Eliminating the arbitary constants B and C in the expression
` y = 2/(3C) (Cx-1)^(3//2) +B` , we get

Options

A. `x [1 + ((dy)/(dx)^(2)] = (d^(2)y)/(dx^(2))`
B. ` 2x((dy)/(dx))(d^(2)y)/(dx^(2)) =1 + ((dy)/(dx))^(2)`
C. `((dy)/(dx))(d^(2)y)/(dx^(2))=1`
D. `((dy)/(dx))^(2) +1 = (d^(2)y)/(dx^(2))`

Answer

C

Explanation

` y = 2/(3C) (Cx -1)^(3/2)+B`
On differentiating both sides w.r.t, we get
` (dy)/(dx) = 2/(3C) .3/2 (Cx -1) ^(1/2) .C + 0= (Cx -1)^(1/2)`
` (dy)/(dx) = (Cx -1)^(1/2)`
On squaring both sides, we get
`((dy)/(dx))^(2) = Cx -1`
` Rightarrow ((dy)/(dx))^(2) +1 = Cx`
Now, on differentiating w.r.t we get
` 2 ((dy)/(dx)) . (d^(2)y)/(dx^(2)) =C`
From eq. (i)
`((dy)/(dx))^(2) +1 =2x ((dy)/(dx))(d^(2)y)/(dx^(2))`