Question

Consider the initial value problem for function y given by, y'' - 6 y' + 9 y = -3 ?(t-3), y(0) = 0, y'(0) = 0. (a) Find the Laplace Transform of the source function, F(s) = L[-3 ?(t-3)]. F(s) = -3 e^{-3s} (b) Find the Laplace Transform of the solution, Y(s) = L[y(t)]. Y(s) = -3 e^{-3s}/(s-3)^2 (c) Find the solution y(t) of the initial value problem above. y(t) = -3 (t-3) e^{3(t-3)} u(t-3) Recall: If needed, the step function at c is denoted as u(t-c).

          Consider the initial value problem for function y given by,
y'' - 6 y' + 9 y = -3 ?(t-3),    y(0) = 0,    y'(0) = 0.
(a) Find the Laplace Transform of the source function, F(s) = L[-3 ?(t-3)].
F(s) = -3 e^{-3s}
(b) Find the Laplace Transform of the solution, Y(s) = L[y(t)].
Y(s) = -3 e^{-3s}/(s-3)^2
(c) Find the solution y(t) of the initial value problem above.
y(t) = -3 (t-3) e^{3(t-3)} u(t-3)
Recall: If needed, the step function at c is denoted as u(t-c).

Consider the initial value problem for function y given by,
y” - 6 y' + 9 y = -3 ?(t-3), y(0) = 0, y'(0) = 0.
(a) Find the Laplace Transform of the source function, F(s) = L[-3 ?(t-3)].
F(s) = -3 e^-3s
(b) Find the Laplace Transform of the solution, Y(s) = L[y(t)].
Y(s) = -3 e^-3s/(s-3)^2
(c) Find the solution y(t) of the initial value problem above.
y(t) = -3 (t-3) e^3(t-3) u(t-3)
Recall: If needed, the step function at c is denoted as u(t-c).

Added by Matthew P.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

03/16/2023

Step 1

The source function is given by F(t) = -3(t - 3). (a) To find the Laplace Transform of F(t), we can use the formula: $$ F(s) = \mathcal{L}\{-3(t - 3)\} = -3\mathcal{L}\{(t - 3)\} $$ Now, we can use the property of Laplace Transform for Show more…

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