form PDE by eliminating the arbitrary constants $f$ and $g$ in:

Question

$$z=x^2f(y)+y^2g(x)$$

Please help , not very sure how to differentiate this with respect to $x$ then $y$

Answer 1

Take the derivative according to x and consider function of y as constants. You get:

$$z=x^2f(y)+y^2g(x)$$ $$\partial_yz=\partial_y(x^2f(y)+y^2g(x))=x^2f'(y)+2yg(x)$$ $$\partial_xz=\partial_x(x^2f(y)+y^2g(x))=2xf(y)+y^2g'(x)$$ differentiating twice according to x first then according to y $$\partial_{yx}z=2xf'(y)+2yg'(x)=\partial_{xy}z$$