Find the Laplace transform Y(s) = mathcalLy of the solution...

Find the Laplace transform $Y(s) = mathcal{L}{y}$ of the solution of the given initial value problem. $y'' + 4y = egin{cases} 1, & 0 le t < pi \ 0, & pi le t < infty end{cases}$, $y(0) = 3$, $y'(0) = 6$ $Y(s) = $

Transcript

00:01 So we are given from the question that the laplace transform of y of s is equal to the laplace of y is basically we had to find by using initial problem of y double dash plus 4y is equal to the function is varying to 1 when value of t is greater than or equal to or less than 2 pi.

00:26 And else it is 0 for pi less than equals to t less than to infinity.

00:34 So from here y0 is equals to 3 and y dash of 0 is equals to 6.

00:41 These are the values we have been given.

00:44 So first of all we apply the concept of laplace transform of any function f of t.

00:51 This can be represented as integration 0 to infinity e to the power minus st f of t dt.

00:58 So this should become equals to c what we have been given that that is integral of so first of all the function is varying to 0 to pi in that case it should become e to the power minus st dt and for pi to z infinity the function is defined as 0 so that's not need to calculate calculate so this should become equals to minus of 1 by s e to the power minus of st and that limit will go from 0 to pi.

01:32 So this should become equals to minus 1 by s e to the power minus 5 pi minus of e to the power 0 is going to be 1 here.

01:43 So taking laplace transforms transforms on both sides of the given equation.

01:50 So we get laplace transform of y double dash plus 4 laplace transform of y.

01:58 This should be equals to the laplace transform of the function f of t as we have computed above.

02:06 So this can be equals to the s square laplace transform of y y minus s y naught minus of y dash naught plus 4 the laplace transform of y.

02:22 This should be equals to minus 1 by s e to the power minus 5 minus of 1.

02:30 So we just have to do the common the laplace transform of y.

02:35 So it will remain s square plus 4 into the laplace transform of y.

02:42 Then minus of...

Question

Find the Laplace transform $Y(s) = mathcal{L}{y}$ of the solution of the given initial value problem. $y'' + 4y = egin{cases} 1, & 0 le t < pi \ 0, & pi le t < infty end{cases}$, $y(0) = 3$, $y'(0) = 6$ $Y(s) = $

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Find the Laplace transform $Y(s) = mathcal{L}{y}$ of the solution of the given initial value problem. $y'' + 4y = egin{cases} 1, & 0 le t < pi \ 0, & pi le t < infty end{cases}$, $y(0) = 3$, $y'(0) = 6$ $Y(s) = $