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Determine the Laplace transform of the given function.$$f(t)=\left\{\begin{array}{ll}2 t / \pi, & 0 \leq t<\pi / 2 \\\sin t, & \pi / 2 \leq t<\pi\end{array}\right.$$ where $f(t+\pi)=f(t)$.

$\frac{\left[e^{-\pi / 28}(s+1)\right]}{s^{2}}+\frac{e^{-\pi 8}}{s^{2}+1}+\frac{8 e^{-}}{s^{2}+1}$

Calculus 3

Chapter 10

The Laplace Transform and Some Elementary Applications

Section 3

Periodic Functions and the Laplace Transform

Second-Order Differential Equations

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Okay, so we have this function that's defined piece wise from zero to pi over two and then from pi over to two pi And it's also periodic in pie. So we've got that Big T is equal to pi so we want to figure out the LaPlace transform. So we use theorem 10.3 point three in the book. This gives us that it is 1/1 minus e to the minus see into from zero. So, um okay, so into So this is big t so big tease pie now. So it's that zero to pi of e to the minus s t times f of t d t. Of course, we'll have to split this integral up from going from zero to pi over two in them from pirate to two pi. So let's call this I So I is equal to zero to pi over two e to the minus esti times fft fft is to t over pie and then into grown from pi over to two pi of e to the minus s t times fft But fft now it's sine of t So this is okay now we've got to integrate this. So let me call this guy. I won. And then we call this I tune. Let's figure out what I want and I two r so as an aside, I won. We're going to do this by integration by parts. So we're gonna differentiate the tea and integrate the exponential. So that's gonna leave us with this for the boundary term, right? I So it's you Devi. So now this is U V. The minus integral. So v. So the integral of this is gonna be is going to give us plus one over us E to the minus esti. And then the derivative. This would just be one. And so what do we get? So let me swap the limits around to get rid of this minus sign. So we're going to get minus one over us times by pi over two times by e to the minus pi over two times s and thats plus one over us. Okay? And so that's minus pi over two. Yes, e to the minus pious, over to And then stop the limits here again just to get rid of the minus sign. That's plus one over r squared into one minus E to the minus pi t over. Pi s over to okay. And then we can combine this all into home and denominators two s squared that IHS So it's minus pi e to the minus pious over to okay. Plus, um oh, after multiplied by ass course minus pi s. And then this is plus to the minus two eats the minus pi over two. Now for integral to that's this. So I'm gonna different cheat this and integrate this so that's going to be minus caused t and then leave this alone for the boundary term. And then that's minus. So now I have to have to integrate this indifferent sheet. This so that's plus S e to the minus s t then caused t so that picks up another minus. Sign DT this boundary term right here. Zero because I know it isn't sorry. Careful. So this boundary term is eat the minus pious and the minus will cause a pirate too is zero. So this is just e to the minus pious, okay, and then this is minus asked times. Now we do integration by parts again on this guy. So that's integrate this and differentiate. This minus s multiplying. So this becomes e to the minus s sine of t evaluated between pi Over to two pi on the minus into go from minus pirate to two pi of plus s e to the minus Esty. Yes. Can come out Sign T DT, right. You ve minus V D u. Okay, And now let's simplify this. So sign of pi zero. So that's just gonna give us minus e to the minus pi the pie s remember where substituting for T values and then sign of Pirated is just one that's plus s. So notice that this is exactly I to again. So I two is equal to this. Okay, so let's just simplify, okay? And so I to into one plus s squared is equal to this right hand side. And so I two is equal to one plus s over one plus x squared times e to the minus pious. Okay? No, I made a mistake. And I hope you caught that pious over to right. Because I'm substituting teas equals pi over to the t easy with the pie thing. The teasing The pie part is zero. Because sign up. I zero. So let's track back. Okay on. And then this is just gonna be okay. So that's what I two is. So I want is this I've slightly simplified it, and I to is this. Now let's go back to original problem. Right? So we're trying to figure out I write, we split up into these two into girls. So So eyes equal to this. This is equal to two over pi times by I one. But I want is this business right here? So adding them together we get this the twos council. Well, I could bring the pie over here, so that's what I is equal to. And so our answer is going to be whatever we got for I that must miss expression times by this factor. Okay, so therefore the LaPlace transform of f is going to be 1/1, minus e to the minus. Pi s into into all of this business. All right, let me let me write it back so that you write it out again. If you want, you can combine the denominators. I didn't bother doing that. Okay, so that should be the LaPlace transform back

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