Question

08(1分)求下列微分方程的通解(x2+y2)(ydx-xdy)=2y2(xdx+ydy)

Answer 1

(x^2+y^2)(ydx-xdy) = 2y^2(xdx+ydy) 为齐次方程,
令 y = xu， 则 dy = xdu+udx , 微分方程化为
-x^4(1+u^2)du = 2x^3u^2[(1+u^2)dx + xudu]
x ≠ 0 时， 微分方程化为
-x(1+u^2)du = 2u^2[(1+u^2)dx + xudu]
-x(1+u^2+2u^3)du = 2u^2(1+u^2)dx
(1+u^2+2u^3)du/[u^2(1+u^2)] = -2dx/x
[1/u^2 + 2u/(1+u^2)]du = (-2/x)dx
-1/u + ln(1+u^2) = -2lnx + lnC
ln[(1+u^2)x^2] = lnC + 1/u
(1+u^2)x^2 = Ce^(1/u)
x^2+y^2 = Ce^(x/y)