Ay"+B(1+y')^(1/2)=0 求原函数Y~

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A*y"+B*(1+y'')^(1/2)=0求原函数Y~A*y"+B*(1+y'')^(1/2)=0求原函数Y~A*y"+B*(1+y'')^(1/2)=0求原函数Y~设y''=dy/dx=zy''''=dz/d

A*y"+B*(1+y')^(1/2)=0 求原函数Y~
A*y"+B*(1+y')^(1/2)=0 求原函数Y~

A*y"+B*(1+y')^(1/2)=0 求原函数Y~
设
y'=dy/dx = z
y''=dz/dx
=>
Adz/dx + B(1+z)^(1/2)=0
=>
dz/(1+z)^(1/2) = -B/A*dx
=>同取积分
2(1+z)^(1/2) = -B/A*x + C1
=>
dy/dx = BB/4AA*x^2 -BC1/2A*x + C1^2/4 - 4
=>
y = BB/12AA*x^3 - BC1/4A*x^2 + (C1C1-4)/4*x + C2

你学过拉普拉斯变换没有嘛，学过的话你可以做的，你可以参考数学物理方法第六章，或者复变函数

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