# IVP With Laplace

• Nov 16th 2009, 10:17 PM
Smac
IVP With Laplace
y"+4y'+29y = <delta>(t-2); y(0) = 0 ; y'(0) = 0
.....
Do laplace of both sides and end up with...
e^-2s * inv laplace [ 1 / (s^2 +4s +29) ]
I tried to comp the square, but it didnt seem to work out quite like i needed...

1 / ( (s+2)^2 + 25)

it looks nice bec 25 is a square too, but i dont know what to do if i did that... i could +4 -4 to top, but i still feel like i am stuck? any help (Hi)
• Nov 17th 2009, 12:28 AM
CaptainBlack
Quote:

Originally Posted by Smac
y"+4y'+29y = <delta>(t-2); y(0) = 0 ; y'(0) = 0
.....
Do laplace of both sides and end up with...
e^-2s * inv laplace [ 1 / (s^2 +4s +29) ]
I tried to comp the square, but it didnt seem to work out quite like i needed...

1 / ( (s+2)^2 + 25)

it looks nice bec 25 is a square too, but i dont know what to do if i did that... i could +4 -4 to top, but i still feel like i am stuck? any help (Hi)

You should have:

$Y(s)=\frac{e^{-2s}}{s^2+4s+29}$

Then:

$y(t)=u(t-2) \mathcal{L}^{-1} \left[ \frac{1}{s^2+4s+29} \right]$

CB
• Nov 17th 2009, 03:17 PM
Smac
yeah thats what i have, but when i did the CTS for the inverse laplace, i got stuck and idk where to go from there.
• Nov 17th 2009, 04:20 PM
mr fantastic
Quote:

Originally Posted by Smac
yeah thats what i have, but when i did the CTS for the inverse laplace, i got stuck and idk where to go from there.

Complete the square, use a shift theorem and recognise a standard transform.