Question #231351

11)Solve the PDE (x+y²)p+yq=z+x².

Expert's answer

this is hyperbolic paraboloid "(x+y\u00b2)p+yq=z+x\u00b2"

"x^2-(x+y\u00b2)p+yq=z"

real solution is;

"0=x^2-px+py^2+yq"

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