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38. Suppose b. and are constants and a # 0. Let as +b Ji =L-1 and J2 = L-1 T) as2 + bs + € as2 + bsShow thatV1 (0) = 1, J() = 0 and Y2(0) = 0, J(0) = 1, HINT:...

Question

38. Suppose b. and are constants and a # 0. Let as +b Ji =L-1 and J2 = L-1 T) as2 + bs + € as2 + bsShow thatV1 (0) = 1, J() = 0 and Y2(0) = 0, J(0) = 1, HINT: Use the Laplace transform to solve the initial value problemsay" +hy' +cy (V +by' +cyJ() = 1, "() = 0 Y() = 0. Y (0) = 1,

38. Suppose b. and are constants and a # 0. Let as +b Ji =L-1 and J2 = L-1 T) as2 + bs + € as2 + bs Show that V1 (0) = 1, J() = 0 and Y2(0) = 0, J(0) = 1, HINT: Use the Laplace transform to solve the initial value problems ay" +hy' +cy (V +by' +cy J() = 1, "() = 0 Y() = 0. Y (0) = 1,



Answers

Use the Laplace transform to solve the initial-value problem $$\begin{aligned}y^{\prime \prime}+\omega^{2} y &=A \sin \omega_{0} t+B \cos \omega_{0} t \\y(0) &=y_{0}, \quad y^{\prime}(0)=y_{1}\end{aligned}$$ where $A, B, \omega,$ and $\omega_{0}$ are positive constants and $\omega \neq \omega_{0}$.

Hi. Today we're going Thio, verify the initial value theorem for a collection of functions. I'm gonna list the functions here. We have a constant. We have an exponential. We have another exponential coast value. A sign T square and T Coast T. Now we can calculate the lack last transforms of these functions quite easily. Using tables left transfer. The first is one of the s the 2nd 1 over X minus t the 2nd 1 over X plus t The third s over s squared plus one that fool one of s squared plus one The fifth, um, to over s cubed. And the final one being one over X squared plus one minus two s square over X squared, plus one squared. Now, if we take a look at the limit off s times L s. We see this is a constant. This is s over s minus T. This is s S plus t. And I think this is a square of escort us one s over a squared plus one, two over X squared And the final one s over. Escape plus one minus. Two s cute. Over s squared, plus one square. Now all of these are rational functions, which makes the limit quite easy to calculate. In the first case to the limit as s tends to infinity is obviously won the second the limit. It's one because the numerator and denominator denominator the same degree. The fur is also one s and s are the same degree. That force is also one. The fit is zero because the denominator has a bigger degree. That many racer, um two of s squared is also zero for the same reasons. And the losses also zero for the same reasons. And if we plug t hold a zero in the original functions, then detained exactly these cells.

So our problem here is why to the four prime last y equals zero. And other than that, why is vehicles one problem? 00. Why that promise vehicles one and watch from 00. So they would start with this we're gonna start off taking plus transforms of each term. Are you? As to the 4th? That was the glass transform before I good uh minus s cube. Watch the zero my s squared Why prime of zero my S. S Why double from zero minus y triple prime zero minus applause transform of why equals zero. Sorry to cancel out terms. So we see that why primary zero triple prime zero are both equal to zero. Can cross off all these terms assuming zero And this is going to be zero and we see that these two terms of both people won. So we know that this term it's just going to be equal to s in this term is just going to be equal to s cube. It's not going to rewrite down here simpler to the 4th applause transform of Why Maya's s cute minus s minus laplace transform of why All equal to zero. So now it's much simpler when I separated out now on the left hand side we're gonna lost transform of why? Outside of S to the 4th um minus one. On the right hand side we have s to third plus ass. Sorry uh divide to um get the transform alone. Yeah, as to the third plus S All over us to the 4th. My ass one. All right, so now we have on top um we're going to try to simplify this out so we can do something with it. And so we're gonna separate our denominator first. So we said this is uh one is the other one square. So you see as a perfect square uh It's a product of S Plus one X -1 and fear that's called but this is only S Cubes Plus three of Top. I'm sorry. Plus ass you can go back. So if I was one further it's a little bit frozen right now. one Good. Let me do this thing. All right, we're back. Alright, so we're gonna tractor and that's on top. We're going to S. S squared plus one mm on the bottom where everything else is is the product of S squared plus one and essex where minus one. Now I have this on both top and bottom. She crosses out. Now all of a sudden we have just S over. That's Weird -1. Yeah. Which is a much easier. Um from deal with. So we're gonna give the denial of the same treatment again separated out. So you can see this is a product of S. Plus one & S -1. And as you might guess, we're gonna do a partial fraction decomposition. So take this down here have a over as plus one plus B. Or S -1. So now we're going to take it further and go A. S. My S. A. Plus B. S. Plus B equal to S. Sorry I separate out into like terms yes plus B. S equals S. I'm sorry yes never have negative A plus B equals zero. So from this we get that A. Is equal to be yeah and A plus B To go to one. So this is a pretty easy solved but so we get that A. And B Are both equal to 1/2 mm. Yeah now I plug that back into a partial fractions over here and seeing how one half, Let's have one over s. Plus one plus one half. I have one over S -1. So this is far enough for us to solve so get a little more space right now. Why Equal to 1/2? Has the universal laws transform of one over S plus one plus one half. I was in virtual applause transform of one over S. Uh I'm sorry S minus one. So now we recognize this as the applause transform of the exponential function. Yeah. You know it's going to come out to why is he going to one half, eat a negative T. Plus one half E. C. T. There you go.

Our first step here is going applying the LaPlace transform to ah equation here. And when you do that to get through the class, transform on the second derivative plus two in terms of loss, transform of the first derivative closely plus transform of just why being able to boss transform out without to function, senator for and now we can use our rules for the flash transforms to simplify this a little bit. So this first time will be as squared times capital I minus best friends wide zero minus y prime it zero plus two call times asked times why minus y zero plus capitalize equal to you the class transformer Delta Well, our is for us also be needed the negatives for yes and now we can plug in our initial conditions. So why in zero we were given a zero same with the first derivative. So it was three terms of zero. Therefore, on the left, we're gonna have s squared plus two s plus one of all times capital. Why is equal to eat in the negative for s? No, we can just divide by this polynomial to get that capital y is equal to you to the negative for s times one over a squared plus two s plus one. We can factor this denominator to just be asked. Plus one call squared. Now we're gonna take the inverse of loss transform to get that are why is equal to inverse LaPlace transform. You needed the negative four s times One over X plus one all squared. And here to start, we're going to use the first shifting Durham which says that we haven't We haven't each of the negative any team in our you know, this will be the second shifting fear. Um, because we haven't needed the negative A s in an hour. LaPlace transform for take the inverse will get heavy side at a So this will be every side at four MT. Times Universal boss transform of one over X plus one all square. But this evaluated at team minds four. Try to keep that in mind. This is not a tires is evaluated at T minus four. Now we're gonna evaluate the inverse of Austrians form here and now we're gonna use the first shifting fear. Um, since we have an s minus a, this is really the inverse with last transform of one over s minus negative one all squared. So therefore, this will be e to the negative team times the universal class transform of one over s squared, which is just in to the negative team. And here we can write this as one over s to the one plus one and one is one factorial. This will be t to the first power. So this jeans t and then we have to put it back into original expressions. We're gonna the heavy cited for of tea times. All of this, but evaluated a T minus four. Neither the negative T minus four all times T minus four.


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