Laplace Transform

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

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

Related Threads on Laplace Transform

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
504
  • Last Post
Replies
1
Views
741
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
972
Top