Question

Verify that the given function is a solution of the differential equation, xy = log y + c ; $y'=\frac{y^{2}}{1-xy}, \ \ xy\ne1$

Answer 1

To verify that the given function is a solution of the differential equation, xy = log y + c ; $y'=\frac{y^{2}}{1-xy}, \ \ xy\ne1$

Since the differential equation have first derivative only,thus differentiate given equation once and try to achieve the differential equation

$xy = log \: y + c \\ \\ x \: \frac{dy}{dx} + y = \frac{1}{y} \frac{dy}{dx} \\ \\ \\ x \: \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} \: + y = 0 \\ \\ \frac{dy}{dx} (x - \frac{1}{y} ) + y = 0 \\ \\ \frac{dy}{dx} ( \frac{xy - 1}{y} ) = - y \\ \\ \frac{dy}{dx} = \frac{ - { y}^{2} }{xy - 1} \\ \\ since \: {y}^{'} = \frac{dy}{dx} \\ \\ so \\ \\ {y}^{'} = \frac{ { y}^{2} }{1-xy }$
hence proved