dy/根号（y^2-1）=+-dx,怎么求y两边积分怎么得有=（e^x-1+e^1-x）/2
2019-05-28
dy/根号（y^2-1）=+-dx,怎么求y
两边积分怎么得有=（e^x-1+e^1-x）/2
优质解答
令 y=secu,则 dy=secutanudu,
∫dy/√(y^2-1) = ∫secutanudu/tanu =∫secudu
=∫secu(secu+tanu)du/(secu+tanu)
=∫d(tanu+secu)/(secu+tanu)
= ln(secu+tanu) = ln[y+√(y^2-1)]
故 ln[y+√(y^2-1)] = ±x
y+√(y^2-1)=e^(±x),
√(y^2-1)=e^(±x)-y,
y^2-1=e^(±2x)+y^2-2ye^(±x),
y=[e^(±2x)+1]/[2e^(±x)],
得 y=[e^x+e^(-x)]/2.
令 y=secu,则 dy=secutanudu,
∫dy/√(y^2-1) = ∫secutanudu/tanu =∫secudu
=∫secu(secu+tanu)du/(secu+tanu)
=∫d(tanu+secu)/(secu+tanu)
= ln(secu+tanu) = ln[y+√(y^2-1)]
故 ln[y+√(y^2-1)] = ±x
y+√(y^2-1)=e^(±x),
√(y^2-1)=e^(±x)-y,
y^2-1=e^(±2x)+y^2-2ye^(±x),
y=[e^(±2x)+1]/[2e^(±x)],
得 y=[e^x+e^(-x)]/2.