Question

d^2y/dx^2 of 1-xy=x-y express in terms of x and y

Answer 1

Answer:

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{2(1 + y)}{ {(x - 1)}^{2} }$

Step-by-step explanation:

To find

$\frac{ {d}^{2} y}{d {x}^{2} } \\ \\ of \\ \\ 1 - xy = x - y$

Differentiate it once first,with respect to x

$\frac{d(1 - xy)}{dx} = \frac{d(x - y)}{dx} \\ \\ \because \: \frac{d(uv)}{dx} = u \frac{dv}{dx} - v \frac{du}{dx} \\ \\ 0 - x \frac{dy}{dx} - y \frac{dx}{dx} = \frac{dx}{dx} - \frac{dy}{dx} \\ \\ - x \frac{dy}{dx} - y.1 = 1 - \frac{dy}{dx} \\ \\ \frac{dy}{dx}(1 - x) = 1 + y \\ \\ \frac{dy}{dx} = \frac{1 + y}{1 - x} ....eq1 \\ \\again \: do \: differentiation \\ \\ - x \frac{ {d}^{2} y}{d {x}^{2} } - \frac{dy}{dx} - \frac{dy}{dx} = - \frac{ {d}^{2} y}{d {x}^{2} } \\ \\ ( x - 1)\frac{ {d}^{2} y}{d {x}^{2} } + 2\frac{dy}{dx} = 0 \\ \\ \\ ( x - 1)\frac{ {d}^{2} y}{d {x}^{2} } = - 2 \frac{dy}{dx} \\ \\ put \: value \: of \: \frac{dy}{dx} \: from \: eq1 \\ \\ ( x - 1)\frac{ {d}^{2} y}{d {x}^{2} } = - 2 (\frac{1 + y}{1 - x}) \: \\ \\( x - 1)\frac{ {d}^{2} y}{d {x}^{2} } = 2 (\frac{1 + y}{x - 1}) \\ \\ \frac{ {d}^{2} y}{d {x}^{2} } = \frac{2(1 + y)}{ {(x - 1)}^{2} }$

Hope it helps you.