Question

Given that `v=(sin u)^(1/2)`, Show that `4v^3 (d^2v)/(du^2)+v^4+1=0`?

Answer 1

We have:

`v = (sinu)^(1/2)`

Squaring we get:

`v^2 = sinu` .... [A]

Implicitly differentiating wrt `u`:

`2v \ (dv)/(du) = cos u` .... [B]

Differentiate again wrt `u` (using the product rule):

`(2v)( (d^2v)/(du^2)) + (2(dv)/(du))( (dv)/(du) )= -sin u`

`:. 2v (d^2v)/(du^2) + 2((dv)/(du))^2 = -sin u`

`:. 4v^3 (d^2v)/(du^2) + 4v^2((dv)/(du))^2 = -2v^2sin u`

`:. 4v^3 (d^2v)/(du^2) + (2v(dv)/(du))^2 = -2v^2sin u`

`:. 4v^3 (d^2v)/(du^2) + (cosu)^2 = -2v^2sin u \ \ \ \ \` (from [B])

`:. 4v^3 (d^2v)/(du^2) + (1-sin^2u) = -2v^2sin u`

`:. 4v^3 (d^2v)/(du^2) + (1-v^4) = -2v^2 \ v^2`

`:. 4v^3 (d^2v)/(du^2) + 1+v^4 = 0 \ \ \ \` QED