Question

Use the "mixed partials" check to see if the following differential equation is exact. (4x³ – y)dx +(y³ – x)dy=0. This equation has the form M(x,y)dx + N(x, y)dy = 0. Find: ∂M/∂y (x.y) = 12x² OM (2, y) = 3y² If the equation is exact, find a function F(x,y) whose differential, dF(, y), is the left hand side of the differential equation. That is, the level curves F(x, y) = C are solutions to the differential equation. If the equation is not exact, enter "not exact". F(x, y) = not exact

Answer 1

The differential equation is exact, as the mixed partial derivatives are equal. The potential function is found to be F(x, y) = x^4 - xy + y^4/4.

Step-by-step explanation:

To determine whether the given differential equation (4x³ – y)dx +(y³ – x)dy=0 is exact, we need to check if the mixed partial derivatives of M and N are equal. Here, M(x, y) = 4x³ – y and N(x, y) = y³ – x. The mixed partial derivative of M with respect to y is ∂M/∂y = -1, and the mixed partial derivative of N with respect to x is ∂N/∂x = -1. Since the mixed partial derivatives are equal, the differential equation is exact.

We now need to find a function F(x, y) such that dF = Mdx + Ndy. We integrate M with respect to x and N with respect to y, and then combine the results to get F(x, y). Integrating M gives us F(x, y) = x^4 - xy + g(y), where g(y) is a function of y only. To determine g(y), we differentiate F with respect to y and set it equal to N. This gives us g'(y) = y^3, which when integrated gives g(y) = y^4/4. Thus, F(x, y) = x^4 - xy + y^4/4, which is the potential function for the exact differential equation.