A.Maths

Please see the answer below

圖片參考：http://in.geocities.com/myisland8132/7007052003588.gif

首先兩邊一齊d
dx^(2/3)/dx+dy^(2/3)/dx=da^(2/3)/dx
2/3x^-(1/3)+2/3y^-(1/3)dy/dx=0--------------------(1)

dy/dx
=[-2/3x^-(1/3)]/[2/3y^-(1/3)]
=[-x^(1/2)y^(1/3)]----------------------------------------(2)

用番(1)再d
2/3[dx^-(1/3)/dx]+2/3[dy^-(1/3)/dx](dy/dx)+2/3y^-(1/3)(d^2y)/(dx^2)=0
-(1/3)x^-(4/3)-1/3[y^-(4/3)](dy/dx)^2+[y^-(1/3)](d^2y/dx^2)=0

so...
(d^2y)/(dx^2)
=[(1/3)x^-(4/3)+1/3y^-(4/3)(dy/dx)^2]/[y^-(1/3)]
=1/3[x^-(4/3)+y^(4/3)[-x^(1/2)y^(1/3)]^2]/[y^-(1/3)]-----------put番(2)個result
=1/3[x^-(4/3)y^-(1/3)+y^(-4+1/3)[x^-(1/3)y^(1/3)]^2]
=1/3[x^-(4/3)y^(1/3)+y^-1x^-(2/3)y^(2/3)]
=1/3[x^-(4/3)y^(1/3)+x^-(2/3)y^-(1/3)]
=1/3x^-(4/3)y^-(1/3)[y^(2/3)+x^(2/3)]--------------x^(2/3)+y^(2/3)=a^(2/3)
=1/3x^-(4/3)y^-(1/3)[a^(2/3)]