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Verify that the given function is a particular solution to the specified nonhomogeneous equation. Find the general solution and evaluate its arbitrary constants to find the unique solution satisfying the equation and the given initial conditions.$$\frac{1}{2} y^{\prime \prime}+y^{\prime}+y=4 e^{x}(\cos x-\sin x)$$

Calculus 3

Chapter 17

Second-Order Differential Equations

Section 2

Nonhomogeneous Linear Equations

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This second, over perenitial equation of 1 half of y double prime plus y prime plus y, is equal to 4. It is supposed to of x minus sine x this equation and were given some particular solution that solve this equation is into d x. Force of this is a particular solution, particularly so we want to find that general solution. Giveth satisfies a t. 0 is equal to 0, and l at 0 is equal to 1 point, so the general solution is going to be the solution free, homogeneous equation. As this equation here particular where the particular solves the question itself, this question and the homogeneous is this equation cospatquations in equal to 0, so we can split it into the that is equal to 0 and the part is equal to it, because this is a linear, Linear uminso, first, let's check at this the solution in this in the shops, the question so here, let's see so we have this y prime, that the wet will be differentiated in this pores x, cos 2 tin there so differentiating the now. The first identical sine will have, through p d x times, minus sine sin, so that is why prime it can be written as 15 e 3 x cos minus now, waybilled pitapat differentiate for the exponential ateenth second part, and now are even these constant can differentiate. In this part, so for the vaterchen sine of x, the rations oopathi de going texnotamia and then have minus sine of x, minus sine of x, minus 2 sine of x minus 4. Note x now, so we plug in all this equation. But for the particular you have 1 half over that minus 43 x, sine x, 1, half plus y prime and y prime. Is that so, let's write that here down x, minus sine of x plus the function itself there is 33 x to so. This part here is half of idomprime. This part here is the eratio function, and this is the so. This will be equal to not zeronfactord 3 x on factor, indeed 2 and 3 x 2 to the x, and he is hale minus. So this will be this withdrawn into 2 minus 2 into the x, and so it will be minus sine minus sine of x and here plus cross minus sine o plus cosine. So, as we can see here, there is twice cosine and there is twice minus sine. So this will be 2 times. 23 x cos minus sine, and it is indeed his part. So these soles they a particular equation and now to solve them with genius or genus. We can do undetermined coefficients so for that in propose that the solutions softeform a combination of e to the r x, so we have half y o prime plus y prime plus y. This corresponds to the polynomial for r r squared half well. This is equal to 0. That'S got to solve, plus r plus 1 that equals to 0 and for conclusions. We can multiply by 2 on both sides, so this 0 remains 0 and there here you have r square plus, 2, r, plus 1 and so artifoni logan plus 2 n. This condition is you'll have a septosoyeteso. We have all. The solution is pollination by of this centre form a b c, the statius plus or minus square root of s, a boat of the solution. The fact of this, as r 1 plus case we're going to have her these and then these here with me, 4 minus 4 times in his it was that so this putting on a pandering as her plus 1 plus, i i'm sorry all g minus the thominus. Most nerve was on this. Spargo can be written like that ease multiply these. We have r square plus 1, plus i r plus 1 minus- i r, plus 1 square minus i squared, and this is equal to 2, and these are uses. Some of these solutions given by this is r minus minus 1 minus i under r minus minus 1 plus ci. So that's the hegemony have 1 some constant times minus x, 2, sine plus an ol constant from c to the minus x. So the complex is a trigonometrical function of the pi and then minus is minus 1, which is texnothe solution will be a tame minus x, plus b to minus 6 in 2 m d. L, of course, since we want that, the function at 0 is equal to 0, so at 0 we have in this form it 0. This is not an the only imagine at the half, so he becomes 1 so that this will be a we're going to have. This will be 0, so 00 plus 2 point this be so tobasco or so a is equal to minus 2 and the reality of the function that so tativof. This is going to be at the mines and now differentiate in the cosine minus 3 minus x. I i i i now for this second term retain the first 1. We have minus the tin mines now, the first that an the second 1 will have plus un intimius of cos. There is, for this will be seen to the x cos, plus 2 e t x of x minus 1, so that this etive evaluator at 0 is going to be here. We have minus a minus 8 plus on this part 0. This part will 0 and this would be 1 plus b, so minus a plus b, a the sea. This would be 2 plus 2 and then these other part would 0. So the the vertical 0 has to be equal to 1. So this will be equal to 1, but we know that is minus 2, so this equation imply. We have minus minus 2, that is 2 plus b plus 2 is equal to 1 point. So moving these over there, we have b is equal to 1 minus 4 minus 3 point. So the general solution will be. The general solution will be minus. The minus 3 e to the minus x sine of x minus 3, is equal to minus x, sound x, plus 2 x. So this is the solution of his find these conditions. The function at 00 preterition function of 0 is 1.

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