用降阶方法解微分方程 要具体过程 好的加分 谢谢 麻烦了

1、∵y"+(y')^2=1
==>dy'/dx=1-(y')^2
==>dy'/(1-(y')^2)=dx
==>(1/(1+y')+1/(1-y'))dy'=2dx
==>ln│(1+y')/(1-y')│=2x+ln│C1│ (C1是常数)
∴(1+y')/(1-y')=C1e^(2x)
∵y'(0)=-1,则C1=0
∴(1+y')/(1-y')=0
==>1+y'=0
==>y'=-1
==>y=-x+C2 (C2是常数)
∵y(0)=0,则C2=0
∴原方程满足所给初始条件的特解是y=-x.
2、∵y"+(y')^2=y
==>y'dy'/dy+(y')^2=y
==>y'dy'+(y')^2dy=ydy
==>2y'e^(2y)dy'+2(y')^2e^(2y)dy=2ye^(2y)dy (等式两端同乘2e^(2y))
==>d((y')^2e^(2y))=d(ye^(2y)-e^(2y)/2)
∴(y')^2e^(2y)=ye^(2y)-e^(2y)/2+C1 (C1是常数)
∵y(0)=1/2,y‘(0)=0,则C1=0
∴(y')^2e^(2y)=ye^(2y)-e^(2y)/2
==>(y')^2=y-1/2
==>y'=±√(y-1/2)
==>dy/√(y-1/2)=±dx
==>2√(y-1/2)=C2±x (C2是常数)
∵y(0)=1/2,则C2=0
∴2√(y-1/2)=±x
==>y-1/2=x^2/4
==>y=x^2/4+1/2
故原方程满足所给初始条件的特解是y=x^2/4+1/2.