Y ^2 y prime +2 XY cubed equals 6X. This is 1.6 #20. The first thing we want to do is to get this Y prime all by itself. So we're going to divide everything through by y ^2. We get Y prime +2 XY equaling 6 XY to the -2. Now when we do our V substitution, because we have a single Y here, but we have AY to the -2 here, we're going to leave the single Y portion alone for just a minute. We're going to let our V equal Y to the one minus whatever this exponent is over here. So 1 - -2 is going to give us y ^3. Now when we take the derivative in terms of X of both sides, DVDX is equaling three y ^2 d YDX or DYDX is 1 / 3 Y squared DVDX. So I'm going to take out this Y prime and I'm going to put in what it equal. So we have 1 / 3 Y squared DVDX +2 XY equaling 6 XY to the -2. Now I want to get that DVDX by itself. So I'm going to take each of these terms and multiply it by three y ^2. If I multiply each term by three y ^2, what'll happen on this far right term is the Y to the -2 and the Y ^2 are going to cancel. So we get DVDX plus six XY cubed equaling 18X. Now this y ^3 a moment ago we said was really just our V. So that's why we do that substitution the way we did. So now we have DVDX +6 XV equaling 18X. Next we're going to use our row. So we're going to say row is going to equal E to the. Remember, it's whatever is being multiplied by the V in this case, so 6 XDX. So this is going to be E to the three X squared. And then we're going to take that and we're going to multiply everything through by E to the three X squared DVDX plus six xe to the three X ^2 v equal 18 xe to the three X squared. Now the whole purpose of that was so that this left side turns in to the derivative in terms of X of E to the three X ^2 * v We can double check that by saying the derivative of this V in terms of X is DVDX here. So the derivative of this one times the other plus. Now the derivative here, we have to do the exponent, so we get 6X, which is that piece times E to the three X squared all times the other term. So that's going to equal 18 xe to the three X squared. Now using that fundamental theorem of calculus, we're going to take the integral of each side in terms of DX and the integral and the derivative are going to cancel. So on the left side we're just going to get E to the three X ^2 v, and on the right side, if we needed to, we could think about AU sub letting U equal 3 X squared. So DU would equal 6X DX or 16 DU equal X DX. There's my X and there's my DX. So I'm going to place replace that with 16DU, so 18 * 1/6 du. And that leaves me this E to EU. So this is going to be 3 E to EU + C but I can then replace that U back to three X squared. And over here we're going to replace the V with Y to the third, because remember back here we said V was Y to the third. If I divide everything through by E to the three X squared, I'd get y ^3 equaling 3 plus CE to the -3 X squared. And then if we wanted to solve for Y, we could say Y is the cube root of 3 plus CE to the -3 X squared.