Consider the initial value problem for function y given...

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

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00:01 In this problem, we are given an initial value problem and we are going to solve it using the method of laplace transforms.

00:08 We have y double prime plus y prime minus 6y equal to 5t cube delta t minus 2.

00:17 With the initial conditions, y0 equal to 0 and y prime 0 equal to 0.

00:25 And we are going to do it in three steps.

00:27 Steps.

00:28 So in the first part we are going to find the laplace transform of the source function.

00:33 F of s equal to laplace transform of 5 t cube direct delta t minus 2.

00:41 Let us use the definition of the laplace transform.

00:46 Integer from 0 to infinity dt exponential minus st times the the given function 5t3, drag delta t minus 2.

00:57 So using this drag delta we are going to set t equal to 2 everywhere.

01:02 So we have 5 times exponential minus 2s times 2, this gives us 40 exponential minus 2s.

01:16 So this is the laplace transform of the source function.

01:22 Next we are going to find the laplace transform of the solution.

01:25 Let us indicate by capital y.

01:29 So for the first derivative we have s y minus y zero where y zero is equal to zero using this relation.

01:40 And for the second derivative we have s squared y minus s y zero minus y prime zero and y prime zero is equal to zero.

01:52 And you already have the laplace transform of the right hand side so let us bring everything together...

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