高数--微分方程

令y = u²-x²，则dy = 2udu-2xdx
原式可化为2u du/dx - 2x + x = u
即2udu = x + u
再令t = x/u，则du = d(x/t) = (t - t'x)/t²
代入上式，得到
2(t - t'x)/t² = t + 1
2t - 2t'x = t³ + t²
2xdt/dx = 2t - t² - t³
2dt/(2t - t² - t³) = dx/x
积分得
lnt - 1/3 * ln[(t+2)(t-1)²] = lnx + C
把t = x/u = x/√(x²+y)化简得
[x + 2√(x²+y)][x - √(x²+y)]² = C