If `y^3-y=2x ,t h e n(x^2-1/(27))(d^2y)/(dx^2)+x(dy)/d=` `y` b. `y/3` c. `y/9` d. `y/(27)`

Options

A. y
B. `(y)/(3)`
C. `(y)/(9)`
D. `(y)/(27)`

Answer

C

Explanation

We have
`y^(3)-y=2x`
Differentiating both sides w.r.t. x, we get
`(3y^(2)-1)(dy)/(dx)=2 rArr (dy)/(dx)=(2)/(3y^(2)-1)" (1)"`
Again, differentiating both sides w.r.t. x, we get
`(d^(2)y)/(dx^(2))=(-2.6y((dy)/(dx)))/((3y^(2)-1))`
Using (1), we get
`(d^(2)y)/(dx^(2))=(-2y)/((3y^(2)-1)^(3))" (2)"`
Now,
`(x^(2)-(1)/(27))(d^(2)y)/(dx^(2))+(xdy)/(dx)`
`=(x^(2)-(1)/(27))((-24y)/((3y^(2)-1)^(2)))+(2x)/((3y^(2)-1))`
`=(y^(2)((y^(2)-1)^(2))/(4)-(1)/(27))((-24y)/((3y^(2)-1)^(3)))+(y(y^(2)-1))/(3y^(2)-1)`
`" "(because y^(3)-y=2x)`
`=({27y^(2)(y^(2)-1)^(2)-4})/(108)((-24y))/((3y^(2)-1))+(y(y^(2)-1))/(3y^(2)-1)`
`=(y)/(9){(-54y^(2)(y^(2)-1)^(2)+8)/((3y^(2)-1)^(3))+(9(y^(2)-1))/(3y^(2)-1)}`
`=(y)/(9){(-2(1+alpha)(alpha-2)^(2)+8)/(alpha^(3))}+(3(alpha-2))/(alpha)`
`" where "alpha=3y^(2)-1`
`=(y)/(9)`