Laplace Transform with IVP and completing the square

  • Thread starter sami23
  • Start date
  • #1
76
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Homework Statement


Use the Laplace Transform to solve: y"+2y'+2y=t y(0)=y'(0)=1


Homework Equations


L{y(t)} = Y(s)
L{y'(t)} = sY(s)-y(s)
L{y"(t)} = s2Y(s)-sy(0)-y'(0)
using the laplace transform table: tn = n!/(sn+1) where n=1


The Attempt at a Solution


Take laplace on both sides:
L{y"(t)} + 2L{y'(t)} + 2L{y(t)} = L{t}
s2Y(s) - sy(0) - y'(0) + 2[sY(s)-y(0)] + 2Y(s) = 1/s2

after plugging in the initial conditions I get:
s2Y(s) - s - 3 + 2sY(s) + 2Y(s) = 1/s2

isolate Y(s):
Y(s)[s2+2s+2] = 1/s2 + s + 3
Y(s) = (1/s2 + s + 3) / (s2+2s+2)
I completed the square in the denominator to get: (m+1)2+1

Y(s) = s / [(s2+1)(s+3)2-8]

Take the Laplace inverse
L-1{ s / [(s2+1)(s+3)2-8]}

I add and subtract 3 in the numerator to get:
L-1{ (s+3)-3 / [(s2+1)(s+3)2-8]}

Use linearity property of inverse transform to get from L-1{Y(s)} to y(t):
L-1{ (s+3) / [(s2+1)(s+3)2-8]} - 3L-1{ 1 / [(s2+1)(s+3)2-8}

How do I apply partial fraction decomposition to get y(t)?
 
Last edited:

Answers and Replies

  • #2
16
0
Right after you completed the square in the denominator, I would have split it up there to:
Y(s)=1/(s^2(s^2+2s+2)) + (s+3)/(s^2+2s+2) = u+v
There we can see that
v=(s+1)/(s^2+2s+2) +2/(s^2+2s+2) so
L-1{v}=e^(-t)(cos(t)+2sin(t))
Now partial fractions with just u= 1/(s^2(s^2+2s+2)) is easier than what you tried to do.
Set it up as:
u=A/s+B/s^2+(Cs+D)/1/(s^2+2s+2). Now find A,B,C,and D and you'll be home free.
 
  • #3
16
0
Right after you completed the square in the denominator, I would have split it up there to:
Y(s)=1/(s^2(s^2+2s+2)) + (s+3)/(s^2+2s+2) = u+v
There we can see that
v=(s+1)/(s^2+2s+2) +2/(s^2+2s+2) so
L-1{v}=e^(-t)(cos(t)+2sin(t))
Now partial fractions with just u= 1/(s^2(s^2+2s+2)) is easier than what you tried to do.
Set it up as:
u=A/s+B/s^2+(Cs+D)/(s^2+2s+2). Now find A,B,C,and D and you'll be home free.
 
  • #4
76
1
Thank you so much!!! Made my life easier.
 

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