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

Jan 16, 2018

We have:

$v = {\left(\sin u\right)}^{\frac{1}{2}}$

Squaring we get:

${v}^{2} = \sin u$ .... [A]

Implicitly differentiating wrt $u$:

$2 v \setminus \frac{\mathrm{dv}}{\mathrm{du}} = \cos u$ .... [B]

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

$\left(2 v\right) \left(\frac{{d}^{2} v}{{\mathrm{du}}^{2}}\right) + \left(2 \frac{\mathrm{dv}}{\mathrm{du}}\right) \left(\frac{\mathrm{dv}}{\mathrm{du}}\right) = - \sin u$

$\therefore 2 v \frac{{d}^{2} v}{{\mathrm{du}}^{2}} + 2 {\left(\frac{\mathrm{dv}}{\mathrm{du}}\right)}^{2} = - \sin u$

$\therefore 4 {v}^{3} \frac{{d}^{2} v}{{\mathrm{du}}^{2}} + 4 {v}^{2} {\left(\frac{\mathrm{dv}}{\mathrm{du}}\right)}^{2} = - 2 {v}^{2} \sin u$

$\therefore 4 {v}^{3} \frac{{d}^{2} v}{{\mathrm{du}}^{2}} + {\left(2 v \frac{\mathrm{dv}}{\mathrm{du}}\right)}^{2} = - 2 {v}^{2} \sin u$

$\therefore 4 {v}^{3} \frac{{d}^{2} v}{{\mathrm{du}}^{2}} + {\left(\cos u\right)}^{2} = - 2 {v}^{2} \sin u \setminus \setminus \setminus \setminus \setminus$ (from [B])

$\therefore 4 {v}^{3} \frac{{d}^{2} v}{{\mathrm{du}}^{2}} + \left(1 - {\sin}^{2} u\right) = - 2 {v}^{2} \sin u$

$\therefore 4 {v}^{3} \frac{{d}^{2} v}{{\mathrm{du}}^{2}} + \left(1 - {v}^{4}\right) = - 2 {v}^{2} \setminus {v}^{2}$

$\therefore 4 {v}^{3} \frac{{d}^{2} v}{{\mathrm{du}}^{2}} + 1 + {v}^{4} = 0 \setminus \setminus \setminus \setminus$ QED