5.5 #10 Y double prime +9 Y equal 2 cosine three X + 3 sine 3X So on the left hand side we're going to figure out what our Rs are. R ^2 + 9 = 0 so R equal positive -3 I. So our Y sub C is equal C1 E to the 0X because there was no real portion cosine 3X plus C2 E to the 0X sine 3X. Remember our trick functions always come as conjugates. So cosine 3X and sine 3 xe to the 0X is just one. So this simplifies up to C1 cosine 3X plus C2 sine 3X. Now that's duplicated on the right hand side and we need to not have duplicated terms. So what we're going to do with our Y sub T, instead of having it be a cosine 3X plus B sine 3X, we're going to multiply it by an X so that we don't have any duplicated terms. So this AX cosine 3X is now not repeated here because there was no X in the YC. And the BX sine 3X is not repeated here because there was no X in the C2 term. So we find our YT prime and our YT double prime, and then we substitute that back into the original equation. So once we find our YT prime and our YT double prime, we put it back into the Y double prime +9 Y equal to cosine three X + 3 sine 3X. I'm going to group together all of my sine terms. So I have -6 A -9 BX over here. We have the plus this nine we have to distribute, so nine BX. So the -9 BX and the +9 BXS are going to cancel. So we really end up with -6 a sine 3X equaling 3 sine 3X because we have the three sine 3X on the right hand side. We can see that A is just negative 1/2. We're going to do the same thing with the cosines, so -9 AX cosine 3X6B plus the nine that gets distributed. So then +9 AX and the -9 AX cancel. So we just end up with the 6B cosine 3X equaling the two cosine 3X. So we can see B is 1/3. So our Y of P, our Y particular is just -1 half X cosine 3X plus 1/3 sine 3X. Because remember, our Y of T was this ax cosine 3X plus BX sine 3X. Our YT and our YP are directly related. Our YP is just when we've solved for A and for B.