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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