Question #223860

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.

Expert's answer

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

may be re-written as

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

Eliminating from the equations , we have

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

Putting values of and, the desired integral surface is

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

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