Question #184223

z(x+y)p + z(x-y)q = x2 + y2

1
Expert's answer
2021-04-27T00:45:02-0400

The Lagrange’s equation isdxx+y=dyxy=dzx2+y2(xy)dx=(x+y)dyx22y22xy=Cx2y22xy=Cy=2x2+Cxdzx2+y2=dxx+ydz=3x2+C2x2x2+C2x2+Cdz=322x2+CdxC2dx2x2+C2xdxdz=322x2+CdxC2dx2x2+C2xdxz=38(2xC+2x2+2Clog(2C+2x2+2x))C42(log(2x+2C+2x2)log(2x+2C+2x2))x2+Az=3xC+2x24+3C42(log(2C+2x2+2x))C42(log(2x+2C+2x2)log(2x+2C+2x2))x2+Az=3xC+2x24+C22(log(2C+2x2+2x))C42log(2x+2C+2x2)x2+AThe solution to the PDE isϕ(z3xC+2x24+C22(log(2C+2x2+2x))+x2A+C42log(2x+2C+2x2),y2x2+C+x)\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)


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