# Inverse Laplace transform of $\dfrac{s}{(s + 1)^2 - 4}$

I am trying to find the inverse Laplace transform $\mathcal{L}^{-1} \left\{ \dfrac{s}{(s + 1)^2 - 4} \right\}$. My textbook says that the solution is $e^{-t} \cosh(2t) - \dfrac{1}{2}e^{-t}\sinh(t)$.

But I think this is incorrect. If this is incorrect, can someone please help me find the correct one?

Thank you for any help.

• Just do partial fraction decomposition. $\mathcal{L}^{-1}\{ \frac{1}{s+a}\} = e^{-at}$ for $t>0$. – Jakobian Aug 29 '18 at 22:09
• @Rumpelstiltskin Partial fraction decomposition of $\dfrac{s}{(s + 1)^2 - 4}$ ? – Wyuw Aug 29 '18 at 22:11
• Yes, do partial fraction decomposition of that. – Jakobian Aug 29 '18 at 22:12
• Yes, your book is wrong. Expression written in book is made up of 4 different exponential functions, but our inverse is made out of only 2 – Jakobian Aug 29 '18 at 22:16
• I think it's just a minor typo - the answer you get through partial fraction decomposition is equivalent to $e^{-t} \cosh(2t) - \frac{1}{2} e^{-t} \sinh(2t)$. – Daniel Schepler Aug 29 '18 at 22:16

Hint: $$\frac{s}{(s+1)^2-2^2} = \frac{s}{(s-1)(s+3)} = \frac{1}{4(s-1)}+\frac{3}{4(s+3)},$$ and $$\mathcal{L}(e^{at}) = \frac{1}{s-a}.$$

For the textbook solution, it is useful to write the given fraction as $$\frac{s}{(s+1)^2-2^2} = \frac{s+1-1}{(s+1)^2-2^2} = \frac{s+1}{(s+1)^2-2^2} - \frac{1}{2}\cdot\frac{2}{(s+1)^2-2^2},$$ and note that $$\mathcal{L}(e^{-at}\cosh(bt)) = \frac{s+a}{(s+a)^2-b^2},$$ and $$\mathcal{L}(e^{-at}\sinh(bt)) = \frac{b}{(s+a)^2-b^2}.$$

$$\mathcal{L}^{-1} \left\{ \dfrac{s}{(s + 1)^2 - 4} \right\}$$

$$=\mathcal{L}^{-1} \left\{ \dfrac{s+1}{(s + 1)^2 - 4} \right\}-\mathcal{L}^{-1} \left\{ \dfrac{1}{(s + 1)^2 - 4} \right\}$$

$$=e^{-t} \cosh(2t) - \dfrac{1}{2}e^{-t}\sinh(2t)$$

• Are you sure your last part $\dfrac{1}{2}e^{-t}\sinh(t)$ is correct? – Wyuw Aug 29 '18 at 22:23
• Yes it is correct. The 1/2 takes care of the missing 2 in the numerator – Mohammad Riazi-Kermani Aug 29 '18 at 22:25
• It should be $\sinh(2t)$ instead of $\sinh(t)$. – Math Lover Aug 29 '18 at 22:27
• Oh yes. You are correct. I fixed it in my answer, thanks – Mohammad Riazi-Kermani Aug 29 '18 at 23:28