Question

:y^3y''-1=0的通解,补充（其中y^3是Y的3次方,y''是Y的二阶导数）

:y^3y''-1=0的通解,补充（其中y^3是Y的3次方,y''是Y的二阶导数）高数

Answer 1

let

u= y'
du/dy = d/dx ( y' ) . dx/dy = y''/y' = y''/u
y''= u.du/dy
/
y^3.y''-1=0
y''= 1/y^3
u.du/dy = 1/y^3
∫udu = ∫dy/y^3
(1/2)u^2 = -(1/2)(1/y^2) + C'
u^2 = -1/y^2 + C1
dy/dx = √(-1/y^2 + C1)
∫dy/√(-1/y^2 + C1) = ∫dx
∫y/√(C1.y^2-1) dy= x
[1/(2C1)].∫ d(C1y^2-1) /√(C1.y^2-1) = x
(1/C1).√(C1.y^2-1) = x +C2
√(C1.y^2-1) = C1x +C1.C2
C1.y^2-1 = (C1x +C1.C2)^2
y =√ { [(C1x +C1.C2)^2 +1]/C1}