How do you find (d^2y)/(dx^2) given x+siny=xy?

Sep 5, 2016

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

Explanation:

$x + \sin y = x y \Rightarrow \sin y = x y - x = x \left(y - 1\right) \Rightarrow x = \sin \frac{y}{y - 1}$

$\Rightarrow \frac{d}{\mathrm{dy}} \left(x\right) = \frac{d}{\mathrm{dy}} \left(\sin \frac{y}{y - 1}\right)$

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

$= \frac{\left(y - 1\right) \cos y - \sin y}{y - 1} ^ 2$

Since, $\sin y = x \left(y - 1\right)$, we have,

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - 1}{\cos y - x} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(\star\right)$

$\Rightarrow y ' \left(\cos y - x\right) = y - 1$

$\therefore \frac{d}{\mathrm{dx}} \left\{y ' \left(\cos y - x\right)\right\} = \frac{d}{\mathrm{dx}} \left(y - 1\right)$

$\therefore y ' \frac{d}{\mathrm{dx}} \left(\cos y - x\right) + \left(\cos y - x\right) \frac{d}{\mathrm{dx}} \left(y '\right) = \frac{\mathrm{dy}}{\mathrm{dx}} - 0 = y '$.

$\therefore y ' \left\{\frac{d}{\mathrm{dx}} \cos y - \frac{d}{\mathrm{dx}} \left(x\right)\right\} + \left(\cos y - x\right) y ' ' = y '$.

$\therefore y ' \left\{\left(- \sin y\right) y ' - 1\right\} + \left(\cos y - x\right) y ' ' = y '$.

$\therefore \left(\cos y - x\right) y ' ' = y ' \left(y ' \sin y + 1\right) + y ' = y ' \left(y ' \sin y + 2\right)$.

$= \frac{y - 1}{\cos y - x} \left\{\frac{\left(y - 1\right) \sin y}{\cos y - x} + 2\right\} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left[b y \left(\star\right)\right]$

$\therefore y ' ' = \frac{y - 1}{\cos y - x} ^ 2 \left\{\frac{\left(y - 1\right) \sin y}{\cos y - x} + 2\right\}$

$\Rightarrow \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = y ' ' = \frac{\left(y - 1\right) \left\{\left(y - 1\right) \sin y + 2 \left(\cos y - x\right)\right\}}{\cos y - x} ^ 3$.

Enjoy maths.!