求微分方程一般解
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let
u=y/x
du/dx = (1/x).dy/dx - y/x^2
dy/dx = x[ du/dx + (1/x)u]
//
(2x+y)y'-x=0
(2+y/x)y'-1=0
(2+u)(x[ du/dx + (1/x)u]) -1=0
(2+u) ( xdu/dx + u ) =1
x.du/dx = 1/(2+u) - u
= (1-2u-u^2)/(2+u)
∫(u+2)/(u^2+2u-1) du = - ∫dx/x
ln|u^2+2u-1| = -lnx + C'
u^2+2u-1 = C/x
(y/x)^2 + 2(y/x) -1 = C/x
u=y/x
du/dx = (1/x).dy/dx - y/x^2
dy/dx = x[ du/dx + (1/x)u]
//
(2x+y)y'-x=0
(2+y/x)y'-1=0
(2+u)(x[ du/dx + (1/x)u]) -1=0
(2+u) ( xdu/dx + u ) =1
x.du/dx = 1/(2+u) - u
= (1-2u-u^2)/(2+u)
∫(u+2)/(u^2+2u-1) du = - ∫dx/x
ln|u^2+2u-1| = -lnx + C'
u^2+2u-1 = C/x
(y/x)^2 + 2(y/x) -1 = C/x
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