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Problem (10 points) Answer the following questions pertaining the following initial value problem:FyI 1 € (-1.1) V1-r y (0) = 1(3 points) State the order the ...

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Problem (10 points) Answer the following questions pertaining the following initial value problem:FyI 1 € (-1.1) V1-r y (0) = 1(3 points) State the order the ODE whether it is linear Or nonlinear; and whether it is homogeneous Or inhomogeneous (ignore the initial condition for this classification).points) Find the solution y (r)

Problem (10 points) Answer the following questions pertaining the following initial value problem: FyI 1 € (-1.1) V1-r y (0) = 1 (3 points) State the order the ODE whether it is linear Or nonlinear; and whether it is homogeneous Or inhomogeneous (ignore the initial condition for this classification). points) Find the solution y (r)



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In each of Problems 1 through 10 find the general solution of the given differential equation. $y^{\prime \prime}-2 y^{\prime}+y=0$

Here we have the second audit differential equation to Y. Prime prime plus two I. Prime plus one Is equal to zero. And here to check if this is a complex oh or he has real and distinct routes. We use B squared minus four A. C. Is the determinant. And our B our age go to to our bees go to two and our seas go to one and I'll be square to be two squared minus four by two by one. And he said he was 4 -8 And we have our peace Grade -4 a. c. is -4. So since this is less than zero means that we have a complex we have complex roots and our general solution will be in the form see one E. To the poor lambda T. Of course mm two plus C. To eat the poor lambda T saying new T. In to find the roots complex roots. We use the quadratic equation minus B plus or minus the root of peace, create minus four A. C. Which we have already calculated his -4 over to A. In substituting our values here we have minus to plus or minus. They're right off minus four Over two x 2. And excuse us minus half year Plus or -12 -4 gives us to I Developed by four we get a half day and we have our hot so this is our roots and plugging in the values since our land our here will be equal to minus half in our mute is equal to half. So our solution Y. T. Will be equal to see one. Mhm to the post minus half T. And we have course half of T plus C, two E. To the minus half of T. And sign uh huh. T. And this would be uh general solution to the given differential equation.

So let's start this problem out by solving for the homogeneous solution. So r squared minus two are what's two equals to zero and I do not see any immediate factor income the mind. So I'm just going to use the quadratic formula. So I'm going to have negative B plus or minus A squared B squared and it's four times a time. See all over two times A. And that equals two. One plus or minus the square root of for sorry, negative for divided by two. And that equals to one plus or minus I. Right. And so R equals to one plus or minus I. And with that we can build a homogeneous solution. So I have C one E. The X. Co Synnex plus C two into the X syntax. And our guests for the Hamas or the particular solution is going to be a X plus B plus C mm to the X. And we're gonna take the driver this twice. So the first derivative is going to be a plus C E L E X. And our second derivative is going to be C D. E to the X. So let's plug that into our original equation. Let's remind ourselves that the original equation, what's Y double prime minus two? Y. Prime plus two. Y equals two X plus eating the X. So when we plug that in, Y double prime is C. E. To the X minus two times a minus two. C. E. To the X. Plus to a X. Plus to be plus two C E X at all equals X plus E. The X. And so let's try and create a system of equations here. So we'll have let's actually simplify this down actually. Um let's see. So we have these terms right here, they're like, so we'll have negative C. E. To the x minus two, A plus to a X. Plus to be. And since this is also like that, we're going to have a positive, see the X. Can remove this term completely and that really goes to explicitly the X. So now we can create systems equations. So we'll have C equals one negative to a. What's to be equals to zero and to A equals to one. So A equals to one half. And with the we can solve for B. So he is going to be negative 21 half plus to be equal zero. So that equals two negative one plus two, B equals zero. So if we bring the one to the other side, we end up with the growing one half. So our total solution is the homogeneous solution plus the particular solution. Yeah. In this case or homogeneous solution we said was um C one E. To eat the X co sign X plus C two E X. Synnex. And our particular solution is going to be E X plus X over two plus one half

So let's start this problem off by solving for the modular solution. So r squared minus one equals zero. If we bring the one to the other side we have r squared equals one Or r equals to plus or -1. So we can translate that are equaling 2 -1 and one. And with that we can build our homogeneous solution through modular solution is going to be C one E. To the X Plus c. two e. to the negative acts. And we can take a guess at our particular solution. So our guest is going to be a X plus B Times E to the two X. We can simplify that down to be a X. Eat the two X plus B each of the two X. And now we can take the driver this twice. It's our first derivative. It's gonna be A. E. To the two X plus to a X. E. To the two X plus To be each of the two X. And our second derivative is going to be so I have to a eight the two X plus four A. X. Each of the two X. Plus to a 22 X. Plus for B. Eat the two x. And so original equation was Y double prime. And let's simplify this a little bit. So I have four A. Each of the two x. plus four A. X. Each of the two x. Plus four B. Eat the two x minus. Why? In this case why was AX. Each of the two x minus pe 22 X. And that all equals to accede to the two X. Right. So it's some fighters down a little bit. So I have four a. Each of the two x. Plus three A. X. Each of the two x plus three B. Eat the two x. All equals two X. E. To the two X. Right. And with this we can build a system of equations here. So have for a plus three B equals zero and three A equals one Right off the bat. We can solve equals one third and we can plug this into the first equation to get four times one third Plus three B equals zero or four thirds plus three B equals zero. Which if we bring beat that to the other side, we should have negative 4/3 here and we're b equals 2 -4 nights. Right? So now we can build our total solution here. So our total solution consists of a homogeneous solution plus the particular solution. And so we have C. One E. To the X plus C. To each of the negative X. And then we have 1/3 minus 41 3rd x minus four. Nineths each of the two X. And we can use our initial conditions too soft for the C. One and C. Two right now. So we have Y. Of zero equals zero. And why prime of cereal equals one. So let's take the derivative first. So we'll have why Prime equals two. C. One E. To the X minus C. Two. Eat the negative X plus 1 3rd each. The two x plus two thirds X. E. to the two x minus 8/9 E. To the two acts. And we can now plug in zero for the first equation. So when we do that we end up getting C. one plus c. two. This term right here cancels out goes to zero Sort of -4 nights And that all equals to zero. And for Y. Prime equation if we plug in zero for X this term goes to zero And then we have C1 minus C. two plus one, third minus eight nights All equals to one. Right? And so we can simplify that a little bit. So our first equation stays the same but let's bring the constant to the right hand side. And in our second equation we can rewrite this 1/3 to be 3/9. So we have negative five nights. This whole thing simplifies down to negative five nights. Let's bring up to the right hand side. We can do in the first time. So have five nights here And of course this all equals one. So we'll have fine nights plus one. Okay We can simplify that little bit further so that's gonna be great right now has 9/9. So the first equation remains the same but her second equation is going to be 14 divided by nine. And let's add these two so we can cancel the C2 term. We'll have to see one equals two, 18/9 or C one equals to one. and now we can solve for the C2 term. So we know that C one plus C two -4 nights equals zero And we just figured out that C1 is equal to one. So we have that one plus E to minus four nights equals to zero. So C two plus five nights equals zero, which gives us C 22 equal negative finance. Right? So now we can actually write our total solution which consists again of a homogeneous solution plus our particular solution. So our total solution is going to be E to the X -5/9. Eat the negative X plus one third X minus four nights. Yeah. E to the two x.

So start the problem off by solving for the homogeneous solution. So have R squared minds are equals zero. We in fact are on our terms, we have AR -1 times r equals zero. And that gives us our values of zero. And one with that we can build our particular homogeneous solution. So why of H equals to see one plus C. Two E to the X. And we can take a guess at our particular solution. So a particular solution, guest is going to be a X plus B times E to the X. Now I notice that there's a term that's matching the motion solution right here and this right here. So if I some fight down a little more might become a little more evident. Right? So these two terms right here are matching. So we have to multiply by a factor of X. So we'll have a X squared E. To the X plus B, X E to the X. And that's now going to be your actual guests for the particular solution. So we're going to take over this twice. So the first derivative, it's going to be to a X. Each of the X plus A X. Where each of the X plus B. E. To the X plus the X. E. To the X. And our 2nd derivative. Why Double Promise P. It's going to be to a each of the X. Plus to a X. Each of the X. Plus to a X. E. To the X plus A X squared E. To the X plus the E. To the X. Plus the E. To the X plus E. X. E. T. X. Sorry. And let's see if we can simplify this down a little bit. So uh Y P double front Equalling to A. E. To the X. These two terms are the same. So have four a. x. e. to the x. Um A X squared eat at the X. Do you see in terms of the same? So I have to be each of the X. What the X. E. To the X. Right? And now we can build or actually plug it into our original equations. We have Y double prime, which we happen right? Right here minus Y. Prime. And what crimes? Right here. So I have minus two A X. E to the X minus A. X squared each of the X minus B. E. To the X minus B. X. B to the X. And this all equals two X. E. To the X. Right? So let's see if we can do a little bit simplification to make her life easier. So I have X. E to the X. Actually, yeah, let's do it. So let's see if we can find the terms that are similar. So we have here and here. So please cancel out We have a term right here and here the same. So they cancel out. We have a term right here and here. But they're not the same in size so they won't cancel out. And these two are the same. So we have to A B to the X. Remember these two are the same. So we'll have to A. X. E. To the X. And these two are the same. So we'll have plus E to the X. Thanks. And so with that we can create a system of equations. So have to april's B equals zero. And we also have to a course to one. So with that we get a equals to 1/2 and B equals two. Rather plug that into the first equation. So we have that two times 1 half plus B. Close to zero. So people's one equals zero, Or B equals 2 -1. And with that we can build our total solution which consists of our modular solution plus a particular solution. So in this case will have C1 plus C two. Each of the X plus one half x squared minus x. E. To the X. Right. And with this we can solve for initial or rather proceed. Want to see to using the initial value conditions. So initial value conditions are Y0 equals to two And why? Prime of zero equals to one. So let's start off by taking the derivative. So we'll have that white prime equals two. C two B to the X plus one half X squared B to the X plus x E. To the x minus E. To the x minus x. E to the X. These terms right here cancel out. So you can simplify this down to being C2 B to the X. Um plus one half X squared B to the x minus E to the X. Right? And so let's plug in zero for the first equation. So when we do that we get C one Plus C two. This entire thing cancels out to zero. Yeah, so that's equals to two. And for the second equation we plug in zero. This cancels out. So we end up with C2 minus one equals two one. Right? So With that we get that C2 Get that C2 equals to two and see one puts C two must equal to two. NFC one, C two equals to two. That means that C one has to equal to zero. So with this we can build our total solution, which again, because this is our homogeneous solution plus a particular solution, it's a total solution. It's kind of equal, you know, the X. We do a little bit of factoring one half expert, plus rather minus X plus two.


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