How do you determine if the differential equation 5x + 4y dx + (4x + 8y^3) dy = 0 is exact, and if it is, how do you solve it?

To determine whether the given differential equation is exact, we start by expressing it in the standard form:

M(x, y) dx + N(x, y) dy = 0

where:

A differential equation is said to be exact if the following condition holds:

∂M/∂y = ∂N/∂x

Now, we will calculate the partial derivatives:

Since ∂M/∂y = 4 and ∂N/∂x = 4 are equal, the differential equation is exact.

Next, we will solve the exact differential equation:

We need to find a function Ψ(x, y) such that:

To find Ψ, we first integrate M with respect to x:

Ψ(x, y) = ∫(5x + 4y) dx = (5/2)x² + 4xy + h(y)

Here, h(y) is an arbitrary function of y that we will determine later.

Next, we take the derivative of Ψ with respect to y:

∂Ψ/∂y = ∂/∂y [(5/2)x² + 4xy + h(y)] = 4x + h'(y)

Setting this equal to N(x, y):

4x + h'(y) = 4x + 8y³

From this, we see that h'(y) = 8y³. Thus, we can integrate to find h(y):

h(y) = ∫8y³ dy = 2y⁴ + C

where C is a constant. Now we can substitute h(y) back into Ψ:

Ψ(x, y) = (5/2)x² + 4xy + 2y⁴ + C

Setting Ψ(x, y) equal to a constant (say K), we get:

(5/2)x² + 4xy + 2y⁴ = K

This equation represents the solution to the differential equation. Thus, the final answer is:

(5/2)x² + 4xy + 2y⁴ = C

where C is a constant determined based on initial or boundary conditions given in specific problems.