On #13 XY prime equal Y plus the square root X ^2 + y ^2. Remember we're going to let Y equal VX and dy DX is then going to equal by the product rule v + X DVDX. So we're going to have X instead of Y prime. We're going to get V + X DVDX equaling instead of YVX plus the square root X ^2 + v ^2 X ^2. So if we keep going with this, we're going to distribute, we're going to get XV plus X ^2 d VDX equal VX plus. We can pull out an X, so we get X square root 1 + v ^2. So the VXS on each side or XVS are going to cancel and we're going to end up putting all the XS on one side and all the VS on the other. So we're going to get DV divided by sqrt 1 + v ^2, and that's going to equal 1 / X or DX over X. So now when we integrate each side, we're going to take and integrate DV over sqrt 1 + v ^2 and we're going to integrate DX over X. Now our right side is pretty simplistic. It's our left side that's going to cause us some problems. So let's think about doing a trig substitution. So if we take our triangle, because it's a plus, we're going to have V here and one here. So tangent Theta is going to equal VS secant squared Theta D Theta equal DV. So when we continue on this, we're going to end up with secant squared Theta D Theta over sqrt 1 plus tangent squared Theta equaling the integral of DX over X is the natural log of X + C. So sqrt 1 plus tangent squared Theta is actually going to give us just plain secant Theta because one plus tangent squared is secant squared and the square root and square cancel. So we get secant Theta D Theta equal the natural log of X + C. Now, we actually may not recall, but we know the integral of secant Theta D Theta. But let's pretend like we don't remember. The way we're going to get it is we're going to think about taking the secant Theta and multiplying the top and the bottom by secant Theta plus tangent Theta. When we actually figured out what this was equal to, this is the steps that we used. So if we multiply by secant Theta plus tangent Theta on each side or not each side but top and bottom because it's plus 1 * 1, then when we distribute we get secant squared Theta plus secant Theta tangent Theta all over secant Theta plus tangent Theta. And this is actually going to be another U sub. If we thought about letting U equal secant Theta plus tangent Theta, then DU is secant squared secant Theta, tangent Theta plus secant squared Theta. That's all times D Theta. So now we're going to get this to turn into one over udu equaling the natural log, the absolute value of X + C One over U we know is the natural log of the absolute value of U. So this is going to turn into the natural log of secant Theta plus tangent Theta. And we could think of that C as some natural log of let's call it C1 perhaps. So now we could think about the natural log of secant Theta plus tangent Theta equaling the natural log of C1X. So we could drop the natural log of each side and then secant Theta plus tangent Theta. Well the secant Theta from our triangle back here secant Theta, the third side would have been sqrt 1 + v ^2. So secant Theta is going to be sqrt 1 + v ^2 / v no over one. And then the tangent Theta we already knew was V. So when we come back here, we're going to get sqrt 1 + v ^2 + v equaling C1X. And then the V is going to get replaced with y ^2 / X ^2 + y / X = C One X. So now over here we could get a common denominator X ^2 + y ^2 / X ^2 + y / X = C One X square root would come out of square root of the X ^2 would come out. So we get sqrt X ^2 + y ^2 over plane X + y / X = C one over X. So if we multiplied everything through by the X to get rid of the denominator sqrt X ^2 + y ^2 + y = C One X squared, we can't do much more than that As far as trying to get the YS together. We could possibly take this over and foil it, but then we would end up doing the quadratic, so that's actually an OK answer right there.