Question #231351

11)Solve the PDE (x+y²)p+yq=z+x².


1
Expert's answer
2021-09-21T02:06:55-0400


this is hyperbolic paraboloid (x+y2)p+yq=z+x2(x+y²)p+yq=z+x²


x2(x+y2)p+yq=zx^2-(x+y²)p+yq=z


real solution is;


0=x2px+py2+yq0=x^2-px+py^2+yq




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