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25Solve the differential equalion Ihat 9(0)=3 and Y(o)by usirIg Laplace Iransfom and givenouLol(b) Obtain Ihe Fourier series (0 roprosent Uhe |urction f6)= (T2) 0&l...

Question

25Solve the differential equalion Ihat 9(0)=3 and Y(o)by usirIg Laplace Iransfom and givenouLol(b) Obtain Ihe Fourier series (0 roprosent Uhe |urction f6)= (T2) 0<*<2nJ questian

25 Solve the differential equalion Ihat 9(0)=3 and Y(o) by usirIg Laplace Iransfom and given ouLol (b) Obtain Ihe Fourier series (0 roprosent Uhe |urction f6)= (T2) 0<*<2n J questian



Answers

Use the result of Problem 28 to find the Laplace transform of the given function. $$ \begin{array}{l}{f(t)=\sin t, \quad 0 \leq t<\pi} \\ {f(t+\pi)=f(t)}\end{array} $$

Hello and Welcome to Probleble 32 of Chapter six. Section 3 he was asked to find the applause transform of a given function after two. Equal scientific given that its periodic on the interval. Uh every pie unit. And if we look at the regular graph of sine function sign it is periodic every pie unit. So this should be interesting. Um Uh test to see if the periodic laplace transformation is the same as the regular one. So given an problem 24 when we have a periodic function that's periodic. Every capital T. Units in this case it's pie. Uh It's laplace transform will be equal to 50. Is equal to the integral from 0 to capital T. E. To the minus s lower case too. Fft duty All over one minus E. To the minus S. Times capital too. All right in this case capital T. Is pie because that's the period and the rest of this weekend just uh fill in. So we'll have oh Let's be the integral from 0 to Pi E. to the -4. Sign of T. D. T. Over one minus E. To the minus s. Pie. This is an interesting inter role. I think we'll have to use integration by parts to do this. Another another option equally valid is to write sign in terms of its exponential form To the E. to the -80. Either the I 80 or E. To the I. T minus eats the negative I. T. All over 22 I can solve like that. I'm gonna use the integration by parts method and this will have you equal to sign of t. What do you the co sign of T T of course then tv E to the minus S T T. T and V. Is equal to You to the -S2 for negative s. Okay. And when I rewrite this, I'm going to leave out this bottom part because it'll make the problem a little bit more confusing. But let's evaluate this. Remember to bring this back, bring back. So this will be equal to the integral. Well UV minus V. D. U. So it's sign of t come to eat the minus S. T over negative s minus and the role of the Eu Well we've been negative there soon. Just add a plus on the outside. We'll have um You to the -4. Times co sign of T D T. Um had that one arrest from on the outside. And we also evaluate this from pi and zero. Oops hi and zero. But note that uh sign when evaluated pie is zero. So is signed one evaluated zero. So this term will cancel that. We'll just be left with one of us time syndrome. From zero to pi of the of the minus S. T. Times co sign of T. D. T. And we are going to infect to integration by parts again. So we'll have you be co sign up T do you is negative sine of T. Um T. V. Is really just the same as above john john arrow. All right. And Let's do this. 2nd integration by parts. So this is equal to one of us times. Cosine of T times use the minus S. T over negative S. So co sign of T E to the minus S. T over negative S minus integral from zero to pi of the D U. All of this case. Viejas, is that negative term on the outside to use another negative term? So the end it will be a negative term. So write this to have that. That's one of us integral of. Well it will just be the integral of that original function and what? So from 0 to Pi mm and E to the minus S. T. Sign of T duty. Okay. Um of course this is actually equal to this thing up there which is the same as in here. So we can rewrite that as a variable. I'm just going to call it X. That's a nice variable. So we'll have X. Is equal to one arrests times. Co sign of T times E to the minus S. T over negative S Evaluated at zero and pi minus one over S times X. Where X. Is just this. Mhm Okay. Uh let's evaluate this at pi. We'll have co sign a pie which is negative one and we have E to the minus s. Pie on the top. We voted with the zero will have um co sign of zero is one times E. To the mine times each of the zeros one. So we'll have one over negative S So let's write that out. We have X. Is equal to one of us. Times. Co sign applies -1. So I have eat the minus S. Pie over S. And when we subtract we'll evaluate this, zero will get one. We got one times one on the top, so B -1 over S. And one more step and subtract this from. That's one of us times X. Okay? Um Now we're just going to try to get X by itself right first let's distribute this one arrests. So turn all these into squares. Okay? And we can add that X term over. So we've got X plus one over S squared groups. one arrest squared x. Is equal to E. To the minus s. Pie minus one over the quantity S squared. We'll have X times one plus one of us squared equals E. To the minus S pi minus one over X squared here. We can just multiply both sides of the square to make this a little bit simpler, you can get rid of the s squared there. This will turn into a one. This will turn into an S squared term there mrs squared. Great. So X. Which is original uh integral which is all the way up here. That is equal to uh E to the minus s pie minus one over S squared plus one. Okay? and now we can bring back but we had on the bottom just one minus E. To the minus s pie. So the final answer which is well placed transform the fft plus transform of fft is equal to E. To the minus s pie minus one over S squared plus one. All over. So what is it one minus E. To the minus s pie, one minus E. To the minus s pie and just looking at this um that should be a a negative term. Um Yeah so the only difference is that there will be a negative one on top and so we can just cross it out, cross it out. Little plus transformation Fft is equal to negative one Uber S squared plus one. And um since the sign is an odd function this should be equal to um it's the applause transform and that concludes the problem.

Hello and Welcome to Probleble 26 of Chapter six. Section 3 we were asked to find the applause transform of a given function which is an infinite series of terms of heavy side functions and we're also told that integration of each component is allowed. So let's just find a little pause transform of uh each term separately. So we'll take capital F F. S. And this is equal to well particular post transform of one that's one of us we subjected from the whole past transform of Heavy side function of one that's minus E to the -1 over us. It will add this to E to the -2 us over us. And likewise we can continue this pattern indefinitely. But what is the end goal you want to write this in terms of a series? Well we can write this in terms of capital F F s is equal to Well I'm just going to have this one restaurant on the outside because the only one without an E and we're gonna subtract it from Actually we'll just add it to its infinite serious. Or we started one we're going to go up to two N plus one just like the original problem and let's see what is coming to all of them. Well we're gonna have a divided by s on the bottom and let's say we're also going to alternate negative and positive every time. So we're gonna have a negative ones at the end and since we started one and that's negative we don't have to worry about 10 plus one. Okay and then what's going to happen to this term? Well, we know every term as an e to a power of a negative number. And really each constant here is just whatever in it is in a series. So have you to the negative N. S. And that concludes the problem.


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