微分方程xdx+((x^2)•y+y^3+y)dy =0 的通解

x^2+y^2+1看成一个整体,y^2+ln(x^2+y^2)=C

0 = xdx + (x^2y + y^3 + y)dy = xdx + x^2ydy + (y^3 + y)dy,
xdx/dy = -x^2y - y^3 - y
x = e^u, dx/dy = e^udu/dy, u = lnx,
e^u*e^udu/dy = -e^(2u)y - y^3 - y,
du/dy = -y -(y^3 - y)e^(-2u),
dt/dy = -y, t = -y^2/2.
z = u - t, u = z + t,
dz/dy = du/dy - dt/dy = -y -(y^3 - y)e^(-2u) - (-y)
= -(y^3 - y)e^(-2u) = -(y^3 - y)e^[-2(z-y^2/2)] = -(y^3 - y)e^[-2z + y^2] = -y(y^2 - 1)e^(-2z)*e^(y^2),
e^(2z)dz = -y(y^2 - 1)e^(y^2)dy = -(y^2 - 1)e^(y^2)d(y^2)/2,
2e^(2z)dz = (1 - y^2)e^(y^2)d(y^2)
d[e^(2z)] = d[(2-y^2)e^(y^2)],
e^(2z) = (2-y^2)e^(y^2) + C. C = const.
2z = ln|(2-y^2)e^(y^2) + C|,
u = z + t = (1/2)ln|(2-y^2)e^(y^2) + C| - y^2/2.
x = e^u = e^[(1/2)ln|(2-y^2)e^(y^2) + C| - y^2/2]
= e^[(1/2)ln|(2-y^2)e^(y^2) + C|]*e^[- y^2/2]
= |(2-y^2)e^(y^2) + C|^(1/2)*e^(-y^2/2)
= |2 - y^2 + Ce^(-y^2)|^(1/2)

方程化为(xdx+ydy)＋y(x^2+y^2)dy＝0,以1/(x^2+y^2)为积分因子,得
(xdx+ydy)/(x^2+y^2)＋ydy＝0
d(ln(x^2+y^2))＋dy^2＝0
d[ln(x^2+y^2)+y^2]＝0
所以,方程的通解是ln(x^2+y^2)+y^2＝C