# How to Calculate the Inverse Laplace Transform

concept

To use the Laplace transform to solve differential equations we turn the differential equation into an algebraic equation with the transform. Then after some algebra we turn it back into a function of time by computing an inverse Laplace transform.
So to get a final answer as a function of time (which is what you want) you'll need to convert some function of \(s\) back into a function of \(t\).
We make heavy use of the Laplace transform table in this process. But more than that, we often need to do some clever algebra on our \(F(s)\) to get it into a form (or a sum of forms) that is in the table. This is where the real test is in the Laplace transform method.

fact

The inverse Laplace transform, written \(\Laplace^{-1}\{F(s)\}\) is defined as:
$$ \Laplace^{-1}\{F(s)\} = f(t) $$
So:
$$ \Laplace^{-1}\{\Laplace\{f(t)\}\} = f(t) $$

fact

To find the inverse Laplace transform of some function \(F(s)\) we find it in the transform table.

A good strategy is to match your function's denominator to the correct denominator in the table. You can always re-arrange things to get the right numerator later on.

example

Find \(\Laplace^{-1}\{\frac{5}{s}\}\)

Looking at our table we can see that a denominator of \(s\) matches up with \(\Laplace\{1\} = \frac{1}{s}\) so it looks like \(f(t) = 5\) looks like a good fit. We can check by indeed seeing that \(\Laplace\{5\} = \frac{5}{s}\) Just as we wanted.fact

The inverse Laplace transform is linear just like the forward transform so:
$$ \Laplace^{-1}\{AF(s) + BG(s)\} = Af(t) + Bg(t) $$

example

Find the inverse Laplace transform of \(\frac{s}{s^2 + 25} + \frac{3}{s-2}\)

We can do the two terms separately, which is nice. So first off let's find the inverse transform of \(\frac{s}{s^2 + 25}\). We're in luck since this matches up very well with \(\Laplace\{\cos(at)\} = \frac{s}{s^2 + a^2}\). So \(\cos(5t)\) looks like it fits the bill perfectly for the first term. Now the second term. That \(s-2\) denominator matches with an exponential term. It looks like \(\Laplace\{3e^{2t}\} = \frac{3}{s-2}\) So our final solution is: \(\cos(5t) + 3e^{2t}\)
practice problems