ODE with Laplace transform

  • Thread starter bosox09
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Homework Statement



The solution to the ODE y''(t) + 4y(t) = 1 + u(t − 2), y(0) = 0, y'(0) = 0 is given by...

The Attempt at a Solution



OK well I figured this one is good to solve with Laplace transforms. So I took the Laplace of both sides to obtain (s2 + 4)Y(s) = [e-2s/s] + 1/s, which equals (e-2s + 1)/s. Isolating Y(s) gave me (e-2s + 1)/s(s2 + 4). I used partial fraction expansion to obtain (1/4) - (1/4)cos2t, but this is apparently only half of the whole answer, given as (1/4)(1 − cos 2t) + (1/4)(1 − cos 2(t − 2))u(t − 2). What am I missing?
 

Answers and Replies

  • #2
HallsofIvy
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How are you expanding in partial fractions? What expansion did you get? In particular shat happened to the [itex]e^{-2s}[/itex]? It's very easy to see that [tex]y= \frac{1}{4}(1- cos(2t))[/tex] satisfies y"+ 4y= 1, y(0)= 0, y'(0)= 0, not the equation you have.
 

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