Answer to Question #138215 in Differential Equations for Nikhil

Question

Solve
dx/y^2+yz+z^2=dy/z^2+zx+x^2=dz/x^2+xy+y^2

Answer 1

$\displaystyle \frac{\mathrm{d}x}{y^2 + yz + z^2} = \frac{\mathrm{d}y}{x^2 + z^2 + xz} = \frac{\mathrm{d}z}{x^2 + xy + y^2} \\ \textbf{\textsf{This is a cyclic differential equation.}}\\ \textsf{To verify the integrablity of the above differential equation},\\ \textsf{we note that}\, \textbf{X} = (y^2 + yz + z^2, x^2 + z^2 + xz, x^2 + xy + y^2) \, \textsf{so that} \\ \mathrm{curl}\, \textbf{X} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^2 + yz + z^2 & x^2 + z^2 + xz & x^2 + xy + y^2 \end{vmatrix}\\ = \textbf{i}\left(\frac{\partial}{\partial y}(x^2 + xy + y^2) - \frac{\partial}{\partial z}(x^2 + z^2 + xz)\right) \\- \textbf{j}\left(\frac{\partial}{\partial x}(x^2 + xy + y^2) - \frac{\partial}{\partial z}(y^2 + yz + z^2)\right) \\+ \textbf{k}\left(\frac{\partial}{\partial x}(x^2 + z^2 + xz) - \frac{\partial}{\partial y}(y^2 + yz + z^2)\right) \\ = (x + 2y - 2z - x)\textbf{i} - (2x + y - y - 2z)\textbf{j} + (z + 2x - 2y - z)\textbf{k}\\ = 2(y - z)\textbf{i} - 2(x - z)\textbf{j} + 2(x - y)\textbf{k} = 2(y - z, z - x, x - y) \\ \textbf{X}\cdot \mathrm{curl}\, \textbf{X}\\ = (y^2 + yz + z^2, x^2 + z^2 + xz, x^2 + xy + y^2)\cdot2(y - z, z - x, x - y) \\ = 2(y^3 + y^2z + yz^2 - y^2z - yz^2 - z^3 - xz^2 - x^2z - x^3 \\ + z^3 + xz^2 + x^2z + x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3) = 0 \\ \textsf{Hence, the differential equation is integrable} \\ \textsf{Let}\, x = uz, \, \textsf{and}\, y = vz\\ \textsf{then}\, \mathrm{d}x = u\mathrm{d}z + z\mathrm{d}u \\ \textsf{and}\, \mathrm{d}y = v\mathrm{d}z + z\mathrm{d}v \\ \frac{u\mathrm{d}z + z\mathrm{d}u}{z^2(v^2 + v + 1)} = \frac{v\mathrm{d}z + z\mathrm{d}v}{z^2(1 + u + u^2)}= =\frac{\mathrm{d}z}{z^2(u^2 + uv + v^2)} \\ \textsf{Which yields} \\ \frac{u\mathrm{d}z + z\mathrm{d}u}{v^2 + v + 1} = \frac{v\mathrm{d}z + z\mathrm{d}v}{1 + u + u^2}= =\frac{\mathrm{d}z}{u^2 + uv + v^2} \\ \textsf{By the first two equations,}\\ \frac{u\mathrm{d}z + z\mathrm{d}u}{v^2 + v + 1} = \frac{v\mathrm{d}z + z\mathrm{d}v}{1 + u + u^2}= \\ \implies z(1 + u + u^2)\mathrm{d}u + u(1 + u + u^2)\mathrm{d}z = z(1 + v + v^2)\mathrm{d}v + v(1 + v + v^2)\mathrm{d}z \\ z((1 + u + u^2)\mathrm{d}u - (1 + v + v^2) \mathrm{d}v) = (v(1 + v + v^2) - u(1 + u + u^2))\mathrm{d}z \hspace{1cm} (1)\\ \textsf{Let}\, z \,\textsf{be a constant}\, \mathrm{d}z = 0\, \textsf{by Natani's method} \\ \therefore (1 + u + u^2)\mathrm{d}u - (1 + v + v^2) \mathrm{d}v = 0 \\ \int (1 + u + u^2)\mathrm{d}u - \int(1 + v + v^2) \mathrm{d}v = f(z) \\ (u - v) + \frac{u^2 - v^2}{2} + \frac{u^3 - v^3}{2} = f(z) \hspace{1cm} (2)\\ \textsf{Equating the first and the last equations and then set}\\v = 0,.\, \textsf{This implies that}\, \mathrm{d}v = 0 \\ \implies \frac{z\mathrm{d}u + u\mathrm{d}z}{1 + v + v^2} = \frac{\mathrm{d}z}{u^2 + uv + v^2} \\ \implies zu^2 \mathrm{d}u + u^3\mathrm{d}z = \mathrm{d}z\\ \mathrm{d}z (1 - u^3) = zu^2\mathrm{d}u \\ \frac{\mathrm{d}z}{z} = \frac{u^2}{1 - u^3}\\ \int \frac{\mathrm{d}z}{z} = \int\frac{u^2}{1 - u^3}\\ -3\ln(z) +\ln(c_1)= \ln(1 - u^3)\\ \ln\left(c z^{-\frac{1}{3}}\right) = \ln(1 - u^3)\\ \therefore u = \sqrt[3]{1 - cz^{-\frac{1}{3}}}\\ \textsf{Substitute}\, u= \sqrt[3]{1 - cz^{-\frac{1}{3}}}\, \textsf{and}\, v = 0\, \textsf{in}\, (2)\\ \therefore f(z) = \sqrt[3]{1 - cz^{-\frac{1}{3}}} + \frac{\left(\sqrt[3]{1 - cz^{-\frac{1}{3}}}\right)^2}{2} + \frac{\left(\sqrt[3]{1 - cz^{-\frac{1}{3}}}\right)^3}{3} \hspace{1cm} (3)\\ (2) \, \textsf{and}\, (3)\, \textsf{implies that}\\ (u - v) + \frac{u^2 - v^2}{2} + \frac{u^3 - v^3}{2} = \sqrt[3]{1 - cz^{-\frac{1}{3}}} + \frac{\left(\sqrt[3]{1 - cz^{-\frac{1}{3}}}\right)^2}{2} + \frac{1 - cz^{-\frac{1}{3}}}{3}\\ \textsf{Substitute}\, u = \frac{x}{z},\, v = \frac{y}{z}\, \textsf{in the above} \\ \therefore \frac{x - y}{z} + \frac{x^2 - y^2}{2z^2} + \frac{x^3 - y^3}{2z^3} = \sqrt[3]{1 - cz^{-\frac{1}{3}}} + \frac{\left(\sqrt[3]{1 - cz^{-\frac{1}{3}}}\right)^2}{2} + \frac{1 - cz^{-\frac{1}{3}}}{3} \\ \textsf{After simplifying this yields},\\(\textsf{i.e multiplying by}\, 6z^3 \, \textsf{and collecting like terms})\\ 6z^2(x - y ) + 3z(x^2 - y^2) + 2(x^3 - y^3) = 6z^3\sqrt[3]{1 - cz^{-\frac{1}{3}}} + 3z^3\left(\sqrt[3]{1 - cz^{-\frac{1}{3}}}\right)^2 +2z^3(1 - cz^{-\frac{1}{3}})\\ \therefore 6z^2(x - y ) + 3z(x^2 - y^2) + 2(x^3 - y^3) = z^3\left(6\sqrt[3]{1 - cz^{-\frac{1}{3}}} + 3\left(\sqrt[3]{1 - cz^{-\frac{1}{3}}}\right)^2 +2(1 - cz^{-\frac{1}{3}})\right)\\ \textsf{is the solution to the PDE.}$