Question

Find the integral surface of the partial differential equation:(x–y) y^2 p+(y–x)x^2 q=(x^2 +y^2)z through the curve xz=a^2, y=0

Answer 1

$dx/((x-y)*y^2)=dy/((y-x)*x^2)=dz/((x^2+y^2)*z)$

$dx/((x-y)*y^2)=dy/((y-x)*x^2)$

$dy/dx=-x^2/y^2$

$dy*y^2=dx*(-x^2)$

$C1=y^3+x^3$

$dx/((x-y)*y^2)=dz/((x^2+y^2)*z)$

$dz/z=dx(x^2+y^2)/((x-y)*y^2)$

$lnz=1/y^2*F(x,y)+C2$

y can't be 0 in this equation that is why there is no intersection with curve xz=a^2, y=0.

Answer: no solution.