Answer to Question #147408 in Differential Equations for Abshar

Question

Find the integral of (y² -1)dx -2dy =0

Answer 1

$\displaystyle (y² -1)\mathrm{d}x -2\mathrm{d}y =0\\ y^2 - 1 - 2y' = 0\\ y^2 - 1 = 2y' \\ \frac{2}{y^2 - 1} \,\mathrm{d}y = \mathrm{d}x\\ \int\frac{2}{y^2 - 1} \,\mathrm{d}y = \int\mathrm{d}x\\ \frac{1}{2}\cdot 2 \ln\left(\frac{y - 1}{y + 1}\right) = x + C\\ \frac{y - 1}{y + 1} = e^{x + C}\\ \frac{y - 1}{y + 1} = Ae^{x}\\ y - 1 = A(y + 1)e^{x}\\ y(1 - Ae^{x}) = Ae^{x} + 1 \\ y = \frac{1 + Ae^{x}}{1 - Ae^{x}}$