Laplace Transform Initial Value Problem

  • #1
1. y'' -5y' + 6y = 0, y(0) = 1, y'(0) =2






3.
[s^2 F(s) - s f(0) - f'(0)] -5 [F(s) - f(0)] + 6[F(s)] = 0
(s^2 +1)F(s) - (s -5)f(0) - f'(0) = 0 (s^2 + 1)F(s) - (s-5)(1) - 2 = 0
(s^2 + 1)F(s) = s -3 F(s) = (s-3)/(s^2 + 1)
Here's where I'm stuck. I can't factor the denominator to get partial fractions, so I feel like I should just divide up the fraction like this: s/(s^2 + 1) -3/(s^2 + 1) And then use cost - 3sint, but that's not the answer in my book.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
35,653
7,522
1. y'' -5y' + 6y = 0, y(0) = 1, y'(0) =2






3.
[s^2 F(s) - s f(0) - f'(0)] -5 [F(s) - f(0)] + 6[F(s)] = 0
Mistake in the line above. It should be
[s^2 F(s) - s f(0) - f'(0)] -5 [sF(s) - f(0)] + 6[F(s)] = 0
(s^2 +1)F(s) - (s -5)f(0) - f'(0) = 0 (s^2 + 1)F(s) - (s-5)(1) - 2 = 0
(s^2 + 1)F(s) = s -3 F(s) = (s-3)/(s^2 + 1)
Here's where I'm stuck. I can't factor the denominator to get partial fractions, so I feel like I should just divide up the fraction like this: s/(s^2 + 1) -3/(s^2 + 1) And then use cost - 3sint, but that's not the answer in my book.
Your answer is way off - I get y = e2t as the solution.
 

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