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

Oct 14, 2016

Deleted, because it was incorrect

Oct 14, 2016

I get $- \frac{5}{4 {y}^{3}}$

#### Explanation:

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

$4 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{2 y}$

$\frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{d}{\mathrm{dx}} \left(\frac{x}{2 y}\right)$

$= \frac{\left(1\right) \left(2 y\right) - x \left(2 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right)}{2 y} ^ 2$

$= \frac{2 y - 2 x \left(\frac{x}{2 y}\right)}{4 {y}^{2}}$

$= \frac{y - x \left(\frac{x}{2 y}\right)}{2 {y}^{2}}$

$= \frac{y - x \left(\frac{x}{2 y}\right)}{2 {y}^{2}} \cdot \frac{2 y}{2 y}$

 = (2y^2-x^2)/(4y^3

We started with $5 = {x}^{2} - 2 {y}^{2}$, so we have

$2 {y}^{2} - {x}^{2} = - 5$, making the second derivative,

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{5}{4 {y}^{3}}$

Oct 14, 2016

$y ' ' = - \frac{5}{4} \frac{1}{y} ^ 3$

#### Explanation:

$2 y {\left(x\right)}^{2} - {x}^{2} + 5 = 0 \to \frac{d}{\mathrm{dx}} \left(2 y {\left(x\right)}^{2} - {x}^{2} + 5\right) = 4 y y ' - 2 x = 0$

$\frac{d}{\mathrm{dx}} \left(2 y y ' - x\right) = 2 \left({\left(y '\right)}^{2} + y y ' '\right) - 1 = 0$ so

$y ' ' = \frac{\frac{1}{2} - {\left(y '\right)}^{2}}{y}$ but $y ' = \frac{1}{2} \frac{x}{y}$ so

$y ' ' = \frac{2 {y}^{2} - {x}^{2}}{4 {y}^{3}} = - \frac{5}{4} \frac{1}{y} ^ 3$