来自陈小雷的问题
微分方程求解.yy''-(y')^2=1
微分方程求解.yy''-(y')^2=1
1回答
2020-01-23 17:12
微分方程求解.yy''-(y')^2=1
微分方程求解.yy''-(y')^2=1
y'=p
y''=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)|+C1x=(-1/√C)ln|√Cy+√(Cy^2-1)|+C1
∫dy/√(Cy^2-1)
√Cy=secudy=(1/√C)secutanudu
=(1/√C)∫secudu=(1/√C)ln|secu+tanu|=(1/√C)ln|√Cy+√(Cy^2-1)|