Find the Laplace inverse of the following.

Question

$$ \frac{2s+5}{s^2+6s+34} $$

I am stuck on this part:

Wolfram has the step by step showing that you simply split up the original fraction into

$$ \frac{2s}{s^2+6s+34} + \frac{5}{s^2+6s+34} $$

and then it solves it. But that doesn't help. Could someone please help me understand how to do this problem?

Answer 1

Use the method of completing the square to see that the denominator is $(s+3)^2+5^2$. Then we have $$\frac{2s+5}{(s+3)^2+5^2}=2\frac{s+3}{(s+3)^2+5^2}-\frac{1}{5}\frac{5}{(s+3)^2+5^2}.$$ These functions are now in a standard form to be inverse-transformed.

Answer 2

Consider the following. \begin{align} \frac{2s+5}{s^2+6s+34} = \frac{2 s + 5}{ (s+3)^{2} + 5^{2}} = 2 \, \frac{(s+3)}{(s+3)^{2} + 5^{2}} - \frac{1}{5} \, \frac{5}{(s+3)^{2} + 5^{2}} \end{align} Now, for \begin{align} f(s) \doteqdot \int_{0}^{\infty} e^{-st } \, f(t) \, dt, \end{align} then \begin{align} \frac{2s+5}{s^2+6s+34} \doteqdot 2 e^{-3t} \, \cos(5 t) - \frac{1}{5} \, e^{-3t} \, \sin(5t). \end{align}