Question

x(y^2+z)p-y(x^2+z)q=z(x^2-y^2)

Answer 1

$$Solution$$

Given

$$x(y^2+z)p-y(x^2+z)q=z(x^2-y^2)--------->(1)$$

Lagrange's auxiliary equation of (1) are;

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

Choosing $(\frac{1}{x}, \frac{1}{y}, \frac{1}{z})$ as multipliers, we get

$$\frac{dx}{x}, \frac{dy}{y}, \frac{dz}{z}=0$$

Integrating $\frac{dx}{x}+ \frac{dy}{y}+\frac{dz}{z}=0$ we get $xyz=c_1$

Choosing $(x,y,-1)$ as multipliers, we have

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

Integrating $xdx+ydy-dz=0$

$$\frac{x^2}{2}+ \frac{y^2}{2}-z=c \implies x^2+y^2-2z=c_2$$

$\therefore$ the solution is the system of equations:

$$xyz=c_1$$

$$x^2+y^2-2z=c_2$$

Taking $t$ as a parameter, the given equation of the straight line $x+y=0,z=1$ can be put in parametric form $x=t,y=-t,z=1$

Using this,

$$xyz=c_1$$

$$x^2+y^2-2z=c_2$$

May be written as,

$$-t^2=c_1$$

$$2t^2-2=c_2$$

Eliminating $t$ from the equation we get,

$$2(-c_1)-2=c_2 \implies 2c_1+c_2+2=0$$

Putting values of $c_1$ and $c_2$ , the desired integral surface is,

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