Answer to Question #236272 in Chemical Engineering for Lok

Question #236272
11)Solve the PDE (x+y
1
Expert's answer
2021-09-22T00:25:20-0400

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 the straight line is



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

Substitute



t^2+(-t)^2-2(1)=C_1t(-t)(1)=C_2

Eliminate t



2t^2-2=C_1t^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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