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

Find $\Laplace^{-1}\{\frac{1}{({\left(- 5 + s\right)}^{2} + 16)} \cdot s\}$

$f(t) = $

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