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The given equation is:

\[
(4x^3y + 3) dx + (x^4 - 8) dy = 0
\]

This is a first-order differential equation, and we will try to find the general solution.

### Step 1: Check if the equation is exact
An exact differential equation has the form:

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

where \( M(x, y) = 4x^3y + 3 \) and \( N(x, y) = x^4 - 8 \).

The equation is exact if:

\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]

1. \(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(4x^3y + 3) = 4x^3\)
2. \(\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^4 - 8) = 4x^3\)

Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.

### Step 2: Find the potential function \( \psi(x, y) \)

Since the equation is exact, there exists a function \( \psi(x, y) \) such that:

\[
\frac{\partial \psi}{\partial x} = M(x, y) = 4x^3y + 3
\]
\[
\frac{\partial \psi}{\partial y} = N(x, y) = x^4 - 8
\]

#### Integrating \( M(x, y) \) with respect to \( x \):

\[
\psi(x, y) = \int (4x^3y + 3) dx = x^4y + 3x + g(y)
\]

where \( g(y) \) is a function of \( y \) only.

#### Differentiating \( \psi(x, y) \) with respect to \( y \):

\[
\frac{\partial \psi}{\partial y} = x^4 + g'(y)
\]

We know that \( \frac{\partial \psi}{\partial y} = x^4 - 8 \), so:

\[
x^4 + g'(y) = x^4 - 8
\]

This simplifies to:

\[
g'(y) = -8
\]

Integrating this with respect to \( y \), we get:

\[
g(y) = -8y
\]

### Step 3: Write the general solution

Thus, the potential function \( \psi(x, y) \) is:

\[
\psi(x, y) = x^4y + 3x - 8y
\]

The general solution is:

\[
\psi(x, y) = C
\]

or:

\[
x^4y + 3x - 8y = C
\]

### Step 4: Compare with the options

This matches the third option:

\[
x^4y + 3x - 8y = C
\]

### Final Answer:
The correct answer is the third option.

Would you like any further details or have any questions?

### Related Questions:
1. How do you determine if a differential equation is exact?
2. What are the steps to solve an exact differential equation?
3. How do you find the potential function \( \psi(x, y) \)?
4. What if the differential equation is not exact? How can you proceed?
5. Can you solve this type of differential equation using an integrating factor?

### Tip:
Always check if the differential equation is exact by comparing the partial derivatives of \( M(x, y) \) and \( N(x, y) \). If they are equal, the equation is exact, and you can proceed to find the potential function.

Step-by-step solution of the exact differential equation (4x^3y + 3) dx + (x^4 - 8) dy = 0, explaining key concepts such as exact equations, partial derivatives, and the general solution for calculus students.

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