Answer to Question #161506 in Differential Equations for Rida

Question

dx/x(y²-z²⁾=dy/-y(z²+x²)=dz/z(x²+y²)

Answer 1

$\frac{dx}{x(y^2-z^2)}=\frac{dy}{-y(z^2+x^2)}=\frac{dz}{z(x^2+y^2)}$

Solution:

Let's make an integral combination using the property of equal fraction:

$\frac{xdx+ydy}{x^2(y^2-z^2)-y^2(z^2+x^2)}=\frac{dz}{z(x^2+y^2)}$

$\frac{\frac12d(x^2+y^2)}{-z^2(x^2+y^2)}=\frac{dz}{z(x^2+y^2)}$

$\frac12d(x^2+y^2)=-zdz$

$\frac12(x^2+y^2)=-\frac{z^2}{2}+C$

$x^2+y^2+z^2=C_1$

$\phi_1(x,y,z)=x^2+y^2+z^2=C_1$ - first integral of the system.

Another integral combination:

$\frac{ydx-xdy}{xy(y^2-z^2)+yx(z^2+x^2)}=\frac{dz}{z(x^2+y^2)}$

$\frac{y^2d\frac{x}{y}}{xy(x^2+y^2)}=\frac{dz}{z(x^2+y^2)}$

$\frac{d\frac{x}{y}}{\frac{x}{y}}=\frac{dz}{z}$

$\ln|\frac{x}{y}|=\ln|z|+\ln{C_2}$

$\frac{x}{zy}=C_2$

$\phi_2(x,y,z)=\frac{x}{zy}=C_2$ - another integral of the system.

Since

rank$\begin{pmatrix} \frac{\partial\phi_1}{\partial x} & \frac{\partial\phi_1}{\partial y} & \frac{\partial\phi_1}{\partial z} \\ \frac{\partial\phi_2}{\partial x} & \frac{\partial\phi_2}{\partial y} & \frac{\partial\phi_2}{\partial z} \end{pmatrix}= \begin{pmatrix} 2x & 2y & 2z \\ \frac{1}{zy} & - \frac{x}{zy^2} & -\frac{x}{yz^2} \end{pmatrix}=2$ ,

then $\phi_1$ and $\phi_2$ are independent.

General solution:

$u=\varPhi(\phi_1, \phi_2)=\varPhi(x^2+y^2+z^2, \frac{x}{yz})$ , where $\varPhi$ is any continues differentiable function of arguments $\phi_1$ and $\phi_2$ .

Answer: $u=\varPhi(x^2+y^2+z^2, \frac{x}{yz})$ .