Question

考研数学，常微分方程

考研数学，常微分方程求解例5……

Answer 1

dy/dx = (1-y)/[(y-1)-(x-1)] , 令 u = x-1, v = y-1，
则原微分方程变为 dv/du = -v/(v-u) = -(v/u)/(v/u-1)
令 p= v/u， 微分方程变为 p+udp/du = -p/(p-1)
(p-1)dp/(p^2+2p) = -du/u
[3/(p+2) - 1/p] dp = -2du/u
3ln(p+2) - lnp = -2lnu + lnC
(p+2)^3/p = C/u^2
u^2(P+2)^3 = -Cp
(x-1)^2[(y-1)/(x-1)+2]^3 = -C(y-1)/(x-1)
通解 (2x+y-3)^3 = -C(y-1)

追问

这个答案不对(°_°)…

追答

dy/dx = (1-y)/[(y-1)-(x-1)] , 令 u = x-1, v = y-1，
则原微分方程变为 dv/du = -v/(v-u) = -(v/u)/(v/u-1)
令 p= v/u， 微分方程变为 p+udp/du = -p/(p-1)
udp/du = -p^2/(p-1), (p-1)dp/p^2 = -du/u
[1/p - 1/p^2] dp = -du/u
lnp + 1/p = -lnu + lnC
pe^(1/p) = C/u
upe^(1/p) = C
通解 (y-1)e^[(x-1)/(y-1)] = C

追问

换了这么多元…其实我是希望你用公式法做的

追答

化为 dx/dy - x/(y-1) = -y/(y-1)
是 x 对 y 的 一阶线性微分方程。通解是
x = e^[∫dy/(y-1)]{-∫[y/(y-1)] e^[-∫dy/(y-1)]dy + C}
= (y-1){-∫[y/(y-1)^2]dy + C}
= (y-1){-∫[(y-1+1)/(y-1)^2]dy + C}
= (y-1){-∫[1/(y-1)+1/(y-1)^2]dy + C}
= (y-1){-ln(y-1) + 1/(y-1) + C}
= -(y-1)ln(y-1) + 1 + C(y-1)

Answer 2

书上不是有答案吗！这种基础题做多了就知道了。

追问

我做的答案跟标准答案有一点点不同，他又没写详细步骤，思路我是知道的