Answer to Question #272374 in Differential Equations for Supriya

Question

Solve (1+2xy)ydx+(1-2xy)xdy=0 by using inspection method

Answer 1

$(1 + 2xy)ydx + (1-2xy)xdy = 0 \, \, .\, .\, . \,(1) \\ \text{By inspection, multiply eq(1) by} \, 1/(4x^2y^2) \\ \text{and divide through by 2. We have},\\ (\frac{1}{x} + \frac{1}{2x^2y})dx + (\frac{1}{2xy^2} - \frac{1}{y})dy = 0 \\ \text{Since} \, \frac{\partial{M}}{\partial{y}} = \frac{\partial{N}}{\partial{x}} = -\frac{1}{2x^2y^2}\\ \text{Hence, the solution is}\\ \int({\frac{1}{x} + \frac{1}{2x^2y}})dx + \int{-\frac{1}{y}dy} = C \\\text{where C is an arbitrary constant}\\ lnx - \frac{1}{2xy} - lny = C \\ ln(\frac{x}{y}) - \frac{1}{2xy} = C \, \text{where C is an arbitrary constant}$