Question

Using Jacobi's method find the complete integral of the equation
2u₁xz + 3u₂y² + u₂²u₃ = 0

Answer 1

Using Jacobi's method find the complete integral of the equation  

$$2u_1xz+3u_2z^2+u_2^2u_3=0$$

This equation is already in the form of 

$$F(x,y,z,u_1,u_2,u_3)=0$$

 Its auxiliary equations are 

$${dx \over F_{u_1} }={dy \over F_{u_2} }={dz \over F_{u_3} }={du_1 \over -F_x }={du_2 \over -F_y }={du_3 \over -F_z }$$

$${dx \over 2xz }={dy \over 3z^2+2u_2u_3 }={dz \over u_2^2 }={du_1 \over -2u_1z }={du_2 \over 0 }={du_3 \over -2u_1x-6u_2z }$$

From the first and fourth ratios, we get

$${dx \over 2xz }={du_1 \over -2u_1z }=>-{dx \over x }={du_1 \over u_1 }=>\ln{x}+\ln{u_1}=\ln{a}$$ $$u_1x=a, a=const$$

From the fifth ratio, we get

$${du_2 \over 0 }=>du_2=0=>u_2=b, b=const$$

With the above forms of $u_1$ and $u_2,$ we obtain from the given equation 

$$2az+3bz^2+b^2u_3=0$$ $$u_3=-{2az+3bz^2 \over b^2}$$

The equation

$$du=u_1dx+u_2dy+u_3dz$$

obtained for these values of $u_1,u_2,u_3,$ is integrable. 

$$du=({a \over x})dx+bdy+(-{2az+3bz^2 \over b^2})dz$$

On integrating the above equation, we arrive at the following solution of the given equation 

$$u=a\ln x+by-{az^2 \over b^2}-{z^3 \over b}+c,$$ $$a=const, b=const, c=const$$