# How do you find (d^2y)/(dx^2) for x^3=2y^2+5?

Sep 1, 2016

Taken you to a point where you can take over. Hopefully my workings are correct.

#### Explanation:

$2 {y}^{2} = {x}^{3} - 5$

${y}^{2} = \frac{1}{2} {x}^{3} - \frac{5}{2}$

$\implies y = \pm {\left(\frac{1}{2} {x}^{3} - \frac{5}{2}\right)}^{\frac{1}{2}}$

Let $u = \frac{1}{2} {x}^{3} - \frac{5}{2} \text{ " =>" } \frac{\mathrm{du}}{\mathrm{dx}} = \frac{3}{2} {x}^{2}$

Consider:

$y = \pm {u}^{\frac{1}{2}} \text{ " =>" } \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2} {u}^{- \frac{1}{2}}$

But $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \times \frac{\mathrm{dy}}{\mathrm{du}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {u}^{- \frac{1}{2}} \times \frac{3}{2} {x}^{2}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(\frac{1}{2} {x}^{3} - \frac{5}{2}\right)}^{- \frac{1}{2}} \times \frac{3}{2} {x}^{2}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{4} {x}^{3} {\left(\frac{1}{2} {x}^{3} - \frac{5}{2}\right)}^{- \frac{1}{2}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Second differential

Let $u = \frac{3}{4} {x}^{3} \text{ " =>" } \frac{\mathrm{du}}{\mathrm{dx}} = \frac{9}{4} {x}^{2}$

Let $v = {\left(\frac{1}{2} {x}^{3} - \frac{5}{2}\right)}^{- \frac{1}{2}}$
$\implies \text{ } \frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{1}{2} {\left(\frac{1}{2} {x}^{3} - \frac{5}{2}\right)}^{- \frac{3}{2}} \left(\frac{3}{2} {x}^{2}\right)$
$\implies \text{ } \frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{3}{4} {x}^{2} {\left(\frac{1}{2} {x}^{3} - \frac{5}{2}\right)}^{- \frac{3}{2}}$

So for the second differential we have:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \to v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

$\textcolor{w h i t e}{.}$

(d^2y)/(dx^2) ->[(1/2x^3-5/2)^(-1/2)xx9/4x^2] + [3/4x^3xx(-3/4x^2(1/2x^3-5/2)^(-3/2)]

$\textcolor{w h i t e}{.}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \to \frac{9 {x}^{2}}{4 \times \sqrt{\frac{1}{2} {x}^{3} - \frac{5}{2}}} - \frac{9 {x}^{5}}{16 \times {\left(\sqrt{\frac{1}{2} {x}^{3} - \frac{5}{2}}\right)}^{3}}$

I will let you take it from this point