When trying to find this original equation from the differential, we can see that we have YY squared and dy DX, but we only have an X. And if we thought about solving this in terms of DXDY, we would have a DXDY and then an X with some sort of a coefficient. So we're going to start by multiplying each side by DXDY, and then we're going to take all of the XS to one side. So we get the quantity of 1 + y ^2 DX dy -2 Y times our X equaling one. Now we're going to try to make this left side into a derivative in terms of Y that we can use the product rule or the quotient rule. This case we're going to start by dividing everything 3 by 1 + y ^2. So we get A1 coefficient on that DX dy. So 1D XD y -, 2 Y over 1 + y ^2 X equaling 1 / 1 + y ^2. We're going to let our row be the coefficient on the X term. So row is going to equal E to the integral of -2 Y over 1 + y ^2 dy. We could do AU sub here if we don't see that evaluation easily. If we let U equal 1 + y ^2, our du is going to equal 2Y dy, so we're going to get E to the integral of -1 / U du. So that would turn into E to the negative lnu. Taking that negative up and top into the exponent, we'd get E to the lane of U to the -1, which would be 1 / U. But remember, U is just 1 + y ^2. So now we have 1 / 1 + y ^2. Going back to our equation, we're going to multiply all three terms by 1 / 1 + y ^2. So we get 1 / 1 + y ^2 d XD y -, 2 Y over the quantity 1 + y ^2 ^2 X equaling the quantity 1 / 1 + y ^2 ^2. The left hand side is now the product of a derivative. So the derivative in terms of Y of 1 / 1 + y ^2 X, and that's going to equal the quantity 1 / 1 + y ^2 ^2. By the fundamental theorem of calculus, we're going to take the integral, and if we integrate one side, we're going to integrate the other and the integral and a derivative are going to cancel. So the left side's just going to turn into 1 / 1 + y ^2 * X. The right side, we have the integral of the quantity 1 / 1 + y ^2 ^2 d Y. That may not be one that we know off the top of our heads, but we can see that that's really in the form of a Pythagorean identity of a ^2 + b ^2 = C ^2. So we can do a trig substitution. If we think about our Y being the opposite side and our one being the adjacent side, we'd get tangent Theta equaling Y. Taking the derivative of each side, we get secant squared Theta D Theta equaling dy. Now we can see by Pythagorean theorem that our third side would be sqrt 1 + y ^2. So substituting N instead of this y ^2, we're going to put in tangent squared Theta. So 1 / 1 plus tangent squared Theta. That's quantity squared. Instead of DY we're going to put in secant squared Theta D Theta. Now remember one plus tangent squared Theta is really secant squared Theta. So we get secant to the 4th Theta in the denominator and a secant squared Theta in the numerator, which would live as one over secant squared Theta or just cosine squared Theta. We don't know how to do an integral of cosine squared Theta, D Theta. But from our Pythagorean identities, actually from our trigonometric identities, we know that cosine squared Theta is really just one plus cosine of twice the angle two Theta over 2D Theta. So we'd get 1/2 Theta plus 1/4 sine of the quantity 2 Theta plus C this Theta. Going back to our substitution, our Theta is going to be tangent inverse of Y, so 1/2 tangent inverse y + 1/4. We don't have a two Theta angle, so we need to rewrite sine of two Theta as an equivalent and sine of two Theta once again is just a trig identity. It's two sine Theta, cosine Theta, the sine Theta on our triangle opposite over hypotenuse y / sqrt 1 + y ^2 cosine adjacent over hypotenuse 1 / sqrt 1 + y ^2 + C. So then when we multiply these two terms together, we can see that the square roots are going to go away and we're going to end up with y / 2 times the quantity 1 + y ^2. Now remember that really equaled 1 / 1 + y ^2 X. So to solve in terms of X, we're going to multiply everything 3 by 1 + y ^2 and we get X equaling 1 + y ^2 / 2 tangent inverse y + y / 2 plus C times the quantity 1 + y ^2.