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y''+(y')^2=1,y(0)=1,y'(0)=0求满足初始条件的特解
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y''+(y')^2=1,y(0)=1,y'(0)=0 求满足初始条件的特解
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答案和解析
令y'=p
dp/dx=1-p^2
dp/(1-p^2)=dx (1/2)[1/(1+p)+1/(1-p)]dp=dx
积分,得:(1/2)[ln(1+p)-ln(1-p)]=x+c1
将y'(0)=0代人得:c1=0
(1+p)/(1-p)=e^2x
dy/dx=[e^x-e^(-x)]/[e^x+e^(-x)]
积分得:y=ln[e^x+e^(-x)]+c2
将y(0)=1代人得:c2=1-ln2
∴y=ln[e^x+e^(-x)]+1-ln2
dp/dx=1-p^2
dp/(1-p^2)=dx (1/2)[1/(1+p)+1/(1-p)]dp=dx
积分,得:(1/2)[ln(1+p)-ln(1-p)]=x+c1
将y'(0)=0代人得:c1=0
(1+p)/(1-p)=e^2x
dy/dx=[e^x-e^(-x)]/[e^x+e^(-x)]
积分得:y=ln[e^x+e^(-x)]+c2
将y(0)=1代人得:c2=1-ln2
∴y=ln[e^x+e^(-x)]+1-ln2
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