#8 on section 55Y double prime -4 Y equal hyperbolic cosine of 2X. So the first thing we're going to do is we're going to take that left hand side and figure out what our Rs would be or the exponents on our E term. So we have r ^2 -, 4 = 0. We can see that that's positive. And -2 so our Y sub C is just C1 E to the negative 2X plus C2 E to the 2X. Now in this example, we can see that these YS here don't have any duplication with what was on the right hand side of the equation. Hyperbolic cosine of 2X doesn't repeat with the C1 E to the negative 2X or C2 E to the 2X. So we're going to have our Y sub T equaling a hyperbolic cosine 2X plus B hyperbolic sine 2X. Remember, just like with our regular trig functions, that they're always going to come with conjugates. So hyperbolic cosine has to be paired with hyperbolic sine. We're going to find our YT prime and our YT double prime. And then we're going to stick that back into the original equation to try to solve for our A and our B. So once we do this, we're going to get, let's Scroll down. We're going to get that this is going to actually have zero on the left side because 4A hyperbolic cosine two X -, 4 a hyperbolic cosine 2X and 4B hyperbolic sine two X -, 4 B. That's not going to actually be able to equal the hyperbolic cosine 2X. Remember that the definition of a hyperbolic cosine is E to the two X + e to the -2 X all over two. Actually, whatever is here is what's in the exponent and then the opposite. So if we thought about solving that, we could see that we could cross multiply by two. We could make this E to the -2 X 1 / e to the two X. We could multiply everything through by E to the 2X, subtract the one, and we can see that over the real numbers. We can't take the natural log of a -1. So that's not going to be possible. So now we're going to try if it was a second term or duplicate term of just the hyperbolic cosine and hyperbolic sine. So we're going to do it with a power. In this case, we're going to have AX hyperbolic cosine 2X plus BX hyperbolic sine 2X. If this one doesn't work, we would go AX squared hyperbolic cosine 2X plus BX squared hyperbolic sine 2X. So when we're doing this, we get our YT prime, our YT double prime. We can see that there are some terms that we can combine. 2A hyperbolic sine 2X and 2A hyperbolic sine 2X is going to give us four of them. Same thing with the hyperbolic cosine 2X's. So when we substitute that back into the original equation again, we can see that our Y double prime -4 Y equal hyperbolic cosine 2X. Literally just taking our Y double prime from a moment ago. Let's see if we can get it all on here. So here's our Y double prime that I'm going to put in right here and then -4 ry so here. And we have it equaling hyperbolic cosine 2X. So when we do this, we can see that this +4 AX hyperbolic cosine 2X is going to cancel with this -4 AX hyperbolic cosine 2X. And this +4 BX hyperbolic sine 2X is going to cancel with this -4 BX hyperbolic sine 2X. So that's going to leave us this 4A hyperbolic sine 2X equaling no hyperbolic sines over on the right, so zero. And 4B hyperbolic cosine 2X is going to equal hyperbolic cosine 2X. So now we're going to solve for A&B. If I divide each side by hyperbolic sine 2, XI can see that and divide by 4. I can see that a is 0 and for this one the hyperbolic cosine 2 XS are going to divide out. I'm going to get 4B equal 1 or B is 1/4. So our Y particular is going to be 1/4 X sine hyperbolic sine 2X. Because if you recall, we had B being with the hyperbolic sine term, so BX hyperbolic sine 2X so 1/4 X hyperbolic sine 2X.