Let "f(x,y,z,u_x,u_y,u_z) = 0" (1)
The Fundamental idea of the Jacobi method is to introduce two first order PDE involving two arbitrary constants a and b of the following form,
"h_1(x,y,z,u_x,u_y,u_z,a) = 0" (2)
"h_2(x,y,z,u_x,u_y,u_z,b) = 0" (3)
such that "\\frac{\\partial(f,h_1,h_2)}{\\partial(u_x,u_y,u_z)} \\neq 0" (4)
and
Equations (1),(2) and (3) can be solved for "u_x,u_y,u_z"
and equation "du = u_xdx + u_ydy+u_zdz" is integrable
If "h_1 = 0" and "h_2=0" are compitable with"f =0" then "h_1,h_2" satisfy,
"\\frac{\\partial(f,h)}{\\partial(x,u_x)}+\\frac{\\partial(f,h)}{\\partial(y,u_y)}+\\frac{\\partial(f,h)}{\\partial(z,u_z)} =0" for "h = h_i", i =1,2
Last equation will lead to the PDE of the form,
"f_{u_x}\\frac{\\partial h}{\\partial x}+f_{u_y}\\frac{\\partial h}{\\partial y}+f_{u_z}\\frac{\\partial h}{\\partial z} - f_x\\frac{\\partial h}{\\partial u_x}- f_y\\frac{\\partial h}{\\partial u_y}- f_z\\frac{\\partial h}{\\partial u_z} = 0"
It's auxiliary equation will be,
"\\frac{dx}{f_{u_x}}=\\frac{dy}{f_{u_y}}=\\frac{dz}{f_{u_z}} = \\frac{du_x}{-f_x}= \\frac{du_y}{-f_y}= \\frac{du_z}{-f_z}"
After getting this we need to those fractions which will lead to the solution.
Since given equation is
"f(x,u_x,u_z) = g(y,u_y,u_z)"
it can be simplified as
"f(x,u_x,u_z) - g(y,u_y,u_z) = 0"
or it can be written as
"F(x,y,u_x,u_y,u_z) = 0"
Then auxiliary equation will be
"\\frac{dx}{F_{u_x}}=\\frac{dy}{F_{u_y}} =\\frac{dx}{F_{u_z}}= \\frac{du_x}{-F_x}= \\frac{du_y}{-F_y}"
Solve it to obtain the solution.
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