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10. Solve the initial value problem for the nonhomogeneous linear system of differential equationsd"u ~2 11 -32teAt+ 3t2 e4t+4t8e2 1 ~139 8u(0)...

Question

10. Solve the initial value problem for the nonhomogeneous linear system of differential equationsd"u ~2 11 -32teAt+ 3t2 e4t+4t8e2 1 ~139 8u(0)

10. Solve the initial value problem for the nonhomogeneous linear system of differential equations d"u ~2 11 -3 2teAt + 3t2 e4t +4t8e2 1 ~1 39 8 u(0)



Answers

$1-10$ Solve the differential equation or initial-value problem
using the method of undetermined coefficients.
$$9 y^{\prime \prime}+y=e^{2 x}$$

Were given there 2nd order differential equation. Full wipe prime plus 12 Y. Prime plus nine Y secret is you. And Find the General Solution 1st. We have to right the characteristic equation which would be given by four hour squared plus of our plus nine Is what is zero. And we are going to find the roots of this equation to do that. We need to numbers that give us a product of A C. Where are for is a RPS 12 or 6 9. So our A C. Will be four x 9 with his 36. And some of the we used to go off And this will be six and 6 which To give us a sum of 12. So replacing to come up with six hour plus six hour Plus nine here is equal to zero. We can factor Our equation. And here we picked out a two hour Left. two plus 3 plus. Mhm. To her plus three is equal to zero and two plus three squared. He's going to see you. So here we have the case of the repeated roots Which is 3/2 and it's repeated or double route. Now we know that if we have a type of route are general solution will be in the form of whitey. Is you got to see one E. To the power of our tea Plus c. two T. E. To the power of our T. And substituting our route On Arab we have C1 E. to the power of -3 of what to T plus C two T. He to the power of -3/2 T. And this will be our final general solution to this differential equations.

So it starts problem off by solving for the homogeneous solution. So we should get r squared minus four. R plus five equals to zero. And I don't see any obvious factoring here. Some use the quadratic formula, so negative B plus or minus the square root of B squared 16 minus four times a time. See all divided by two times A. And you should end up with two plus or minus square uh negative four divided by two, which is two plus or minus two. Sorry, Plus your mind's eye. And so with that we can build our homogeneous solution which is going to be C one each of the two X co sign of X plus C two. Either two X. Sign facts. And our guests for particular solution is going to be A. E. To the negative X. And we have to take the driver that twice. So we should end up with negative A. E. To the negative X. The second derivative is gonna be A. E. To the negative X. And so we can plug that into our original equation. And let's remind ourselves that is Y double prime minus four. Y. Prime plus five. Y equals E. To the negative acts. So we end up with A. E. The negative X minus. Sorry. Plus for a. You hear the negative X plus five A. Into the native X equals to eat the negative acts. And so can do a little bit of simplification here. We should end up with 10 A. Eat the negative X. Because even the negative acts, we can cancel these terms. So we should end up with 10 A. Equaling one Andy equaling 1/10. And with that we can build our total solution which is a combination of a particular solution are homogeneous solution. And we said initially that are homogeneous solution was C one E. To the two X co sign X plus C, two, each of the two X, sign X. And just now we found our particular solution to be 1/10 E to the negative X.

So let's begin this problem by substituting out the Y prime, Y. Prime. And why? So we'll get our square plus. You are minus eight equals zero. So this is for the homogeneous solution and we're gonna do a little bit of factoring here and we should get our plus four and our minds to equals zero. That gives us our roots of negative four two. So we can build our homogeneous solution here. Why homogeneous equals two C, one E. To the negative four X plus C. Two E. To the two X. And now, well, you can start finding a particular solution. So I guess for the particular solution it's going to be in the form X squared plus bx. Let's see. And when you take the derivative this twice. So first there would be to A X plus B and Y double prime will be to A. And now we can plug it into original equation. And so in doing so we get to a of course two times to a X plus B minus eight times A X squared plus bx or C equals to one minus two X. Word. So let's do some algebra here we're gonna put to A was for A X. Plus two B minus eight A. X squared plus eight. The X plus eight. See sorry these are minus is actually that equals to one minus two X. Squared. And so we will have to a plus to be minus eight C equals to one for A X. We're sorry for a minus eight P. Okay equals to zero. And we'll also have negative eight A. Equals two negative two. And so we'll start off the last equation we can get a equals 1/4 and we can plug into our second equation we'll have the equal zero this is one. So we can bring this to the other side and be should be 18 And then using these two, we can solve for C. We should have Mhm this and that should give us a value of C equaling negative one or 32. And so our total solution is equal to the homogeneous solution plus the particular solution. So it's going to be see one each of the two X plus C. To eat the negative four X plus 1/4 X squared plus 1/8 x minus 1/32.

Here we have a second order differential equation four, y prime prime -4, y prime minus three, Y Is equal to zero. And characteristic equation will be given by four. R squared minus for uh minus three Is equal to zero. And here to figure this equation, we have to look for two numbers that give us a product of -12 which we have found from applying a N C and giving us a sum of P which here is minus four. Now we know that -6 and positive to give us a product of minus wolf In the summer of four. So using those Here we have four hours squared. My last six are Plus two AR -3, score to zero and putting some preliminary practice here and featuring out the highest common factor We have here, we left to R -3 Plus one, figuring out 1, 2 -3 is equal to zero and we have to ara plus one to minus three is go to zero and I will be equal to minus half and our is go to 3/2. Now these our roots and uh general solution will be in the form Y T. Since these tools are real and distinct, we'll see one E. To the power of the pharaoh won t Plus C2 E to the power of Arab too two. And our final solution will be a general solution With C one E to the power of era which is minus half T plus C two E to the power of 3/2, two. And this is our final solution.


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