The quantity X ^2 + 1 Y prime plus three X ^3 y = 6 X eat in the -3 halves X ^2 given the initial condition of Y of 0 equaling 1. So we want to get our dy DX and our single Y all on one side, and we want our dy DX to have A1 coefficient. So we're going to divide everything 3 by X ^2 + 1. Then we're going to take the coefficient on the Y and we're going to make that be our row. So our row is going to equal E to the integral of three X ^3 / d / X ^2 + 1 DX. If we let U equal X ^2 + 1, then we can see du is 2X DX or 1/2 du equal X DX. We can also see that X ^2 is really U -, 1. So if we pull out the three in front on the integral and we take that X ^3 and split it up into X ^2 times XDX because I needed that XDX to be 1/2 du. Then when we do our substitution, we can see we have E to the three halves. The integral instead of X ^2, we're going to put in U -, 1. Instead of the XDX we're going to put in du with the 1/2 we already put in front. All of that over U. Simplifying our U / U is one, and our 1 / U is going to just end up being our -1 / U. Doing the integral, we get three halves U -, 3 halves natural log of U. Taking out EU and putting in what it equaled, we'd get E to the three halves X ^2 + 1 -, 3 halves the natural log of X ^2 + 1. Now when I have two things in the exponent that are being subtracted, I can rewrite them as each being to the base with a division. So we're going to get E to the three halves X ^2 + 1 / e to the three halves natural log of X ^2 + 1, and we're going to take that three halves coefficient up into the exponent by the power rule. So the E to the L1 is going to cancel and we get X ^2 + 1 to the three halves in the denominator. But to move it up to the numerator, we would just make it to a -3 halves. So we end up with X ^2 + 1 to the -3 halves, E to the three halves times the quantity X ^2 + 1. So now we're going to go back to that original equation here, and I'm going to multiply everything through by this X ^2 + 1 to the -3 halves E to the three halves X ^2 + 1. And that's going to get multiplied by the dy DX. When we do it times the next term, we're going to get all of this. And then when we do it over to the right, we're going to end up with the E to the -3 halves X ^2 and E to the three halves X ^2 cancelling, leaving us just an E to the three halves times one. So now the left hand side is really going to turn into the derivative of a product. So the derivative of the quantity X ^2 + 1 to the -3 halves, E to the three halves, the quantity X ^2 + 1 all times Y, remember it's really our row here times the Y. Our right hand side simplified to 6E to the three halves. Remember those are all constants X times the quantity X ^2 + 1 to the -5 halves. So now we're going to take, by the fundamental theorem of calculus, we're going to take the integral of each side. The integral of a derivative is going to cancel. So we're going to get left whatever was on the inside of that derivative on the left. Now we're going to take the integral on the right, the six E to the three halves a constant. So I'm going to pull it out in front. We get the integral of X / X ^2 + 1 to the five halves DX. If I let EU be the inside piece again, so U is X ^2 + 1, du is 2X DX, 1/2 du is X DX. So the X DX in the numerator is going to turn into 1/2 du. So 1/2 * 6 is 3 out here, the du here. Then we get U to the five halves in the denominator evaluating the center girl, IE taking one more to the exponent, we're going to end up with U to the -3 halves multiplying by its reciprocal. So negative 2/3. And then we had the three E to the three halves is our constant in front a moment ago plus C. The threes are going to cancel. So we get -2 E to the three halves, the quantity X ^2 + 1 to the -3 halves plus C, getting the Y by itself. We're going to divide everything through by that X ^2 + 1 to the -3 halves, E to the three halves X ^2 + 1. And we're going to get lots of this. Now, if I simplified it here, I could realize that this E to the three halves and this E to the -3 halves times one is going to cancel, leaving me just an E to the X ^2 in a moment, E to the -3 halves X ^2, which is what I put down here and here. We're going to then put in zero for our XS everywhere and one for our Y. So when we put in zero for our XS and one for our Y, we can see that we end up with E to the three halves and E to the -3 halves, which is going to cancel, leaving us just a -2. If I add the negative to the other side, we're going to get three. And this term down here is going to turn into CE to the -3 halves. So to get C by itself, I'm going to multiply each side by E to the three halves, and we get C being 3E to the three halves. So now I get Y equal to the -2 E to the -3 halves X ^2 and plus the C But the C is now 3 E to the three halves X ^2 + 1 to the three halves times E to the -3 halves X ^2 + 1. So this E to the three halves now, and this E to the -3 halves plus the one is going to cancel. So our final equation is going to be Y equal -2 E to the -3 halves X ^2 + 3, E to the -3 halves X ^2 down here times the quantity X ^2 + 1 to the three halves.