# Find inverse laplace transform of $f(s) = \frac{1}{(s-2)^2+9}$

Find inverse laplace transform of $$f(s) = \frac{1}{(s-2)^2+9}$$

Here is what I have gotten from partial fractions. I observed that $$s^2-4s + 13$$ is irreducible (doesn't have real roots). $$\frac{1}{(s-2)^2 + 9} = \frac{1}{s^2-4s + 13} = \frac{As + B}{s^2-4s + 13}$$A=0, B=1

The corresponding value in my table is $$e^{at}\sin(bt)$$ for the corresponding laplace transform $$\frac{b}{(s-a)^2 + b^2}$$.

so plugging in A and B, I get $$e^{0t}\sin(1\cdot t) = \sin(t)$$. However my textbook answer is $$\frac{1}{3}e^{2t}\sin{3t}$$.

I am not sure how to trouble shoot this problem. I am pretty sure I did the partial fractions correctly, but the textbook answer implies that my approach to finding the inverse laplace transform is very wrong because I have no idea where the $$\frac{1}{3}$$ comes from in the solution.

Remember that

$$\color{blue}{\mathcal{L}^{-1}\{F(s-a)\} = e^{at} \mathcal{L}^{-1} \{F(s)\}}$$ and

$$\color{blue}{\mathcal{L}^{-1}\left\{\frac{b}{s^2+b^2}\right\} = \sin(bt)}$$

So,

$$\mathcal{L}^{-1}\left\{\frac{1}{(s-2)^2+9}\right\} = \frac13\mathcal{L}^{-1}\left\{\frac{3}{(s-2)^2+3^2}\right\} = \frac13e^{2t}\sin(3t)$$

• I don't recall what you're saying to remember. Or at least I don't recognize it in its entirety Apr 10 '20 at 17:37
• Do you remember this formula $\mathcal{L}^{-1}\{F(s-a)\} = e^{at} \mathcal{L}^{-1} \{F(s)\}$? Apr 10 '20 at 17:38

$$F(s)=\frac{1}{(s-2)^2 + 3^2} \implies 3F(s)=\frac{3}{(s-2)^2 + 3^2}$$ Substitute $$s-2=s'$$: $$3F(s'+2)=\frac{3}{(s')^2 + 3^2}$$ $$3\mathcal{L^{-1}}F(s'+2)=\mathcal{L^{-1}} \left (\frac{3}{(s')^2 + 3^2} \right )$$ $$3\mathcal{L^{-1}}F(s'+2)=\sin (3t)$$ Finally $$\mathcal {L^{-1}}(F(s+a))=e^{-at}f(t):$$ $$3e^{-2t}f(t)=\sin (3t) \implies f(t)= \dfrac {e^{2t}}3\sin (3t)$$

You've used the values of $$A,\,B$$ where you should use the values of $$a,\,b$$. Use $$\mathcal{L}^{-1}\frac{1}{(s-a)^2+b^2}=\frac1be^{at}\sin bt$$ with $$a=2,\,b=3$$.