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Let y(t) be the solution of the initial value problem y" - y = g(t) and if 0 < t < 1 g(t) = {1 y(0) = y (0) = 3 if 1<t Then y(t) A (t +1) + B (et-a ...

Question

Let y(t) be the solution of the initial value problem y" - y = g(t) and if 0 < t < 1 g(t) = {1 y(0) = y (0) = 3 if 1<t Then y(t) A (t +1) + B (et-a ~ t) ua (t) with A, B, integers, respectively equal to

Let y(t) be the solution of the initial value problem y" - y = g(t) and if 0 < t < 1 g(t) = {1 y(0) = y (0) = 3 if 1<t Then y(t) A (t +1) + B (et-a ~ t) ua (t) with A, B, integers, respectively equal to



Answers

Solve the initial value problem explicitly.
$\frac{d y}{d t}=\frac{1}{1+t^{2}}+2^{t} \ln 2$ and $y=3$ when $t=0$

16 12. Problem number 17 This is a first order linear, ordinary, different journal equation. We have a standard form, but before I do that, I mean, you might get your You've done enough of these. You recognize what is the derivative with respect to theta? Why times data? If I'm differentiating with their sector fate, I would have to use the product rule. This would be you take Why them's the derivative fatal with respect to fade as one plus the derivative. Or why, with respect to theta time Stada. So this is what theta de y de Theda Plus why you notice that that is the incomplete left side. So I could take this and say OK, well, I can go straight to the derivative with respect to the data of y theta is equal to sign data. I can go straight there and start solving this problem. Okay, if you are getting toward that is not jumping out at you. You're not recognizing that we have a very methodical formula for the way we've solved all the differential equations in this section. That is to get the problem in standard form, find the integrating factor and then proceed. So let's just say that we did not recognize that shortcut. Standard form in this case is going to be I need to divide everything by theta so d Y d theta plus one over theta times. Why is equal to sign of Fada over theta? The integrative factor is going to be e to the anti derivative one of earth. Ada di Data This is e to the natural log Absolute value Fada. Since that is positive, I can remove the absolute value signed. This is E to the natural log of theater which is theta so everything has to be multiplied by theta. You would have the derivative with respect to theta of Fada. Why equal multiple Beth data and you get sign of failure? So this is where we were a moment ago. Now integrate both sides of this equation and you end up with why Time stada is equal to minus coastline data plus a constant of integration. So we've got wife ADA is equal to minus co sign data plus a constant of integration. The initial value problem was that why at pi over two is equal to one. So if why a pi over two equals one that tells me that in this equation So first of all, let's just solve it for theta. Why is equal to minus co sign data or with eight A plus CIA Rotana? And then if y harbor to is one that tells me one is equal to minus co sign higher or two over pi over two see a repair were too co side of PIRA to is zero. And so this tells me that one is equal to to see over pie. Therefore, C is equal to pi over two. So this final answer is why equal minus co sign data over data plus pi over two data.

Section 16 dot to problems. 16. We have a first order, ordinary first order differential equation, and this is called a boundary problem because we have an initial condition that must be met. So easiest way to solve this guy is to put it in standard form. So standard form is going to be D Y d t plus some function of tea terms. Why is equal to some other function of T? Because if I can get it in this form, I know that the key will be to multiply both sides by an integrating factor, which is just the E to the anti derivative of pft. So standard form in this case would be to divide everything by t so D Y d t plus two over tea. Why is equal to a T Square? That would tell me that my integrating factor it's going to be e to the integral of what to over tea T t that's e to the to. After a log of absolute value of tea, we know that T is positive so I can get rid of the absolute value signs. This is going to be e to the natural log of T squared, which is T Square. So my key will be to multiply everything by t squared in this particular problem. So I've got d Y d t plus two over t Why is equal to t squared? Everything has to be multiplied by t squared and that will give me the derivative with respect to t of t squared. Why is equal to t to the fourth now? We integrate both sides and we have t squared. Y is equal to t to the fifth over five, plus a constant of integration to solve this for why you gonna wise equal to t cubed over five plus see over t squared now, my initial condition was that why I have to is equal to one. So why I've to is equal to one. That means that one is equal to 8/5 plus c over four. Um, to solve this guy, Let's just multiply everything by 20. So 20 is equal to 32 plus five c so that tells me that see is equal to negative 12 5th So my final answer, um, plugs into this equation. So I just plug in for C. And in doing that, I end up with the final answer. Why's he couldn t cubed over five plus? And so it will be what, minus 12/5 T squared minus 12 over five T Square, and that's my final answer.

The first step to our solution will be taking the applause transform of both sides of our equation. Such that will get the boss transform on the second derivative minus three times of the lost transforming the first derivative plus two times of boss transform of why single to the class transform on Delta of T minus one. Now we can use what we know of the class transforms to simplify this equation. So we're gonna get s squared times of the blast transformer. Why? Minus s times Y zero minus y prime. And zero minus three times s times wireless minus y of zero was two times Capital Iris being able to little boss transform of Delta T minus one. Well, a is one in this case, was, will you to the negative s. Now we can gather all of our terms here to get as squared minus three s plus two all times capital. Why minus why have zero times s and r y of zero was one so minus s minus y primer and zero, which is zero minus three times negative one. So be positive. Sorry. Being able to you to the native s now we can solve for our capital. Why which will be equal to you need to the negative s plus s minus three, all divided by s squared. Minus three s plus two. And we, in fact what we have in the bottom in the denominator to be you need to Negative s plus s minus three. All divided by s minus two times asked my response. Now we can separate this into two separate fractions yet you need to the negative s not by s minus two. Tens of slyness. One plus s minus three. All invited by s minus two times X minus one. And now we're gonna partial fraction decomposed. Both of these, starting with the fraction on the left. Really get one over X minus two times X minus one Unical to a over X minus two plus being over X minus one. Multiply the denominator on both sides to get to This is one music with you. A times as minus one plus B times X minus two benefits. It s You were the one. We're going together. This is one is equal. That a time zero plus de times one minus two, which is negative. One therefore be tested people. The negative one never said s equal to to really it. That one is equal to a times two minus one, which is one plus any time. Zero. So therefore, a has to meet with one. It's therefore, this is one over. X minus two minus one over X minus one. And now we're gonna look at the other fraction that we had, which was s minus three. All divided by s minus two times X minus one. We want to separate this into another fraction. A over s minus two plus being over s minus one again. Multiply the denominator through together. This is s minus three is equal to a times X minus one plus being times X minus two And again, if we said ask being eagle, the one we're going to get one minus three is negative. Two on the left you mingle, Teoh. A time. Zero plus being times one minus two, which is negative one. So therefore, being passed a vehicle to to never said s being able to to really a to minus three, which is negative. One on the left times A times two minus one is one plus being time zero. So therefore, a has to be equal to negative one. So therefore, this is going to be negative. One over X minus two plus two over X minus one. Now we can take all this, plug it back in for a capital. Why? Expression, which will be our first fraction, which will be eating the negative s turned all of this. So using the negative s times one over s minus one minus each of the negative s times won't ever asked. Minus two. The answer is backwards as minus two is positive in this minus one is negative. Plus the second fraction, which is negative. One of Rs minus two plus one plus two over s minus one. Now we're gonna take the inverse applause transform of everything to get back our Why. So we'll have the inverse laplace transform Negative musical negative s times one of us minus one plus in versatile plus of means of negative s times won't ever asked. Minus two minus in verse the plus of one over X minus two plus two times in verse, The plus of one over X minus one. And now we can look at the inverse or fact that we know from our rules for applause transforms. That's a plus. Transformer one over s minus A in Bristol class of one of rest mind say, is key to the 80 here is too. So this is just e to the negative to teach. And this was the to be a positive too. And this will be easy to the positive t. And we can use the second shift serum to write this. Since we haven't e in our inverse LaPlace transform as negative heavy side function of you raised to negative A and A's won don t times the inverse LaPlace transform of one Arrest minds one which Richard Sermon was e to the T minus the base of the part of the heavy side function wishes one and same thing. Over here. This will be every side of one. I'm t times e to the two times t minus one. Now we can write our general solution as factoring out the heavy side function new one of t times e to the two terms T minus one minus E to the T minus one minus a to the to change plus two into the team

So in this problem we are given The differential equation d. s. d two is equal to two, three T -2. And Were given the initial conditions as physical to zero when she is equal to one. Now integrating both sides with respect to T. On our left we get S. Is equal to the integral of tea by this we can see that we can multiply through multiply needs through the brackets. We get three T squared minus two T. T. T. And now integrating explicitly here we have uh three T squared. So we have the tea. The new power being three, so three T cubed over. The new power three -2 T. to the power of two Developed by the new part two plus the constant of integration. See And here we get T. Cute minus T squared plus C. And this is you got to s substituting our initial conditions, we have zero is equal to one, cute minus one squared plus C. And here we get zero plus C. So I'll see here is equal to zero. Therefore substituting into plugging that into our integral. We get S. Is equal to t cubed minus t squared plus zero. So this is our solution to the given to financial equation


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