# 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

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

Differentiate again wrt

# (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