Answer to Question #284264 in Differential Equations for John

Question #284264

F. 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. [CO1] *


1
Expert's answer
2022-01-18T09:15:04-0500

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"

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