Define a solution the given initial-value problem. 4'' + 4y'4y=delta(t - 4) y(0) = 1 y'(0) = 2

chillywilly12a 2020-12-17 Answered
Define a solution the given initial-value problem.
4 + 4y4y=δ(t  4)
y(0)=1
y(0)=2
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Expert Answer

tafzijdeq
Answered 2020-12-18 Author has 92 answers

To solve, we use the properties of the Laplace transform and the Laplace table
The Laplace transform satisfies the linearity properties
L[f=g]=L[f] + L[g]
L[cf]=cL[f]
For all transfotmable functions f and g and constants c.
L[y + 4y + 4y]=L[δ(t  4)]
L[y] + 4L[y] + 4Ly=L[δ(t  4)]
s2Y(s)  sy(0)  y(0) + 4(sY(s)  y(0)) + 4Y(s)=e4s
Y(s)(s2 + 4s + 4)  y(0)  y(0)(s + 4)=e4s
Substitute initial conditions,
y(0)=1
y(0)=2
Solve for Y(s).
Y(s)(s2 + 4s + 4)  2  s  4=e4s
Y(s)=e4s + s + 6(s + 2)2
=e4s(s + 2)2 + 1s + 2 + 4(s + 2)2
Inverse transform:
L1[Y(s)]=L1[e4s(s + 2)2 + 1s + 2 + 4(s + 2)2]
y(t)=L1[e4s 1(s + 2)2] + L1[1s + 2] + 4L1[1(s + 2)2]
=L1[e4sL[te2t]] + e2t + 4te2t
=u4(t)(t  4)e2(t  4) + e2t(1 + 4t)

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