Answer to Question #283921 in Differential Equations for Amit

Question

Find the integral surface of the linear partial differential equation x(x^2+z)p - y(y^2+z)q = (x^2-y^2)z; p, q has their usual meaning , which contains the straight line

Answer 1

Given question is incomplete and incorrect. We assume it as:

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 .

Solution:

Auxiliary equations are

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

By Choosing multipliers $x, y, -1,$ we get

$${xdx+ydy-dz\over x^2y^2+x^2z-x^2y^2-y^2z-x^2z+y^2z}={xdx+ydy-dz\over 0}$$

Then

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

By Choosing multipliers $1/x, 1/y, 1/z,$ we get

$${{dx \over x}+{dx \over x}+{dz \over z}\over y^2+z-x^2-z+x^2-y^2}={{dx \over x}+{dx \over x}+{dz \over z}\over 0}$$

Then

$$\ln(xyz)=\ln (C_2)$$

Or

$$xyz=C_2$$

Parametric equation of straight line is

$$x=t, y=-t, z=1$$

Substitute

$$t^2+(-t)^2-2(1)=C_1$$ $$t(-t)(1)=C_2$$

Eliminate $t$

$$2t^2-2=C_1$$ $$t^2=-C_2$$

Then

$$-2C_2-2=C_1$$

Or

$$C_1+2C_2+2=0$$

Hence, the integral surface, which contains the straight line

$$x^2+y^2-2z+2xyz+2=0$$