The Lagrange’s equation is d x x + y = d y x − y = d z x 2 + y 2 ( x − y ) d x = ( x + y ) d y x 2 2 − y 2 2 − x y = C x 2 − y 2 − 2 x y = C y = 2 x 2 + C − x d z x 2 + y 2 = d x x + y d z = 3 x 2 + C − 2 x 2 x 2 + C 2 x 2 + C d z = 3 2 2 x 2 + C d x − C 2 d x 2 x 2 + C − 2 x d x ∫ d z = ∫ 3 2 2 x 2 + C d x − ∫ C 2 d x 2 x 2 + C − ∫ 2 x d x z = 3 8 ( 2 x C + 2 x 2 + 2 C log ( 2 C + 2 x 2 + 2 x ) ) − C 4 2 ( log ( 2 x + 2 C + 2 x 2 ) − log ( − 2 x + 2 C + 2 x 2 ) ) − x 2 + A z = 3 x C + 2 x 2 4 + 3 C 4 2 ( log ( 2 C + 2 x 2 + 2 x ) ) − C 4 2 ( log ( 2 x + 2 C + 2 x 2 ) − log ( − 2 x + 2 C + 2 x 2 ) ) − x 2 + A z = 3 x C + 2 x 2 4 + C 2 2 ( log ( 2 C + 2 x 2 + 2 x ) ) − C 4 2 log ( − 2 x + 2 C + 2 x 2 ) − x 2 + A ∴ The solution to the PDE is ϕ ( z − 3 x C + 2 x 2 4 + C 2 2 ( log ( 2 C + 2 x 2 + 2 x ) ) + x 2 − A + C 4 2 log ( − 2 x + 2 C + 2 x 2 ) , y − 2 x 2 + 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) The Lagrange’s equation is x + y d x = x − y d y = x 2 + y 2 d z ( x − y ) d x = ( x + y ) d y 2 x 2 − 2 y 2 − x y = C x 2 − y 2 − 2 x y = C y = 2 x 2 + C − x x 2 + y 2 d z = x + y d x d z = 2 x 2 + C 3 x 2 + C − 2 x 2 x 2 + C d z = 2 3 2 x 2 + C d x − 2 C 2 x 2 + C d x − 2 x d x ∫ d z = ∫ 2 3 2 x 2 + C d x − ∫ 2 C 2 x 2 + C d x − ∫ 2 x d x z = 8 3 ( 2 x C + 2 x 2 + 2 C log ( 2 C + 2 x 2 + 2 x ) ) − 4 2 C ( log ( 2 x + 2 C + 2 x 2 ) − log ( − 2 x + 2 C + 2 x 2 ) ) − x 2 + A z = 4 3 x C + 2 x 2 + 4 2 3 C ( log ( 2 C + 2 x 2 + 2 x ) ) − 4 2 C ( log ( 2 x + 2 C + 2 x 2 ) − log ( − 2 x + 2 C + 2 x 2 ) ) − x 2 + A z = 4 3 x C + 2 x 2 + 2 2 C ( log ( 2 C + 2 x 2 + 2 x ) ) − 4 2 C log ( − 2 x + 2 C + 2 x 2 ) − x 2 + A ∴ The solution to the PDE is ϕ ( z − 4 3 x C + 2 x 2 + 2 2 C ( log ( 2 C + 2 x 2 + 2 x ) ) + x 2 − A + 4 2 C log ( − 2 x + 2 C + 2 x 2 ) , y − 2 x 2 + C + x )
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