Inverse Laplace transform

1. Sep 12, 2014

Mutaja

1. The problem statement, all variables and given/known data

Find the inverse Laplace transform of the expression:

F(S) = $\frac{3s+5}{s^2 +9}$

2. Relevant equations

3. The attempt at a solution

From general Laplace transforms, I see a pattern with laplace transforming sin(t) and cos(t) because:

L{sin(t)+cos(t)} = $\frac{s+1}{s^2 +1}$

All I'm missing here is a couple of constants(?).

I know that the laplace transform works like this:

L{Asin(Bt)+Ccos(Dt)} = $\frac{Cs}{s^2 +D}$ + $\frac{A*B}{s^2 +B^2}$

Looking at my original problem, I can see that I need B to equal D, $D^2$ + $B^2$ = 9, C = 3 and A*B = 5.

If I set A = 5, B = 3, C = 9 and D = 3 I get $3^2$ + $3^2$ = 18, 5
3 = 15 (putting numbers into my equations above).

Therefore I need to divide the whole expression by 3 to get my correct answer:

L{$\frac{1}{3}($5sin(3t)+9cos(3t))} = $\frac{3s+5}{s^2 +9}$

Is there a simpler way to do this? Or do I just have to use the trial and error method until I find the correct factors/constants?

Any input on this will be greatly appreciated. Thanks

2. Sep 12, 2014

HallsofIvy

What? One of the equations above said A*B= 5 which is NOT satisfied by A= 5, B= 3. You have only three equations for the four values A, B, C, and D. And there is no reason to assume that A, B, C, and D must be integers.

Use the fact that the Laplace transform, and so its inverse, is linear.
$$\frac{3s+ 5}{s^2+ 9}= 3\frac{s}{s^2+ 9}+ 5\frac{1}{s^2+ 9}$$
and look up their inverse transforms separately.

The laplace transform of $cos(\omega x)$ is $\frac{s}{s^2+ \omega^2}$ so the inverse transform of $3\frac{s}{s^2+ 9}$ is $3 cos(3x)$

The Laplace transform of $sin(\omega x)$ is $\frac{\omega}{s^2+ \omega^2}$ so the inverse transform of $5\frac{1}{s^2+ 9}= (5/3)\frac{3}{s^2+ 9}$ is $(5/3)sin(3x)$.