5.5 #13 Y double prime +2 Y prime +5 Y equal E to the X sine X. So the first thing we need to do is figure out if there are any duplicate terms. So we're going to take that left side substitute in our Rs and figure out what that would be for our Y sub C. So when we do this we can see R is -1 ± 2 I by using the quadratic formula. So our Y sub C is C1 E to the negative X cosine 2X plus C2 E to the negative X sine 2X. Remember that when we have imaginaries they always come in conjugate. So we have cosine and sine. The real number is the E to the real number portion times X. The imaginary is the cosine and sine portions with the coefficient. So there's no duplicate terms When I look over here at E to the X sine X because this is sine 2X, not sine X. So we're going to have our Y of T equaling AE to the X sine X + b E to the X cosine X. So then I'm going to find my YT prime and my YT double prime. And I'm going to realize that with my YT double prime, a couple of my terms are going to cancel my aex sine X and also my bex cosine XS are going to cancel. So then I take the original equation and I substitute back into it my Y double prime +2 of my Y primes plus five of my YS. And that has to equal my right hand side, which is E to the X sine X. So from there we're going to match up our coefficients and we're going to realize that we have two AE to the X cosine X and another two of them here if we distribute. So we have 4 AE to the X cosine X, and then I have +2 BE to the X cosine X and another 5B E to the X cosine X for seven of those. Then I'm going to do the same thing for my E to the X sine XS. And if I group them, I can see I have 7AE to the X sine X -, 4 BE to the X sine X equaling in this case the E to the X sine X because we have to have the same terms equaling the same terms. So on the top equation we can we can divide everything through by EX cosine X and end up with 4A plus 7B equals zero. On the second one, we're going to divide everything by EX sine X and we're going to get seven A -. 4 B equal 1. I'm just going to do elimination method. So if I start by getting rid of the B's, I'm going to multiply the top one by four and the bottom 1 by 7, add them together, and I can see that A is going to be 760 fifths. If I come back, I don't want to substitute 760 fifths in for A because that just doesn't look nice. So I'm going to do elimination again, this time getting rid of the A's. So I'm going to multiply the top equation by 7 and the bottom 1 by -4 and I get B equal -460 fifths. So we're going to get our Y of T equaling -460 fifths E to the X cosine X + 760 fifths E to the X sine X. We usually put these in alphabetical order, so cosine comes before sine.