Take the laplace transform of the following initial value...

Take the Laplace transform of the following initial value problem and solve for $Y(s) = \mathcal{L}\{y(t)\}$: $y'' + 6y' + 18y = T(t)$ $y(0) = 0$, $y'(0) = 0$ Where $T(t) = \begin{cases} t, & 0 \le t < 1/2 \\ 1-t, & 1/2 \le t < 1 \end{cases}$, $T(t+1) = T(t)$. $Y(s) = \frac{1}{(1 - e^{-s})(s^2 + 6s + 28)} \left\{ \frac{1}{s^2} - \frac{2e^{-\frac{s}{2}}}{s^2} + \frac{e^{-s}}{s} + \frac{2e^{-\frac{s}{2}}}{s} - \frac{e^{-s}}{s^2} \right\}$ Graph of $T(t)$ (a triangular wave function):

Transcript

00:01 So we have given here the differential equation that is y' ' plus 4y ' plus 19y is equal to t of t.

00:18 And given that we have here that is at 0 y gives us a 0 value and at y ' at 0 gives us here again 0 value.

00:29 So first we need to find here the laplace transformation of t of t.

00:34 So that is equal to then we know that we have a formula 1 upon 1 minus e to the power minus of s integration 0 to 1 e to the power minus st into t of t.

00:52 Multiply with respect to we need to integrate here with respect to dt.

00:56 Now we can write this function 1 upon 1 minus e to the power minus of s and we can write this integration a partial integration 0 to 1 upon 2 and then our function become e to the power minus of st and we know that we can write this t of t plus integration 1 by 2 to 1 e to the power minus of st into t of t with respect to dt.

01:35 So now we can write from here that is 1 upon 1 minus e to the power minus of s and integration 0 to 1 by 2 e to the power minus of st and t and we know that it is given here.

01:55 We have given here when t is greater than to 0 and less than to 1 by 2 our function gives us t value tt value gives us here t value and now we need to integrate this with respect to dt plus integration 1 by 2 to 1 e to the power minus of st and function gives us in this interval 1 minus t e to the power minus st into t with respect to dt.

02:31 Now that is equal to then we know that that is t into e to the power e to the power minus of st upon minus s and we need to put here limit 0 to 1 by 2 minus integration now 0 to 1 by 2 and then it is become e to the power minus of st upon minus of s dt.

03:11 So we are applying here the integration product in which this is our first function this is our second function and this is our first function.

03:21 Now after solving this we will get from here that is after putting the limit we will get minus 1 upon 2s e to the power minus s upon 2 and minus after integrating this function it becomes 1 upon s square that is e to the power minus of st and it becomes from here that is 0 to 1 by 2.

03:55 Now after solving this after putting the limits it becomes minus 1 by 2s e to the power minus of s by 2 and minus 1 upon s square e to the power minus s by 2 plus 1.

04:17 Let's say this is our equation number first now we need to integrate again our second integrating function that is 1 by 2 to 1 e to the power minus of st multiply with 1 minus t with respect to dt...

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