作业帮yy''-(y')^2=y^2y'
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yy''-(y')^2=y^2y'
yy''-(y')^2=y^2y'
yy''-(y')^2=y^2y'
YY频道7450 免费签约 秒过直播间
设y=xt,则t=y/x,y'=xt'+t
代入原方程得xt'+t+t=1/t
==>xt'=(1-2t2)/t
==>tdt/(1-2t2)=dx/x
==>d(1-2t2)/(1-2t2)=-4dx/x
==>ln│1-2t2│=-4ln│x│+ln│C│ (C是积分常数)
==>1-2t2=C/x^4
==>(1-2(y/x)2)x^4=C ...展开全文设y=xt,则t=y/x,y'=xt'+t
代入原方程得xt'+t+t=1/t
==>xt'=(1-2t2)/t
==>tdt/(1-2t2)=dx/x
==>d(1-2t2)/(1-2t2)=-4dx/x
==>ln│1-2t2│=-4ln│x│+ln│C│ (C是积分常数)
==>1-2t2=C/x^4
==>(1-2(y/x)2)x^4=C
==>x2(x2-2y2)=C
故原微分方程的通解是x2(x2-2y2)=C (C是积分常数)收起
∵yy''-(y')^2=y^2y'
==>(yy''-(y')^2)/y^2=y'
==>(y'/y)'=y'
==>y'/y=y+C1 (C1是常数)
==>y'=y(y+C1)
==>dy/(y(y+C1))=dx
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∵yy''-(y')^2=y^2y'
==>(yy''-(y')^2)/y^2=y'
==>(y'/y)'=y'
==>y'/y=y+C1 (C1是常数)
==>y'=y(y+C1)
==>dy/(y(y+C1))=dx
==>ln│y/(y+C1)│=C1x+ln│C2│ (C2是常数)
==>y/(y+C1)=C2e^(C1x)
==>y=C1C2e^(C1x)/(1-C2e^(C1x))
∴原方程的通解是y=C1C2e^(C1x)/(1-C2e^(C1x))。
收起