Laplace transform IVP 2nd order

  • Thread starter Pi Face
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  • #1
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Homework Statement



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

Homework Equations





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
 

Answers and Replies

  • #2
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[tex]L^{-1}\frac s{(s+b)^2+a^2}=e^{-bx}(\cos ax-\frac ba\sin ax)[/tex]
 
  • #3
vela
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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##.
 

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