Question

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

Answer 1

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.

Answer 2

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
`\(f(x, y)\)`:

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