0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Just do partial fraction decomposition. $\mathcal{L}^{-1}\{ \frac{1}{s+a}\} = e^{-at}$ for $t>0$. $\endgroup$ – Jakobian Aug 29 '18 at 22:09
  • $\begingroup$ @Rumpelstiltskin Partial fraction decomposition of $\dfrac{s}{(s + 1)^2 - 4}$ ? $\endgroup$ – Wyuw Aug 29 '18 at 22:11
  • 1
    $\begingroup$ Yes, do partial fraction decomposition of that. $\endgroup$ – Jakobian Aug 29 '18 at 22:12
  • 1
    $\begingroup$ 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 $\endgroup$ – Jakobian Aug 29 '18 at 22:16
  • 2
    $\begingroup$ I think it's just a minor typo - the answer you get through partial fraction decomposition is equivalent to $e^{-t} \cosh(2t) - \frac{1}{2} e^{-t} \sinh(2t)$. $\endgroup$ – Daniel Schepler Aug 29 '18 at 22:16
2
$\begingroup$

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 $$\mathcal{L}(e^{-at}\sinh(bt)) = \frac{b}{(s+a)^2-b^2}.$$

$\endgroup$
0
$\begingroup$

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

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.