# How do you determine (d^y)/(dx^2) given 1-xy=x-y^2?

Aug 7, 2016

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{4 {\left(y + 1\right)}^{3}}{{y}^{2} + 2 y - 1} ^ 3$

#### Explanation:

$1 - x y = x - {y}^{2} \Rightarrow 1 + {y}^{2} = x + x y = x \left(1 + y\right)$

$\Rightarrow \frac{1 + {y}^{2}}{1 + y} = x$

Diff.ing both sides of this eqn. w.r.t. y , we have,

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

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

$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{{y}^{2} + 2 y - 1}{1 + y} ^ 2 , h e n c e , \frac{\mathrm{dy}}{\mathrm{dx}} = {\left(y + 1\right)}^{2} / \left({y}^{2} + 2 y - 1\right) \ldots . . \left(\star\right)$, i.e.,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} + 2 y + 1}{{y}^{2} + 2 y - 1} = \frac{{y}^{2} + 2 y - 1 + 2}{{y}^{2} + 2 y - 1}$

=(y^2+2y-1)/(y^2+2y-1)+2/(y^2+2y-1)=1+2/((y^2+2y-1). Therefore,

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

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

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

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

$= - \frac{4 \left(y + 1\right)}{{y}^{2} + 2 y - 1} ^ 2 \cdot {\left(y + 1\right)}^{2} / \left({y}^{2} + 2 y - 1\right) \ldots \ldots \ldots \ldots \ldots \left[b y , \left(\star\right)\right]$

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

Hope, this will be of Help! Enjoy Maths.!