2(1) 微分方程即 dy/dx = 2xy/(3x^2-y^2) 是齐次方程。
令 y = xu, 则 u+xdu/dx = 2u/(3-u^2) ,
xdu/dx = (u^3-u)/(3-u^2)
(3-u^2)du/(u^3-u) = dx/x
[1/(u+1)+1/(u-1)-3/u]du = dx/x
ln[(u^2-1)/u^3] = lnx + lnC
(u^2-1)/u^3 = Cx
通解 y^2-x^2 = Cy^3
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