Question #93312
Find the integral surface of the linear PDE
x(y²+z)p - y(x²+z)q = (x²-y²)z , which contains the straight line x+y =0 , z=1 .
1
Expert's answer
2019-08-30T09:57:10-0400

Auxiliary equations are


dxx(y2+z)=dyy(x2+z)=dzz(x2y2){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,x, y, -1, we get


xdx+ydydzx2y2+x2zx2y2y2zx2z+y2z=xdx+ydydz0{xdx+ydy-dz\over x^2y^2+x^2z-x^2y^2-y^2z-x^2z+y^2z}={xdx+ydy-dz\over 0}

Then


x2+y22z=C1x^2+y^2-2z=C_1

By Choosing multipliers 1/x,1/y,1/z,1/x, 1/y, 1/z, we get


dxx+dxx+dzzy2+zx2z+x2y2=dxx+dxx+dzz0{{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(C2)\ln(xyz)=\ln (C_2)

Or


xyz=C2xyz=C_2

Parametric equation of straight line is


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

Substitute


t2+(t)22(1)=C1t^2+(-t)^2-2(1)=C_1t(t)(1)=C2t(-t)(1)=C_2

Eliminate tt


2t22=C12t^2-2=C_1t2=C2t^2=-C_2

Then


2C22=C1-2C_2-2=C_1

Or


C1+2C2+2=0C_1+2C_2+2=0

Hence, the  integral surface, which contains the straight line


x2+y22z+2xyz+2=0x^2+y^2-2z+2xyz+2=0

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