55 number 5 so Y double prime plus Y prime plus Y equals sine squared X. So if I let my left side r ^2 + r + 1 = 0 and solve for the RI get R equal negative 1/2 ± sqrt 3 / 2 * I. So my Y of C would be C1 E to the -1 half X cosine root 3 / 2 X. And remember, because it's conjugate with the imaginary, if I have cosine, I have to have sine. So plus C2 E to the negative half X sine root 3 / 2 X. Now also recall from our trig days, sine squared X is really just one minus cosine 2X over two or 1/2 -, 1/2 cosine 2X. So the 1/2 here doesn't repeat because there's not a single constant in RYC. And the cosine 2X doesn't repeat because this is cosine root 3 / 2 X so there's no duplication. So our Y of T is going to be a because we had a constant plus B cosine 2X. But if we have a cosine because of the conjugate, we have to also have the sine. So B cosine 2X plus C sine 2X. Now if we find our YT prime and our YT double prime and then stick it back into the original equation, then we're going to group together our constants. So A equal 1/2. Then we're going to group together all of our cosines. So -4 B cosine 2X2C, cosine 2 XB cosine 2X, equal a negative 1/2, cosine 2X. So then if I take that whole equation, divide by cosine 2 XI, get -3 B +2 C equal negative 1/2 grouping together my signs, I don't have any signs on the right hand side, so that would equal 0. If I divide that whole equation through by sine 2X, I'd get -2 B -3 C equals 0. If I multiply the top equation by two, negative C6B plus 4C equal -1. Multiply the bottom equation by negative. 36B plus 9C equals zero. Add them together. 13C is -1 C is -113. Then I just substituted it back in to solve for B so -2 B -3 * -113 negative times negative is a positive, but when I take it to the other side it's subtracted. Divide by -2 negative. Divide by negative is positive. The two double S the denominator. So my Y of P equal the a term 1/2 plus the B term 326 cosine 3X plus the C term or -113 sine 3X.