Question

急，跪等答案！ 求微分方程y''+(2/1-y)*(y')^2=0的通解

Answer 1

解：令y'=p，则y''=pdp/dy 代入原方程，化简得 p[dp/dy-2p/(y-1)]=0 ==>p=0，或dp/dy-2p/(y-1)=0 显然，p=0是dp/dy-2p/(y-1)=0的解 又，由dp/dy-2p/(y-1)=0，得 dp/dy=2p/(y-1) ==>dp/p=2dy/(y-1) ==>ln│p│=2ln│y-1│+ln│C1│ ==>p=C1(y-1)² (∵p=0是一个解，∴C1是任意常数) ==>y'=C1(y-1)² ==>dy/(y-1)²=C1dx ==>1/(1-y)=C1x+C2 (C2是任意常数) ==>(C1x+C2)(1-y)=1 故原方程的通解是 (C1x+C2)(1-y)=1 (C1,C2是任意常数)。
这才对好么？

Answer 2

y'+x^2y=x^2
dy/dx=x^2(1-y)
dy/(1-y)=x^2dx
-d(1-y)/(1-y)=x^2dx
-ln|1-y|=x^3/3+C1
(1-y)=Ce^(-x^3/3)
通解是y=1-Ce^(-x^3/3)
特解C=0
y=1