Solving Tricky Laplace Transforms with S-Shifting

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.

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: \( \Laplace\{e^t\sin(2t)\} = \frac{2}{s^2 + 4} \bigg\rvert_{s=s-1} \) \( = \frac{2}{(s-1)^2 + 4} = \frac{2}{s^2 - 2s + 5} \)

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