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x(lnx-lny)dy-ydy=0 如何求其通解
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若是 x(lnx-lny)dy - ydx = 0 , 必须有 x > 0, y > 0
则 -ln(y/x)dy/dx = y/x 是齐次方程, 令 y/x = u, 则 y = xu
-lnu(u+xdu/dx) = u, xdu/dx = -u/lnu-u = -u(1+lnu)/lnu
lnudu/[u(1+lnu)] = -dx/x, lnudlnu/(1+lnu) = -dx/x,
[1- 1/(1+lnu)]dlnu = -dx/x, lnu - ln(1+lnu) = -lnx + lnC,
ln[u/(1+lnu)] = ln(C/x), u/(1+lnu) = C/x,
xu/(1+lnu) = C, y/[1+ln(y/x)] = C, y = C[1+ln(y/x)].
则 -ln(y/x)dy/dx = y/x 是齐次方程, 令 y/x = u, 则 y = xu
-lnu(u+xdu/dx) = u, xdu/dx = -u/lnu-u = -u(1+lnu)/lnu
lnudu/[u(1+lnu)] = -dx/x, lnudlnu/(1+lnu) = -dx/x,
[1- 1/(1+lnu)]dlnu = -dx/x, lnu - ln(1+lnu) = -lnx + lnC,
ln[u/(1+lnu)] = ln(C/x), u/(1+lnu) = C/x,
xu/(1+lnu) = C, y/[1+ln(y/x)] = C, y = C[1+ln(y/x)].
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