Here is a little table of Laplace transforms 1 1nE t 1a 8 C

Here is a little table of Laplace transforms: {1, 1/n})E^* t, 1/-a + 8} {Cos[a t], E/a^2 + E^2} {Sin [a t], a/a^2 + E^2} {E*^t cos [b t], -a + E/a^2 + b^2 - 2 a E + E^2} {E*^t Sin [b t], b/a^2 + b^2 - 2 a s + E^2} {f\' [t], -f[0] + s Laplace Transform[f[t], t, a]} {f\'[t], -s f[0] + s^2 Laplace Transform [f[t], t, s] - f\'[0]} Given y\"[t] + 4y[t] = 0.5 with y[0] = 2 and y\'[0] = -1, Write down a formula for the Laplace transform of the solution y[t]. Explain the steps that you would go through to solve this differential equation for y[t] by using Laplace transforms. You do NOT need to actually solve it.

Solution

Applying the Laplace transform in the given equation in both sides

L[y^\'\' (t)]+4 L[y(t)] = 0.5 L[1].                                                                                                                               ------(1)

L[y^\'\' (t)] = s² L[y(t),t,s] - s y(0) - y^\' (0) = s² L[y(t),t,s] - 2s + 1, because y(0) = 2 and y^\' (0) = -1.                             ------(2)

L[y(t)]=L[y(t),t,s]                                                                                                                                   ------(3)

Substituting the value of L[y^\'\' (t)] and L[y(t)] from equation (2) and (3) in equation (1), we get

s² L[y(t),t,s] - 2s + 1+ 4 L[y(t),t,s] = 0.5/s, because L[1]= 1/s. Solving for L[y(t),t,s],

(s² +4) L[y(t),t,s] = 1/(2s) +2s -1

L[y(t),t,s] = 1/[2s(s² +4)] +2s/(s² +4) -1/(s² +4)                                                                                                  ------(4)

Applying inverse Laplace trnsformation in equation (4),

y(t) = L^-1[1/{2s(s² +4)}] +2 L^-1[s/(s² +4)] - L^-1[1/(s² +4) ]

y(t) = (1/8)L^-1[1/(s)] - (1/8)L^-1[s/(s² +4)] +2 L^-1[s/(s² +4)] - L^-1[1/(s² +4) ]

y(t) = (1/8)L^-1[1/(s)] + (15/8)L^-1[s/(s² +4)] - (1/2)L^-1[2/(s² +4) ]

y(t) = (1/8)+ (15/8)cos(2t) - (1/2)sin(2t)