Question

Find the integral surface of x(y^2+z)p-y( x^2+z)q=(x^2-y^2)z through the straight through the x+y = 0,z = 1.

Answer 1

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

Lagrange’s auxiliary equations of are

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

thus, the solution is the system of equations:

$\begin{cases} xyz=c_1 \\ x^2+y^2-2z=c_2 \end{cases}$

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

Using this,

$\begin{cases} xyz=c_1 \\ x^2+y^2-2z=c_2 \end{cases}$

may be re-written as

$\begin{cases} -t^2=c_1 \\ 2t^2-2=c_2 \end{cases}$

Eliminating from the equations , we have

$\begin{cases} 2(-c_1)-2=c_2 \\ 2c_1+c_2+2=0 \end{cases}$

Putting values of and, the desired integral surface is

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