Question

Find the integral surface of the linear PDE
x(y²+z)p - y(x²+z)q = (x²-y²)z , which contains the straight line x+y =0 , z=1.

Answer 1

Auxiliary equations are

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

$\frac{xdx+ydy-dz}{x^2(y^2+z)-y^2(x^2+z)-z(x^2-y^2)}=\frac{xdx+ydy-dz}{0}$ (1)

$\frac{\frac{dx}{x}+\frac{dy}{y}+\frac{dz}{z}}{y^2+z-(x^2+z)+(x^2-y^2)}= \frac{\frac{dx}{x}+\frac{dy}{y}+\frac{dz}{z}}{0}$ (2)

from (1)

$x^2+y^2-2z=C_1$ (3)

from (2)

$log(xyz)=C_2,\ or \ xyz=C_3$ (4)

parametric equation of straight line is

$(x=t,y=-t,z=1)$ (5)

On substituting (5) in (3) and (4) we get

$2t^2-2=C_1,\ -t^2=C_3$

Eliminating t

$-2C_3-2=C_1,\ or F(C_1,C_3)=C_1+2C_3+2=0$

Therefor

$x^2+y^2-2z+2xyz+2=0$ is integral surface, which contains the straight line.

Answer: x^2+y^2-2z+2xyz+2=0