求微分方程y✀✀=-(1+(y✀)∧2)∧(3/2)的通解

2024-11-18 20:39:39
推荐回答(2个)
回答1:

dy'/(1+(y')^2)^(3/2)=-dx
y'/(1+(y')^2)^(1/2)=-x+C
(y')^2/(1+(y')^2)=(-x+C)^2
(y')^2=1/(1-(x-C)^2)-1
y'=(x-C)/((x-C)^2-1)^(1/2)
y=((x-C)^2-1)^(1/2)+C2

回答2:

如图所示: