Answer to Question #184223 in Differential Equations for Samdisha

Question

z(x+y)p + z(x-y)q = x² + y²

Answer 1

$\displaystyle \textsf{The Lagrange's equation is} \\ \frac{\mathrm{d}x}{x + y} = \frac{\mathrm{d}y}{x - y} = \frac{\mathrm{d}z}{x^2 + y^2} \\ (x - y)\mathrm{d}x = (x + y)\mathrm{d}y\\ \frac{x^2}{2} - \frac{y^2}{2} - xy = C\\ x^2 - y^2 - 2xy = C \\ y = \sqrt{2x^2 + C} - x \\ \frac{\mathrm{d}z}{x^2 + y^2} = \frac{\mathrm{d}x}{x + y} \\ \mathrm{d}z = \frac{3x^2 + C - 2x\sqrt{2x^2 + C}}{\sqrt{2x^2 + C}} \\ \mathrm{d}z = \frac{3}{2}\sqrt{2x^2 + C} \mathrm{d}x - \frac{C}{2}\frac{\mathrm{d}x}{\sqrt{2x^2 + C}} - 2x \,\, \mathrm{d}x \\ \int \mathrm{d}z = \int\frac{3}{2}\sqrt{2x^2 + C} \mathrm{d}x - \int\frac{C}{2}\frac{\mathrm{d}x}{\sqrt{2x^2 + C}} - \int2x \,\, \mathrm{d}x \\ z = \frac{3}{8}\left(2x\sqrt{C + 2x^2} + \sqrt{2}C\log\left(\sqrt{2}\sqrt{C + 2x^2} + 2x\right)\right) - \\ \frac{C}{4\sqrt{2}}\left(\log\left(2x + \sqrt{2}\sqrt{C + 2x^2}\right) - \log\left(-2x + \sqrt{2}\sqrt{C + 2x^2}\right)\right) - x^2 + A\\ z = \frac{3x\sqrt{C + 2x^2}}{4} + \frac{3C}{4\sqrt{2}}\left(\log\left(\sqrt{2}\sqrt{C + 2x^2} + 2x\right)\right) -\\ \frac{C}{4\sqrt{2}}\left(\log\left(2x + \sqrt{2}\sqrt{C + 2x^2}\right) - \log\left(-2x + \sqrt{2}\sqrt{C + 2x^2}\right)\right) - x^2 + A\\ z = \frac{3x\sqrt{C + 2x^2}}{4} + \frac{C}{2\sqrt{2}}\left(\log\left(\sqrt{2}\sqrt{C + 2x^2} + 2x\right)\right) - \\ \frac{C}{4\sqrt{2}}\log\left(-2x + \sqrt{2}\sqrt{C + 2x^2}\right) - x^2 + A\\ \therefore\textsf{The solution to the PDE is}\\ \phi\left(z - \frac{3x\sqrt{C + 2x^2}}{4} + \frac{C}{2\sqrt{2}}\left(\log\left(\sqrt{2}\sqrt{C + 2x^2} + 2x\right)\right) + x^2 - A\right. + \\ \left.\frac{C}{4\sqrt{2}}\log\left(-2x + \sqrt{2}\sqrt{C + 2x^2}\right), y - \sqrt{2x^2 + C} + x\right)$