with $e^{5t}\sin(2t)$ in it? Well the real value of this trick is in doing the inverse laplace transform, where we take some $F(s)$ and find the $f(t)$ that it came from. This is, in general, very tricky so tools like this are great.

example

Find $\Laplace\{e^t\sin(2t)\}$

We can immediately recognise this as an s-shifting example. So we'll set $g(t) = \sin(2t)$. Now once we find $G(s)$ we replace the $s$'s with $s-1$'s and we'll be done. So since $g(t) = \sin(2t)$ $G(s) = \frac{2}{s^2 + 4}$ Now, finally: $ \Laplace\{e^t\sin(2t)\} = \frac{2}{s^2 + 4} \bigg\rvert_{s=s-1} $ $ = \frac{2}{(s-1)^2 + 4} = \frac{2}{s^2 - 2s + 5} $

example

Find $\Laplace^{-1}\{\frac{1}{(s-1)^3}\}$

Here's an example where this technique shows its value. Expanding this out in partial fractions would be tedious but if we can see that $\Laplace^{-1}\{\frac{1}{s^3}\}$ is simple to solve we'll save ourselves a whole lot of work. So since we know that $\Laplace^{-1}\{\frac{1}{s^3}\} = \frac{1}{2}t^2$ we can see that: $\Laplace^{-1}\{\frac{1}{(s-1)^3}\} = e^t \cdot \frac{1}{2}t^2 = \frac{1}{2}e^t t^2$ Which is much, much simpler than partial fractions.

practice problems