Question

高阶微分方程求通解

高阶微分方程求通解下一步怎么做？

Answer 1

y'' = d/dx ( dy/dx) = [d/dy( y')] / dx/dy = y'(dy'/dy)

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y^3.y''+1 =0
y'' = -1/y^3
y'dy' = (-1/y^3 ) dy
(1/2)y'^2 = 1/(2y^2) + C'
dy/dx =√( 1/y^2 + C)

追问

追答

这样
dy/dx =√( 1/y^2 + C)
dy/dx|x=1 =0, y|x=1 =1
√( 1 + C) =0
C=-1
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dy/dx =√( 1/y^2 - 1)
∫ √[y^2/(1-y^2)] dy = ∫ dx
x =∫ y/√(1-y^2) dy
x =-√(1-y^2) +C1
y(1) =1
=>C1=1
x =-√(1-y^2) +1
√(1-y^2) = 1-x
1-y^2 =(1-x)^2
y^2 = 1-(1-x)^2
y =√(2x-x^2)
let
y= sinu
dy= cosy du
∫ y/√(1-y^2) dy
=∫ sinu du
=-cosu +C1
=-√(1-y^2) +C1

好了，这样！
dy/dx =√( 1/y^2 + C)
dy/dx|x=1 =0, y|x=1 =1
√( 1 + C) =0
C=-1

------------
dy/dx =√( 1/y^2 - 1)
∫ √[y^2/(1-y^2)] dy = ∫ dx

x =∫ y/√(1-y^2) dy

x =-√(1-y^2) +C1

y(1) =1

=>C1=1

x =-√(1-y^2) +1

√(1-y^2) = 1-x

1-y^2 =(1-x)^2

y^2 = 1-(1-x)^2

y =√(2x-x^2)

let

y= sinu

dy= cosy du

∫ y/√(1-y^2) dy

=∫ sinu du

=-cosu +C1

=-√(1-y^2) +C1