
Bobosharif S. answered 03/11/18
Tutor
4.4
(32)
PhD in Math, MS's in Calulus
x1/3 + y1/3 = 94
(1/3)x1/3-1+(1/3)y1/3-1(dy/dx)=0
(1/3)(-2/3)x1/3-2+(1/3)(-2/3)y1/3-2(dy/dx)2+(1/3)y-2/3(d2y/dx2)=0
(2/3)x-5/3+(2/3)y-5/3(dy/dx)2 - y-2/3(d2y/dx2)=0
y-2/3d2y/dx2=(2/3)x-5/3+(2/3)y-5/3(dy/dx)2
d2y/dx2=(2/3)x-5/3y2/3+(2/(3y))(dy/dx)2.

Bobosharif S.
03/11/18