Question

Problem set 4 I. Find the Laplace transform of the function, given the graph: II. Find the Laplace transform of the following : 1. F(t) = 1/3(t^5e^t + 6e^-2t - 9) 2. F(t) = t cos 4t 3. F(t) = e^-3t - e^-3t ?(t - ?) + cos ?(t - ?/2) - cos ?(t - ?) III. Find the inverse of the following: 1. f(s) = (s + 1)e^-8s / (s^2 - 2s + 5) 2. f(s) = 1 / ((s + 1)^3 e^3s) then evaluate F(2) & F(5) 3. f(s) = (s + 2) / (s^4 + s^2) IV. Solve the differential equation (D^2 + 4)y = f(t) f(t) = { t 0 ? t < 3, 3 t ? 3 y(0) = 0 y'(0) = 3 V. Solve the system of d.e.: 2x''(t) = -6x(t) + 2y(t) y''(t) = 2x(t) - 2y(t) + 40sin 3t subject to the initial condition x(0) = x'(0) = y(0) = y'(0) = 0

          Problem set 4

I. Find the Laplace transform of the function, given the graph:

II. Find the Laplace transform of the following :
1. F(t) = 1/3(t^5e^t + 6e^-2t - 9)
2. F(t) = t cos 4t
3. F(t) = e^-3t - e^-3t ?(t - ?) + cos ?(t - ?/2) - cos ?(t - ?)

III. Find the inverse of the following:
1. f(s) = (s + 1)e^-8s / (s^2 - 2s + 5)
2. f(s) = 1 / ((s + 1)^3 e^3s) then evaluate F(2) & F(5)
3. f(s) = (s + 2) / (s^4 + s^2)

IV. Solve the differential equation
(D^2 + 4)y = f(t) f(t) = { t 0 ? t < 3, 3 t ? 3 y(0) = 0 y'(0) = 3

V. Solve the system of d.e.:
2x''(t) = -6x(t) + 2y(t)
y''(t) = 2x(t) - 2y(t) + 40sin 3t
subject to the initial condition x(0) = x'(0) = y(0) = y'(0) = 0

Problem set 4

I. Find the Laplace transform of the function, given the graph:

II. Find the Laplace transform of the following :
1. F(t) = 1/3(t^5e^t + 6e^-2t - 9)
2. F(t) = t cos 4t
3. F(t) = e^-3t - e^-3t ?(t - ?) + cos ?(t - ?/2) - cos ?(t - ?)

III. Find the inverse of the following:
1. f(s) = (s + 1)e^-8s / (s^2 - 2s + 5)
2. f(s) = 1 / ((s + 1)^3 e^3s) then evaluate F(2) F(5)
3. f(s) = (s + 2) / (s^4 + s^2)

IV. Solve the differential equation
(D^2 + 4)y = f(t) f(t) = t 0 ? t < 3, 3 t ? 3 y(0) = 0 y'(0) = 3

V. Solve the system of d.e.:
2x”(t) = -6x(t) + 2y(t)
y”(t) = 2x(t) - 2y(t) + 40sin 3t
subject to the initial condition x(0) = x'(0) = y(0) = y'(0) = 0

Added by Kenneth L.

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