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y"+2*y'^2/(1-y)=0的通解
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y"+2*y'^2/(1-y)=0的通解
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答案和解析
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).
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).
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