## Solving the Differential Equation (5x+4y)dx + (4x-8y^3)dy = 0

This article will guide you through the process of solving the given differential equation:

**(5x+4y)dx + (4x-8y^3)dy = 0**

This equation is a **first-order, non-linear differential equation** and can be solved using the **method of exact differential equations**.

### 1. Checking for Exactness

To determine if the equation is exact, we need to check if the following condition holds:

**∂M/∂y = ∂N/∂x**

Where:

**M(x, y)**= 5x + 4y**N(x, y)**= 4x - 8y^3

Let's calculate the partial derivatives:

- ∂M/∂y = 4
- ∂N/∂x = 4

Since ∂M/∂y = ∂N/∂x, the given differential equation is **exact**.

### 2. Finding the Potential Function

Now, we need to find a function **u(x, y)** such that:

- ∂u/∂x = M(x, y)
- ∂u/∂y = N(x, y)

Integrating the first equation with respect to x, we get:

u(x, y) = ∫(5x + 4y)dx = (5/2)x^2 + 4xy + g(y)

Where g(y) is an arbitrary function of y.

Now, differentiating this equation with respect to y and equating it to N(x, y), we get:

∂u/∂y = 4x + g'(y) = 4x - 8y^3

This implies that g'(y) = -8y^3. Integrating both sides, we get:

g(y) = -2y^4 + C

Where C is an arbitrary constant.

Substituting the value of g(y) back into the potential function, we get:

u(x, y) = (5/2)x^2 + 4xy - 2y^4 + C

### 3. The General Solution

The general solution of the exact differential equation is given by:

**u(x, y) = C**

Therefore, the general solution of the given differential equation is:

**(5/2)x^2 + 4xy - 2y^4 = C**

Where C is an arbitrary constant.

### Conclusion

By using the method of exact differential equations, we have successfully solved the given differential equation:

**(5x+4y)dx + (4x-8y^3)dy = 0**

The general solution of this equation is:

**(5/2)x^2 + 4xy - 2y^4 = C**

Where C is an arbitrary constant.