# Thread: Inverse laplace transform, help needed!

1. ## Inverse laplace transform, help needed!

Hi. I'm trying to find the inverse laplace transform of the following:

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

Problem is - I'm not sure where to begin. Should I complete the square for
$(s^2 +4s+5)$? And if so, where do I go next from that?

Any help would be greatly appreciated.

2. Start by separating the fraction into partial fractions.

3. Originally Posted by whipflip15
Start by separating the fraction into partial fractions.
Thanks for replying. I tried that earlier, but was unsure about something:

$
(s+2) = A(s^2+4s+5) + B(s+1)
$

Setting $s=-1$ produces $A=\frac{1}{2}$ ,

...but what about $B$? The solution's complex, so how do I deal with that?

4. Originally Posted by dan569
Thanks for replying. I tried that earlier, but was unsure about something:

$
(s+2) = A(s^2+4s+5) + B(s+1)
$

Setting $s=-1$ produces $A=\frac{1}{2}$ ,

...but what about $B$? The solution's complex, so how do I deal with that?
Since the quadratic factor is irreducible over real numbers, the correct partial fraction form to use is $\frac{A}{s+1} + \frac{Bs + C}{s^2 + 4s + 5}$.

When getting the inverse Laplace Transform of the second term you will need to complete the square.

5. Thanks, I think I've got it.

$y(t) = \frac{1}{2}[e^{-t} + (sin(t) - cos(t)).e^{-2t}]$

6. Apologies. The actual wording of the question is:

...find the inverse laplace transform of $\frac{(s+2)}{(s+1)(s^2 +4s+5)}$

Hence, show that it is equal to $\int_0^t e^{-(t-u)}.e^{-2u}.cos(u) du\$

7. Originally Posted by dan569
Apologies. The actual wording of the question is:

...find the inverse laplace transform of $\frac{(s+2)}{(s+1)(s^2 +4s+5)}$

Hence, show that it is equal to $\int_0^t e^{-(t-u)}.e^{-2u}.cos(u) du\$
$LT^{-1}\left[ \frac{(s+2)}{(s+1)(s^2 +4s+5)}\right] = LT^{-1}\left[ \frac{1}{s+1} \cdot \frac{s+2}{(s+2)^2 + 1}\right]$

$= LT^{-1}\left[ \frac{1}{s+1}\right] * LT^{-1} \left[\frac{s+2}{(s+2)^2 + 1}\right]$

$= e^{-t} * ( e^{-2t} \cos t )$

where the second and third lines follow from the usual theorems and standard results.

8. Originally Posted by dan569
Apologies. The actual wording of the question is:

...find the inverse laplace transform of $\frac{(s+2)}{(s+1)(s^2 +4s+5)}$

Hence, show that it is equal to $\int_0^t e^{-(t-u)}.e^{-2u}.cos(u) du\$
Or perhaps all you're expected to do is integrate $\int_0^t e^{-(t-u)}.e^{-2u}.cos(u) du\$ and show that it's equal to $\frac{1}{2}[e^{-t} + (sin(t) - cos(t)).e^{-2t}]$ ....?