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Can there exist a c^2 function f(x,y) with fx=2x-5y and fy=4x+y

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Final answer:

A C^2 function f(x, y) with the given partial derivatives fx = 2x - 5y and fy = 4x + y cannot exist, because the mixed partial derivatives are not equal, which fails to satisfy the Clairaut's theorem.

Step-by-step explanation:

The question asks if a function f(x, y) that is twice differentiable (C^2 function) can exist with partial derivatives fx = 2x - 5y and fy = 4x + y. To determine if such a function exists, we must check if the mixed partial derivatives of f are equal, that is, if the Clairaut's theorem (also known as the Schwarz theorem) holds, which states that for a C^2 function, the mixed partial derivatives are equal no matter the order of differentiation.

For this to be true, the partial derivative of fx with respect to y must be equal to the partial derivative of fy with respect to x. By differentiating fx with respect to y, we get -5. Similarly, by differentiating fy with respect to x, we obtain 4. Since these two values are not equal, the equality of mixed partials does not hold, and therefore, a C^2 function with the given partial derivatives cannot exist.


User Popeye
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4 votes

The function
\(f(x, y) = x^2 - 5xy + (1)/(2)y^2 + C\) satisfies the given conditions, with
\(C\) being an arbitrary constant.

If you want to find a function f(x, y) such that
\(f_x = 2x - 5y\)and \
f_y = 4x + y\), you can integrate each partial derivative with respect to its corresponding variable to obtain \(f(x, y)\).

Starting with
\(f_x\):


\[f(x, y) = \int (2x - 5y) \, dx = x^2 - 5xy + g(y),\]

where
\(g(y)\) is an arbitrary function of \(y\) only.

Now, take the partial derivative of
\(f(x, y)\) with respect to (y) and equate it to
\(f_y\):


\[f_y = (\partial f)/(\partial y) = -5x + g'(y).\]

Compare this with
\(f_y = 4x + y\):


\[-5x + g'(y) = 4x + y.\]

To satisfy this equation, set
\(g'(y) = y\)and integrate to find
\(g(y)\):


\[g(y) = (1)/(2)y^2 + C,\]

where \(C\) is an arbitrary constant.

Substitute this back into the expression for
\(f(x, y)\):


\[f(x, y) = x^2 - 5xy + (1)/(2)y^2 + C.\]

User Sethammons
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