# How to solve this Inverse Laplace Transform $\frac{s}{s^2-s+\frac{17}{4}}$

How would I solve this Inverse Laplace transform?

$$\mathscr{L}_s^{-1} \left\{ \frac{s}{s^2-s+\frac{17}{4}} \right\}$$

The solution is $$f(t) = (1/4 )e^{t/2} (\sin(2 t)+4 \cos(2 t))$$

I know I need to break up $F(s)$ into more common Laplace transforms, but I'm not quite sure how to begin.

• Partial fractions, completing the square, or both. Commented Apr 28, 2015 at 2:06
• Use the derivative property of Laplace transform to get the final result after partial fractions Commented Apr 28, 2015 at 2:09
• Like Chappers said, complete the square in the numerator. You get $$\frac{s}{(s - \frac{1}{2})^{2} + 4}$$ which hopefully looks familiar to you. Commented Apr 28, 2015 at 2:16
• That was it! I can't believe I didn't see it. Thank you.
– WHY
Commented Apr 28, 2015 at 2:49
• @Mattos, do you mean "denominator"?
– JRN
Commented Apr 28, 2015 at 2:54

\begin{align} F(s)&=\frac{s}{(s - \frac{1}{2})^{2} + 4}\\ &=\frac{\frac{1}{2}}{(s - \frac{1}{2})^{2} + 4} + \frac{s-\frac{1}{2}}{(s - \frac{1}{2})^{2}+ 4}\\ \\ f(t) &= (1/4)e^{t/2} (\sin(2 t) + 4\cos(2 t)) \end{align}