5.5 #9 we have Y double prime +2 Y prime -3 Y equaling one plus xe to the X. So we have r ^2 + 2 R -3 equaling 0R plus three r - 1 = 0. So we get our Y of C being C1 E to the negative 3X plus C2 E to the X. So in this case, we have duplicate terms because if we had our Y of T, we could see it would have to be A + B E to the X plus Cxe to the X. And this BE to the X would duplicate the C2 E to the X. And we need to have our YC and our YT not have duplicate terms. So what we're going to do is we're going to take this portion here and we're going to multiply it by another X. So now we're going to get bxe to the X plus CX squared E to the X. So our YT turns into this, and now we don't have any duplicate terms. I don't have just a plain E to the X and my Y of T, and I don't have an E to the -3 X anywhere in our Y to the Ty sub T. So now we're going to take our YT prime and our YT double prime, and we're going to substitute it back into the original. So our Y double prime +2 Y prime -3 Y equal 1 plus xe to the X. So substituting all of that information in then I'm going to start grouping things. So I have a -3 a that equals one that's my only constant on the left and my only constant on the right. When I keep going, I'm going to look for my E to the X terms. So here's a 2B E to the XA2CE to the X. I've got to distribute this to here SO2B E to the X. So I'm going to get 4B E to the XS +2 CE to the X equaling 0 because there were no E to the XS on the right hand side. If I divide that whole thing through by E to the XI, get 4B plus 2C equals zero. Actually, I could have divided by 2 E to the X. Now that I look at, it doesn't matter. So then I'm going to go with the XE to the XS. So Bxe to the X4, Cxe to the X. Once again, remember to distribute the two, two Bxe to the X and two times this 2SO4. And then we have this -3 of those also. So we end up with just eight Cxe to the X equaling on the right hand side, we have one XE to the X. If I just divide each side by xe to the XI, get 8C equal 1. And then if I go in and do the CX squared ex term CX squared EX, and here's distributing the two, two more of them, but now we're subtracting three of them. So those X ^2 e to the X cancel. So from these three equations we can see a is negative 1/3. This 8th C equal 1, so C is 1/8. And then once I get my C, I'm going to substitute it back into here and I get 4B plus 2 * 1/8 = 0 or B equal negative 116th. So putting it back into my Y of T, it was a negative 1/3 + b. But in this case B is negative. So negative 116th XE to the X + C one 8th X ^2 e to the X. So that's our Y of T term.