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

1 Answer
Jan 16, 2018

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