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(2 Marks-5 Minuttes] Jx) = c14F ox In * *> isa solution of tha nonhomogeneous De:Select one: 0 a J Zxy =- #2lx 0 6 ) -2y=--2n*0x~2y=-1 2 In *None of these[2 Mark...

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(2 Marks-5 Minuttes] Jx) = c14F ox In * *> isa solution of tha nonhomogeneous De:Select one: 0 a J Zxy =- #2lx 0 6 ) -2y=--2n*0x~2y=-1 2 In *None of these[2 Marks-5 Minutes] Tne DE xydy =y 2x1 dxSelect one : None of theseNonlinear in Y and Berooulll Homogeneous degree and non-#kuci Homogereous degree 2 and (neal In

(2 Marks-5 Minuttes] Jx) = c14F ox In * *> isa solution of tha nonhomogeneous De: Select one: 0 a J Zxy =- #2lx 0 6 ) -2y=--2n* 0x ~2y=-1 2 In * None of these [2 Marks-5 Minutes] Tne DE xydy =y 2x1 dx Select one : None of these Nonlinear in Y and Berooulll Homogeneous degree and non-#kuci Homogereous degree 2 and (neal In



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In each of the following problems, two linearly independent solutions - $y_{1}$ and $y_{2}-$ are given that satisfy the corresponding homogeneous equation. Use the method of variation of parameters to find a particular solution to the given nonhomogeneous equation. Assume $x > 0$ in each exercise.
$x^{2} y^{\prime \prime}+2 x y^{\prime}-2 y=3 x$, $y_{1}(x)=x,
y_{2}(x)=x^{-2}$

Okay, so just start this problem. We first are going to take a look at the roots of the exhilarate equation here, which is going to be p of our is equal to r squared. Plus, Omega, not squared, is equal to zero. So we're gonna have that r is equal to plus or minus I Omega. Uh, Omega, Not now. This part here has the route b is equal to omega eso if b is able to omega, Uh, but we need to take a look at the or compare the roots here so we'll have plus or minus i omega. So if omega is not equal to Omega, not so let's do that case first. Omega is not equal to Omega. Not so. If this is not Omega, not then our zp of tea is going to be equal to than just a not e to the Omega or Iomega like so. So this is going to have the derivative c p of t is equal to I unless should be Iomega t. So we'll have omega, uh, I or Iomega that a not e to the i omega t the men zP double prime of T is going to be able to the negative omega squared a not e to the I omega t like So let's plug these into our ah differential equation, which is going to solve our plus omega, not squared, is equal to is equal to. And then we're gonna have, um, on the right hand side, this is equal to f not. And then then e to the I omega t like so So Ah, oops, sorry. And then there should also be a zp here. So if we have negative omega squared Ah, so omega squared, um, a not e to the i Omega t and then plus omega not squared a not e to the I Omega T is equal to f not e to the I omega t Then we have omega not squared minus omega squared. A not is equal to f Not so a not is going to be able to f not over omega not squared minus omega squared. So we're gonna have our particular are complex valued particular solutions. E p of t is equal to f not over. Oh, Megan. Not square in minus omega squared. And then times e to the I Omega t so either the I omega t remember, we're gonna replace that, actually with co sign Omega T plus I sine omega t. Since we have the co sign part, we want the real part. So that's going to just be this here. So r y a p of tea is going to be equal to f not mining or divided by omega not squared minus omega squared cosigned omega t And this is for Omega not equal to Omega. Not so. That's for that scenario. Now we need toe Take a look at the case where omega is equal to Omega. Not so for Omega, equal to Omega. Not then That means our particular solution CP of tea. Then it's going to become a not tea e to the I Omega t like So So then zp prime t. Okay, so let's do that. So we're going tohave first times derivative. First time second is gonna be, uh, that and then plus, now we're going to have Iomega tea or so Iomega Times a not ah t e to the I Omega t. So, um, that is here. We took derivative of first time Second, then here we have derivative of second, which is the Iomega E to the Omega T times first, which is Iomega I not or a not tea. So the C double prime of tea this is equal to And then we have a not, uh, or sorry, i omega a not e to the i Omega t and then here we're gonna Ah, the derivative of this is just this. And then we multiply by Iomega. So we have plus Iomega a not e to the I Omega T and then plus I omega squared. That's going to become negative. Omega squared. Sorry. Negative. Omega squared. Okay, negative. Omega squared times a not and then ah t e to the I Omega t. Okay. Ah, we can now combine these two terms into just to be to Iomega a not e to the I Omega t and then minus We have our omega squared a not t e to the i Omega T. Next, we plug it into the complex value difference equation, which is E p uh, double prime plus. Sorry. Plus, and then Omega not is able to omega, so I'm just gonna write it as omega squared, and then zp is equal to R F not and then e to the i Omega t. So now plugging these into our ah, to this here we have to Iomega a not e to the i omega T minus omega squared a not t e to the i Omega T and then plus omega squared a not t e to the i omega T This is evil to f not e to the i omega t the these tube can now cancel out and we just have to Iomega a not it's gonna be able to f not when we also cancel out these terms here. So a not is kind of eagle two f not divided by two Iomega or this is also equal to F Um, sorry, I ah, negative. I f not over to omega. So zp of t is equal to, um negative I f not over to omega times t eats the Iomega t So this e to the Iomega's he I can replace with cosigned omega t plus I sine omega t And again, Since we're still taking the coastline part, we want the real part of this. So the real part of this is when we multiply this by this here. So this negative I becomes ice negative. I squared, which is positive. So why P of tea is gonna end up becoming okay. We have f not divided by to omega and then tea sign Omega t like so. And this is for Omega equals omega, not It's our final solution here.

X were like, You know what? Range eggs and you won. My wise people. My vehicle tax. I want to be one course. The one I'm back. One time axis, Teoh Extending three to one. I'm putting fear. We recognize three times B two, Primary zero. That's your here one. It is naked. What you want? Exit gate. Thanks. Extra 30 Here. So be one of the act is he called the hacks before? Well, tear C 80 access white Axel X squared Well minus X squared and right. Well, Z And then it will be X second, find four. Then why axes equal kill? I want C t X.

Right. So let's sort this problem off by solving for that module this solution. So I have R squared plus ar minus two. R equals to zero. And it looks like we can get our plus two times arm and so on Equals zero with some factory. And so this gives the solutions of our equaling one and -2. And with that we can build our homogeneous solution. So homogeneous solution is going to see one E. To the X plus C. Two E to the negative two X. And we can take a guess at our particular solution. So our guests for a particular solution is going to be X plus B plus see co sign X plus D. Sign Sorry, Co signed two x and decide to X. Right. And we have to take the driver of this particular solution guess twice. So the first drift is going to be a -2 C signed two x plus to the Co signed two x. And the second derivative is going to be negative four C. Co signed two x -4 d. sign two x. And right now we have to plug this into our original equation. Our original equation was wide of a crime plus Y. Prime minus two Y equals two X plus sign to X. And so we substitute into the original equation. So we'll have negative four C. Co sign two x -4 d. sign two X plus Y. Prime which is a minus two C. Signed two X Plus two d. co sign two X minus two times Y. So we're going to have to A X plus to be we're sorry this is a minus. So be a minus here um -2 C. co sign two x -2. The sign two X. And this all equals two X. Plus sign two X. And so let's try and sympathize out so we can have a cleaner system of equations. So I see life terms here and they write this in a different color. Actually you like terms here here here and here. Yeah. So We will end up with negative 60 Co signed two x minus 60. Sign two X plus a -2 C. signed two x plus two D. Co signed two x -2 X -2 B. Equals two X. Plus side two X. And so with that let's build a system of equations here. So let's see if we can get some like terms let's have the co signs first. So we're gonna have negative six C. This term this term contains of co sign so negative 60 plus duty equals to zero. And let's do signs a negative 60 minus two C equals to one. And then let's do this term right here. So we'll have negative to a. He goes to one and finally we'll have a minus to B equals to zero. Right so right off the bat we can solve for a here. So we'll have a equals 2 -1 half. And with that we can solve for the bottom equation. So we'll have a negative one half minus to B equals to zero. Bring the one after the other sides will have negative to be equals to one F. Or b equal to negative 1/4. And now let's try and solve for the top two equations. So let's focus on those um negative 60 plus two D. Equals to zero And -2 C -60. He goes to one. So let me multiply the Top equation by three. We'll have negative 60 plus two D. Equals to zero negative six C minus 18 D. Equals 23 Let's subtract these terms. We can eliminate the seas And we'll have 20 d. Equals 2 -3 or D equals the negative 3/20. And with that we can solve for C. So we have that negative 60 Plus two times negative 3 20 equals to zero. And so then we have negative 60 plus. Actually it's gonna minus six. Or other sorry 3/10 equals to zero. And let's bring the 3/10. The other sides will have negative 60 Equals at 3/10. And if we divide by negative six we get C equals two -1/20. Right? And now we can build um are total solution. So our total solution consists of our particular solution closer homogeneous solution. So we're going to have c. one E. To the X plus C. Two E. To the negative two. X minus one. Fourth minus one half x minus 3 20. It's signed two x minus 1/20. Co sign two X. And with that we can solve for C. One and C. Two using our initial conditions. So our initial conditions are Why have zero equals 1 And why prime of zero go 0. So let's start off by taking the direct route of this equation at the top. So we have that Y. T. Prime equals two. C. One E. To the x minus two C. To eat the negative two X -1 4th. or rather sorry this is a constant. So it cancels out. So I have minus 1/2 minus 3/10. Sign sorry Co sign two X plus 1/10. Sign two X. And so our initial condition for the first equation is Y zero. So it's plug in Y. Of zero in the first equation. So we'll have that see one plus C two minus 1/4. And when you plug in zero for sign it gives you zero. So we'll have that turns to zero and we'll have minus 1/20 That are equals to one. And her second equation is going to be when we plug in zero. So I have C1 -2. C two minus one half minus 3/10. All equals to zero. Right so now we can solve for C1C2 with the system equations. So let's start off by simplifying these constant terms. So we'll have that C one plus C two equals two, 3/10. And for the bottom equation we have C one minus two. C two equals two. 8/10. Uh huh. Right so now Let's subtract this so we can cancel at c. one. So I have 3 C2 Equals, two, or 10. Where that C2 Equals so negative 5/30. Sorry about that. I think I made a mistake here actually in the math this should not equal 2 3/10. In fact It is 1 -3/10. So it's simply going to be 7/10. So with that everything else should work out. So have This being 7/10 And the bottom occasional some fly down to see one -2. C two equals two. Excuse me, I made a mistake again. It's not 7/10 is 3/10 plus 20th or one equals 13 10th. All right. And the bottom occasional simply down to 8/10. So with that let's do some subtraction to cancel. Let's see one. So we'll have that three C two equals two five over 10 and then C2 equals two 5/30 Which sympathize down to one six. So with that we can solve for C. One. So let's use that C one plus C two equals 2 3/10. So I have that C. One plus C two Equals 2/13/10. And let's move the 16 The other sides will have the 13 10th minus 16 equals to see one. And so C1 will equal to 39/30 minus 5/30 which is 34/30 which flies down 17/15. Right? So now we can actually build our total solution which let's remind ourselves is the sum of the homogeneous solution plus the particular solution. So our total solution here it's gonna be 1/6. He to the -2 x plus 17 15. E. To the X plus or rather minus 1/4 minus one half. X minus 3/20. Signed two X minus 1/20. Co sign two X. Right? That's it.


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