Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I approached this problem as follow:

  1. Complete the square in the denominator and obtain $$(s-2)^2 + 1$$

  2. Now break the function into 2 parts.

$$ \frac{2s}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1}$$

Now the inverse laplace transform is straight forward.

$$e^{2t}(2cos(t) + sin(t))$$

However! I am wrong. The solution to this problem is actually

$$e^{2t}(2cos(t) + 5*sin(t))$$

What is wrong with my approach?


share|cite|improve this question
up vote 2 down vote accepted

Remember, the Laplace transform tables have the following kinds of terms

$$ \frac{s-a}{(s-a)^2+b^2},\frac{b}{(s-a)^2+b^2} $$ Your term with an $s$ in the numerator has to look like $s-2$ in order to use that form. So your problem is at this step:

$$ \frac{2s}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1} = \frac{2(s-2)+4}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1} = 2\frac{(s-2)}{(s-2)^2 + 1}+ 5\frac{1}{(s-2)^2 + 1} $$

Basically, you have to have the same form for $s$ in the top and bottom, so you add and subtract $4$ from the numerator of that first term so that you can put it in terms of $(s-2)$. Then the extra $4$ you added goes over to the other term to make a total of $5$, the coefficient of the $\sin$

share|cite|improve this answer
Thank you so much. – 40Plot May 1 '13 at 19:45

$s^2-4 s+5$ has zeroes at $s_{\pm} = 2 \pm i$. The ILT $f(t)$ is simply the sum of the residues of

$$\frac{2 s+1}{s^2-4 s+5} e^{s t}$$

at these poles. Then

$$f(t) = \frac{2 s_+ + 1}{2 s_+-4} e^{s_+ t} + \frac{2 s_- + 1}{2 s_--4} e^{s_- t}$$

Expanding this a bit:

$$f(t) = e^{2 t} \left [ \left (1-i \frac52 \right ) (\cos{t}+i \sin{t}) + \left(1+i \frac52 \right ) (\cos{t}-i \sin{t}) \right ]$$

Simplifying, I get

$$f(t) = e^{2 t} (2 \cos{t} + 5 \sin{t})$$

share|cite|improve this answer

I think if you do the fraction as follows:

$$\frac{2s+1}{(s-2)^2 + 1}= \frac{2(s-2)+5}{(s-2)^2 + 1}=\frac{2(s-2)}{(s-2)^2 + 1}+\frac{5}{(s-2)^2 + 1}$$ and considering very useful fact:


then we get:

$$\mathcal{L}^{-1}\left(\frac{2s+1}{(s-2)^2 + 1}\right)=2\mathcal{L}^{-1}\left(\frac{(s-2)}{(s-2)^2 + 1}\right)+5\mathcal{L}^{-1}\left(\frac{1}{(s-2)^2 + 1}\right)=2e^{2t}\cos t+5e^{2t}\sin t$$

share|cite|improve this answer
You're on a roll today! +1 – amWhy Aug 23 '13 at 11:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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