The solution of the initial value problem y'' + 4y=2delta(t - pi/4), y(0)=0, y'(0)=0 and draw the graph of the solution.

Josalynn 2020-11-23 Answered
The solution of the initial value problem y + 4y=2δ(t  π/4), y(0)=0, y(0)=0 and draw the graph of the solution.
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Expert Answer

Delorenzoz
Answered 2020-11-24 Author has 91 answers

Let the differential equation y + 4y=2δ(t  4π).
Applying the Laplace transform to the differential equation,
Ly+4y=L2δ(t  π/4)
Ly+4Ly=2Lδ(t  π/4)
By using Lfn(t)=snF(s)  sn  1f(0)    fn  1(0) and
Lδ(t  t0)=est0
[s2Y(s)  sy(0)  y(0)] + 4Y(s)=2e(π/4)s
Using the initial conditios y(0)=0 and y(0)=1,
s2Y(s) + 4Y(s)=2e(π/4)s
 Y(s)[s2 + 4]=2e(π/4)s
 Y(s)=2e(π/4)ss2 + 4
Now, applying the inverse Laplace transformation on both sides,
 L1Y(s)=L1{2e(π/4)ss2 + 4}
By using L1ecsG(s)=uc(t)g(t  c) and properties of inverse Laplace,
 y(t)=uπ/4(t)sin(2(t  π/4))
Therefore, the solution of the initial value problem
y + 4y=2δ(t  π/4), y(0)=0, y(0)=0 is
y(t)=uπ/4(t)sin(2(t  π/4))
By using uπ/4={1t  π/40t < π/4
For t < π/4, then y(t)=0
For t  π/4, then y(t)=sin(2(t  π/4))
Thuse, the graph of the solution y(t)=uπ/4(t)sin(2(t  π/4) for t < π/4 and t  π/4 is as below:
image

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