## Laplace Transform Example

Applying the Laplace transform operator to both sides:

And now for the inverse Laplace transform:

The second term is easy:

The first term requires partial fraction decomposition or completing the square and involving complex numbers.  I am going to show the latter.

I am doing this with the Laplace transform for  $\sin(bt)$ , which is  $\frac{b}{s^2+b^2}$.  In this case  $b=\frac{3i}{2}$.

Since we have  $s-\frac{5}{2}$  instead of  $s$ , we get a multiple of  $e^{\frac{5}{2}t}$  in the inverse transform.

One amazing fact about   $\sin(ix)$   is that it is equal to   $i\cdot\sinh(x)$, the hyperbolic sine, which is equal to $\frac{i}{2}\left(e^x-e^{-x}\right)$.

Hence

which equals

Crazy!  Now add on the $e^{4t}$ and we get

Partial fractions will lead to the same solution and is actually a bit easier.  However, this was a wild ride, and well worth the trouble in my opinion.

If you don't have much time to screw around with this, then you could also look up $\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}\right\}$ in a table of Laplace transforms.

### One Response to “Laplace Transform Example”

1. gary says:

Interesting
$e^{i\theta}=\cos(\theta)+i\sin(\theta)$ , so, $e^{i(ix)}=\cos(ix)+i\sin(ix)$, but $e^{i(ix)}=e^{-x}$
$$e^{-x}=\cos(ix)+i\sin(ix)$$
and
$e^{x}=\cos(-ix)+i\sin(-ix)$
so
$$e^{x}=\cos(ix)-i\sin(ix)$$

If we subtract equation $(1)$ from equation $(2)$ we get

Solving for $\sin(ix)$ we get

Beautiful.