Math 325: Differential	Equations    Quiz 2    9/20/96

<<Calculus`LaplaceTransform`

1. Solve the initial value problem and sketch the solution
    y''''[t] + y''[t] = DiracDelta[t - Pi],
    y[0] == 0, y'[0] == 0, y''[0] == 0, y'''[0] == 0

Solution

F[s_] := LaplaceTransform[y[t],t,s];

eqn = LaplaceTransform[y''''[t] + y''[t] == DiracDelta[t-Pi],t,s]

tran = Solve[eqn/.{y[0]->0,y'[0]->0,y''[0]->0,y'''[0]->0}, F[s]]

Apart[1/(s^2(s^2+1))]

soln = InverseLaplaceTransform[F[s]/.First[tran],s,t]

Plot[Evaluate[soln], {t,0,8Pi}]

2. Express the solution to the initial value problem
    y''[t] + 8 y'[t] + 25y[t] == Tan[t],
    y[0] == 0, y'[0] == 0
in terms of a convolution integral.

Solution

F[s_] := LaplaceTransform[y[t],t,s];

eqn = LaplaceTransform[y''[t] + 8 y'[t] + 25y[t] == Tan[t],t,s]

tran = Solve[eqn/.{y[0]->0,y'[0]->0},F[s]]

soln = InverseLaplaceTransform[F[s]/.First[tran],s,t]

We know the inverse Laplace tranform of LaplaceTransform[Tan[t], t, s]/(25 + 8 s + (s^2))is the convolution of Tan[t] and

h[t_] = InverseLaplaceTransform[1/(s^2+8s+25),s,t]

(Note: s^2+8s+25 == (s+4)^2 + 3^2 .) Therefore,

y[t_] = Integrate[Tan[t-u] (1/3)E^(-4u) Sin[3u], {u,0,t}]

Plot[Evaluate[y[t]], {t,0, Pi/4}]