Question

求大佬解一下这道可降阶的微分方程

Answer 1

令 p = y' = dy/dx， 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy
2(y')^2 = y''(y-1) 化为 2p^2 = p(y-1)dp/dy
p ≠ 0 时化为 2p = (y-1)dp/dy， dp/p = 2dy/(y-1),
lnp = 2ln(y-1) + lnC1, p = dy/dx = C1(y-1)^2
dy/(y-1)^2 = C1dx, -1/(y-1) = C1x + C2, (C1x+C2)(y-1) + 1 = 0；
p = 0 时， y = C。

Answer 2

解:∵微分方程为2(y')²=y"(y-1)，化为2y'/(y-1)=y"/y'

∴有2ln|y-1|+ln|a|=ln|y'|(a为任意非零常数)，y'=a(y-1)²，dy/(y-1)²=adx，-1/(y-1)=ax+c(c为任意常数)，y-1=-1/(ax+c)，方程的通解为y=1-1/(ax+c)

请参考

Answer 3

解:∵微分方程为2y'²=y"(y-1)，化为2y'/(y-1)=y"/y'

∴有2ln|y-1|+ln|c|=ln|y'|，c(y-1)²=y'(c为任意非零常数)，dy/(y-1)²=cdx，-1/(y-1)=cx+a，方程的通解

为y=1-1/(cx+a)(a为任意常数)

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Answer 4

解:∵微分方程为2(y')²=y"(y-1)，化为

2y'/(y-1)=y"/y' ∴有2ln|y-1|+ln|a|=

ln|y'|(a为任意非零常数)，y'=a(y-1)²，

-dy/(y-1)²=adx，1/(y-1)=ax+c(c为任意常数)，方程的通解为y=1/(ax+c)+1

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