Question

求微分方程dy/dx=lnx/xy，y（1）＝2满足所给初始条件的特解

Answer 1

dy/dx = lnx/(xy), ydy = (lnx/x)dx, 2ydy = 2(lnx/x)dx
y^2 = (lnx)^2 + C, y(1) ＝ 2 代入， 得 C = 4
特解 y^2 = 4 + (lnx)^2

Answer 2

dy/dx=lnx/(xy)
∫ydy = ∫lnx/x dx
(1/2)y^2 = ∫lnx dlnx
=(1/2)(lnx)^2 + C
y(1)=2 => C=2
(1/2)y^2=(1/2)(lnx)^2 + 2
y^2=(lnx)^2 + 4