d x y 2 + y z + z 2 = d y x 2 + z 2 + x z = d z x 2 + x y + y 2 This is a cyclic differential equation. To verify the integrablity of the above differential equation , we note that X = ( y 2 + y z + z 2 , x 2 + z 2 + x z , x 2 + x y + y 2 ) so that c u r l X = ∣ i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z y 2 + y z + z 2 x 2 + z 2 + x z x 2 + x y + y 2 ∣ = i ( ∂ ∂ y ( x 2 + x y + y 2 ) − ∂ ∂ z ( x 2 + z 2 + x z ) ) − j ( ∂ ∂ x ( x 2 + x y + y 2 ) − ∂ ∂ z ( y 2 + y z + z 2 ) ) + k ( ∂ ∂ x ( x 2 + z 2 + x z ) − ∂ ∂ y ( y 2 + y z + z 2 ) ) = ( x + 2 y − 2 z − x ) i − ( 2 x + y − y − 2 z ) j + ( z + 2 x − 2 y − z ) k = 2 ( y − z ) i − 2 ( x − z ) j + 2 ( x − y ) k = 2 ( y − z , z − x , x − y ) X ⋅ c u r l X = ( y 2 + y z + z 2 , x 2 + z 2 + x z , x 2 + x y + y 2 ) ⋅ 2 ( y − z , z − x , x − y ) = 2 ( y 3 + y 2 z + y z 2 − y 2 z − y z 2 − z 3 − x z 2 − x 2 z − x 3 + z 3 + x z 2 + x 2 z + x 3 + x 2 y + x y 2 − x 2 y − x y 2 − y 3 ) = 0 Hence, the differential equation is integrable Let x = u z , and y = v z then d x = u d z + z d u and d y = v d z + z d v u d z + z d u z 2 ( v 2 + v + 1 ) = v d z + z d v z 2 ( 1 + u + u 2 ) = = d z z 2 ( u 2 + u v + v 2 ) Which yields u d z + z d u v 2 + v + 1 = v d z + z d v 1 + u + u 2 = = d z u 2 + u v + v 2 By the first two equations, u d z + z d u v 2 + v + 1 = v d z + z d v 1 + u + u 2 = ⟹ z ( 1 + u + u 2 ) d u + u ( 1 + u + u 2 ) d z = z ( 1 + v + v 2 ) d v + v ( 1 + v + v 2 ) d z z ( ( 1 + u + u 2 ) d u − ( 1 + v + v 2 ) d v ) = ( v ( 1 + v + v 2 ) − u ( 1 + u + u 2 ) ) d z ( 1 ) Let z be a constant d z = 0 by Natani’s method ∴ ( 1 + u + u 2 ) d u − ( 1 + v + v 2 ) d v = 0 ∫ ( 1 + u + u 2 ) d u − ∫ ( 1 + v + v 2 ) d v = f ( z ) ( u − v ) + u 2 − v 2 2 + u 3 − v 3 2 = f ( z ) ( 2 ) Equating the first and the last equations and then set v = 0 , . This implies that d v = 0 ⟹ z d u + u d z 1 + v + v 2 = d z u 2 + u v + v 2 ⟹ z u 2 d u + u 3 d z = d z d z ( 1 − u 3 ) = z u 2 d u d z z = u 2 1 − u 3 ∫ d z z = ∫ u 2 1 − u 3 − 3 ln ( z ) + ln ( c 1 ) = ln ( 1 − u 3 ) ln ( c z − 1 3 ) = ln ( 1 − u 3 ) ∴ u = 1 − c z − 1 3 3 Substitute u = 1 − c z − 1 3 3 and v = 0 in ( 2 ) ∴ f ( z ) = 1 − c z − 1 3 3 + ( 1 − c z − 1 3 3 ) 2 2 + ( 1 − c z − 1 3 3 ) 3 3 ( 3 ) ( 2 ) and ( 3 ) implies that ( u − v ) + u 2 − v 2 2 + u 3 − v 3 2 = 1 − c z − 1 3 3 + ( 1 − c z − 1 3 3 ) 2 2 + 1 − c z − 1 3 3 Substitute u = x z , v = y z in the above ∴ x − y z + x 2 − y 2 2 z 2 + x 3 − y 3 2 z 3 = 1 − c z − 1 3 3 + ( 1 − c z − 1 3 3 ) 2 2 + 1 − c z − 1 3 3 After simplifying this yields , ( i.e multiplying by 6 z 3 and collecting like terms ) 6 z 2 ( x − y ) + 3 z ( x 2 − y 2 ) + 2 ( x 3 − y 3 ) = 6 z 3 1 − c z − 1 3 3 + 3 z 3 ( 1 − c z − 1 3 3 ) 2 + 2 z 3 ( 1 − c z − 1 3 ) ∴ 6 z 2 ( x − y ) + 3 z ( x 2 − y 2 ) + 2 ( x 3 − y 3 ) = z 3 ( 6 1 − c z − 1 3 3 + 3 ( 1 − c z − 1 3 3 ) 2 + 2 ( 1 − c z − 1 3 ) ) is the solution to the PDE. \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.} y 2 + yz + z 2 d x = x 2 + z 2 + x z d y = x 2 + x y + y 2 d z This is a cyclic differential equation. To verify the integrablity of the above differential equation , we note that X = ( y 2 + yz + z 2 , x 2 + z 2 + x z , x 2 + x y + y 2 ) so that curl X = ∣ ∣ i ∂ x ∂ y 2 + yz + z 2 j ∂ y ∂ x 2 + z 2 + x z k ∂ z ∂ x 2 + x y + y 2 ∣ ∣ = i ( ∂ y ∂ ( x 2 + x y + y 2 ) − ∂ z ∂ ( x 2 + z 2 + x z ) ) − j ( ∂ x ∂ ( x 2 + x y + y 2 ) − ∂ z ∂ ( y 2 + yz + z 2 ) ) + k ( ∂ x ∂ ( x 2 + z 2 + x z ) − ∂ y ∂ ( y 2 + yz + z 2 ) ) = ( x + 2 y − 2 z − x ) i − ( 2 x + y − y − 2 z ) j + ( z + 2 x − 2 y − z ) k = 2 ( y − z ) i − 2 ( x − z ) j + 2 ( x − y ) k = 2 ( y − z , z − x , x − y ) X ⋅ curl X = ( y 2 + yz + z 2 , x 2 + z 2 + x z , x 2 + x y + y 2 ) ⋅ 2 ( y − z , z − x , x − y ) = 2 ( y 3 + y 2 z + y z 2 − y 2 z − y z 2 − z 3 − x z 2 − x 2 z − x 3 + z 3 + x z 2 + x 2 z + x 3 + x 2 y + x y 2 − x 2 y − x y 2 − y 3 ) = 0 Hence, the differential equation is integrable Let x = u z , and y = v z then d x = u d z + z d u and d y = v d z + z d v z 2 ( v 2 + v + 1 ) u d z + z d u = z 2 ( 1 + u + u 2 ) v d z + z d v == z 2 ( u 2 + uv + v 2 ) d z Which yields v 2 + v + 1 u d z + z d u = 1 + u + u 2 v d z + z d v == u 2 + uv + v 2 d z By the first two equations, v 2 + v + 1 u d z + z d u = 1 + u + u 2 v d z + z d v = ⟹ z ( 1 + u + u 2 ) d u + u ( 1 + u + u 2 ) d z = z ( 1 + v + v 2 ) d v + v ( 1 + v + v 2 ) d z z (( 1 + u + u 2 ) d u − ( 1 + v + v 2 ) d v ) = ( v ( 1 + v + v 2 ) − u ( 1 + u + u 2 )) d z ( 1 ) Let z be a constant d z = 0 by Natani’s method ∴ ( 1 + u + u 2 ) d u − ( 1 + v + v 2 ) d v = 0 ∫ ( 1 + u + u 2 ) d u − ∫ ( 1 + v + v 2 ) d v = f ( z ) ( u − v ) + 2 u 2 − v 2 + 2 u 3 − v 3 = f ( z ) ( 2 ) Equating the first and the last equations and then set v = 0 , . This implies that d v = 0 ⟹ 1 + v + v 2 z d u + u d z = u 2 + uv + v 2 d z ⟹ z u 2 d u + u 3 d z = d z d z ( 1 − u 3 ) = z u 2 d u z d z = 1 − u 3 u 2 ∫ z d z = ∫ 1 − u 3 u 2 − 3 ln ( z ) + ln ( c 1 ) = ln ( 1 − u 3 ) ln ( c z − 3 1 ) = ln ( 1 − u 3 ) ∴ u = 3 1 − c z − 3 1 Substitute u = 3 1 − c z − 3 1 and v = 0 in ( 2 ) ∴ f ( z ) = 3 1 − c z − 3 1 + 2 ( 3 1 − c z − 3 1 ) 2 + 3 ( 3 1 − c z − 3 1 ) 3 ( 3 ) ( 2 ) and ( 3 ) implies that ( u − v ) + 2 u 2 − v 2 + 2 u 3 − v 3 = 3 1 − c z − 3 1 + 2 ( 3 1 − c z − 3 1 ) 2 + 3 1 − c z − 3 1 Substitute u = z x , v = z y in the above ∴ z x − y + 2 z 2 x 2 − y 2 + 2 z 3 x 3 − y 3 = 3 1 − c z − 3 1 + 2 ( 3 1 − c z − 3 1 ) 2 + 3 1 − c z − 3 1 After simplifying this yields , ( i.e multiplying by 6 z 3 and collecting like terms ) 6 z 2 ( x − y ) + 3 z ( x 2 − y 2 ) + 2 ( x 3 − y 3 ) = 6 z 3 3 1 − c z − 3 1 + 3 z 3 ( 3 1 − c z − 3 1 ) 2 + 2 z 3 ( 1 − c z − 3 1 ) ∴ 6 z 2 ( x − y ) + 3 z ( x 2 − y 2 ) + 2 ( x 3 − y 3 ) = z 3 ( 6 3 1 − c z − 3 1 + 3 ( 3 1 − c z − 3 1 ) 2 + 2 ( 1 − c z − 3 1 ) ) is the solution to the PDE.
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