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1_ y"x2 + Txy' + (x2 +9_V2)y = 0 Try: y =u/x" with an appropriate choice of n. 2_ Given ODE[40...

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1_ y"x2 + Txy' + (x2 +9_V2)y = 0 Try: y =u/x" with an appropriate choice of n. 2_ Given ODE[40

1_ y"x2 + Txy' + (x2 +9_V2)y = 0 Try: y =u/x" with an appropriate choice of n. 2_ Given ODE [40



Answers

$1-10$ Solve the differential equation or initial-value problem
using the method of undetermined coefficients.
$$y^{n}+y^{\prime}-2 y=x+\sin 2 x, \quad y(0)=1, \quad y^{\prime}(0)=0$$

So start this problem off by solving for the homogeneous solution. So have r squared minus for our plus four equals to zero. And I do not see any obvious factoring. So I'm just going to go ahead and use the quadratic formula here. So negative B plus or minus A squared B squared which is 16 minus four times a. Times C. All divided by two times A. So that equals to two plus or minus the square root of zero, divided by two, that equals two one. So are our values are repeating uh one. We're sorry actually my bed we already divided that term by two initially. So it's just this term that's divided by two. And so that equals to two. So our values are repeating it to and so what this means is and are homogeneous solution? We're going to have see one E. To the two X plus C. Two X E 22 X. And now let's take a guess at our particular solution and that will be A X plus B plus C. Synnex plus the co sign X. Yeah. And we're going to take the driver of that twice. So end up with my prime Equalling two A plus C syntax minus. Do you sign X? Sorry about that, this is Cosine X. And our second derivative is going to be negative. See Synnex minus D. Cosine X. And so let's plug that all into our original equation. So end up putting negative, see Synnex minus D. Cosine X minus for a minus four. Seiko Synnex plus for D syntax when we come up by the next sequence by four. So we're going to end up with for a X. Plus four B plus four C. Synnex plus forward the co syntax. Right? And then we said that all equals two um X minus Synnex. So let's try and create systems of system of equations here. So let's try and get all the sign terms together. So we're gonna have negative C. Us for D. Um plus four C. And that all equals two negative one. And the coastline terms negative D. Plus I'm sorry minus four C. Plus for the that all equals to zero. And all the terms here it's gonna be negative for A. Plus four B. Equal to zero. And the last set is going to be for a. Um That's it for A equals to one. So right off the bat we can solve for A. And we know that A equals to 1/4. With that information we can solve for B. So I have negative four times 1/4 plus four B. Goes to zero and this is negative one. And so if we bring that to the other side we have B. Also he's going to 1/4. And so now let's focus on these top two equations. So it's simplified out first so we'll have four D. Plus three C equals negative one. And her bottom equation supplies down to three D. Minus four C equals to zero. So let's multiply amount. Let's do the top let's multiply it out by three and the bottom let's multiply it out by four. So we get 12 D. Plus nine C. Of course the negative three. And our bottom equation is 12 D minus 16 C equals to zero. So if we do subtraction here we end up with 25 C equals negative three and see equals two negative three divided by 25. And with that we can solve for D. So let's plug it in. We have three D minus four times negative three divided by 25 equals to zero. If we bring that to the other side, we end up with three D. Equalling two negative 12 divided by 25. We divide by three, we end up with negative four divided by 25. And with this we can build our total solution which is the sum of our homogeneous solution and our particular solution. Remember initially we said that are homogeneous solution was C one either two X plus C two X. E. To the T rex. And our particular solution, we found that to be X divided by four plus 1/4 minus three, divided by 25. Synnex minus four, divided by 25. Co sign X.

So let's start this problem out by substituting the white collar crime and lifetime by prime turns we'll have R squared minus three R equals to zero. And we factory get our tents are months, three equals zero. And that gives us our values of zero and three. And so with that we can build a Hamas in a solution. It's going to be C one plus C two either three X. And we can guess our particular solution which is going to be a chosen two X plus B. Sign two X. So let's take a jury to this twice. So I have Y prime equally negative to a sorry, sign two X plus to be co sign two X and Y double crime. It's going to be negative for a co sign two X minus or B. Sign two X. So we can build our particular sedition or rather solve for it. This is the original equation the wife to plug in our values into. So Y double prime. Yes, negative for a co signed two X minus four B. Signed two X. And then we're going to multiply that by negative three times negative to a sign to X. Plus to be coz I. T. Rex. And that all equals signed two X. And so if we multiply this negative three we end up with negative for a co signed two x minus four be signed two X plus six. A sign to x minus six. Uh huh coz I two X. Yeah. And now we can create a system of equations out of this. So we'll have negative four. Hey um minus six P. Equals to zero. And negative for actually gonna right the other way around six A minus four B equals to one. And so let's multiply the top by six and the bottom by four. So you end up with negative 24 A minus 36 B equals to zero and 24 A minus 16 B. Mhm. Because his sorry not zero close to four. Mhm. Now we can add the terms so that this will cancel out and you'll have 52 negative 52 the equals to four. So he's going to be negative 4 52. Which we can simplify, simplify down the negative 1/13. And so with that we can solve for A. So let's use the equation negative for A minus six P equals zero, negative for a minus six B. Of course to zero. And what's substitute in B. So you have negative for a minus six times negative 1 13 equals to zero. And if we move the term to the other side we get negative for A equals two negative 6/13 and A equals two. Sorry? Yes. Okay A equals two positive three divided by 26. And so our total solution is going to equal are homogeneous solution. Plus our particular solution, in this case we set our particular solution was um C one plus C two. Either the three X. And our particular solution is going to be 3 26. Co sign two X minus 1/13. Sign two X.

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.


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