微分方程求解.yy''-(y')^2=1

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y'=py''=dp/dx=dp/dy*dy/dx=pdp/dy
ypdp/dy-p^2=1
pdp/(1+p^2)=dy/y
ln(1+p^2)=lny^2+lnC
1+p^2=Cy^2
p=√(Cy^2-1) 或 p=-√(Cy^2-1)
dy/dx=√(Cy^2-1)
x=(1/√C)ln|√Cy+√(Cy^2-1)|+C1 x=(-1/√C)ln|√Cy+√(Cy^2-1)|+C1
∫dy/√(Cy^2-1)
√Cy=secu dy=(1/√C)secutanudu
=(1/√C)∫secudu=(1/√C)ln|secu+tanu|=(1/√C)ln|√Cy+√(Cy^2-1)|