# Laplace transform IVP 2nd order

1. Feb 16, 2012

### Pi Face

1. The problem statement, all variables and given/known data

y''+6'+10y=0
y(0)=2
y'(0)=1

2. Relevant equations

3. The attempt at a solution

Laplace everything and I get
s^2*Y(s)-2s-1+6s*Y(s)-12+10Y(s)=0

isolate Y(s)
Y(s)=(2s+13)/(s^2+6s+10)

split into 2 terms, bottom can be rearranged by completing the square

2s/[(s+3)^2+1^1] + 13/[(s+3)^2+1^1]

inverse laplace

y(t)=2e^-3t*cost + 13e^-3t*sint

both wolfram and the answer key say the last term should be 7 instead of 13, but i don't see where I made a mistake

2. Feb 16, 2012

### Some Pig

$$L^{-1}\frac s{(s+b)^2+a^2}=e^{-bx}(\cos ax-\frac ba\sin ax)$$

3. Feb 16, 2012

### vela

Staff Emeritus
The Laplace transform of $\cos t$ is $\frac{s}{s^2+1}$, so when you replace s by s+3, you get $\frac{s+3}{(s+3)^2+1}$. Compare that to what you said was the Laplace transform of $e^{-3t}\cos t$.