微分方程求通解

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微分方程求通解

微分方程求通解
微分方程求通解
 

微分方程求通解
∵y''+2y'^2/(1-y)=0
==>y'dy'/dy-2y'^2/(y-1)=0
==>y'[dy'/dy-2y'/(y-1)]=0
∴y'=0.(1)
dy'/dy-2y'/(y-1)=0.(2)
∵由(1),得y=C (C是常数)
由(2),得dy'/dy=2y'/(y-1)
==>dy'/y'=2dy/(y-1)
==>ln│y'│=2ln│y-1│+ln│C1│ (C1是常数)
==>y'=C1(y-1)^2
==>dy/(y-1)^2=C1dx
==>-1/(y-1)=C1x+C2 (C2是常数)
==>(C1x+C2)(1-y)=1
由于y=C包含在(C1x+C2)(1-y)=1
∴原方程的通解是(C1x+C2)(1-y)=1.