The solution of initial value problem y'' + 2y' + 2y=delta(t - pi), y(0)=0, y'(0)=1 and draw the graph of the solution.

generals336 2021-01-04 Answered
The solution of initial value problem y + 2y + 2y=δ(t  π), y(0)=0, y(0)=1 and draw the graph of the solution.
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Expert Answer

Brighton
Answered 2021-01-05 Author has 103 answers

Let the differential equation y + 2y + 2y=δ(t  π)
Applying the Laplace transform to the differential equation,
Ly + 2y + 2y=Lδ(t  π)
Ly+2Ly+2Ly=Lδ(t  π)
By using Lfn(t)=snF(s)  sn  1f(0)    fn  1(0) and
Lδ(t  t0)=est0
 [s2Y(s)  sy(0)  y(0)] + 2[sY(s)  y(0)] + 2Y(s)=eπ s
Using the initial conditios y(0)=0 and y(0)=1,
[s2Y(s)  1] + 2sY(s)=eπ s
 Y(s)[s2 + 2s + 2]  1=eπ s
 Y(s)=1 + eπ ss2 + 2s + 2
 Y(s)=1(s + 1)2 + 1 + eπ s(s + 1) + 1
Now, applying the inverse Laplace transformation on both sides,
 L1Y(s)=L1{1(s + 1)2 + 1 + eπ s(s + 1)2 + 1}
 L1Y(s)=L1{1(s + 1)2 + 1} + L1{eπ s(s + 1)2 + 1}
By using L1ecsG(s)=uc(t)g(t  c) and properties of inverse Laplace,
 y(t)=e1sin t + uπ(t)e(t  π)sin(t  π)
Therefore, the solution of the initial value problem
y + 2y + 2y=δ(t  π), y(0)=0, y(0)=1 is
y(t)=etsin t + uπ(t)e(t  n)sin(t  π).
By using

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