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The problem asks to find an integrating factor for the differential equation 

\[
(2x + 4y^3) \, dx + 4xy \, dy = 0
\]

and then to use it to find the general solution.

### Step 1: Check if the DE is exact
For a differential equation of the form:

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

the equation is exact if:

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

Here, \( M(x, y) = 2x + 4y^3 \) and \( N(x, y) = 4xy \).

Let's compute the partial derivatives:

\[
\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (2x + 4y^3) = 12y^2,
\]

\[
\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (4xy) = 4y.
\]

Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is not exact. We need to find an integrating factor \( \mu(x, y) \) to make it exact.

### Step 2: Finding the integrating factor
A common approach is to assume the integrating factor depends on only one variable, either \( x \) or \( y \). Let's first assume \( \mu = \mu(y) \).

Multiplying the original DE by \( \mu(y) \), we get:

\[
\mu(y) \cdot (2x + 4y^3) \, dx + \mu(y) \cdot 4xy \, dy = 0.
\]

For this to be exact, the condition becomes:

\[
\frac{\partial (\mu(y) \cdot (2x + 4y^3))}{\partial y} = \frac{\partial (\mu(y) \cdot 4xy)}{\partial x}.
\]

Expanding both sides:

\[
\mu(y) \cdot 12y^2 + \mu'(y) \cdot (2x + 4y^3) = 4y \cdot \mu(y).
\]

Let's simplify and equate terms involving \( x \):

\[
\mu(y) \cdot 12y^2 + 2x \cdot \mu'(y) + 4y^3 \cdot \mu'(y) = 4y \cdot \mu(y).
\]

By equating terms, we get the equation for \( \mu(y) \):

\[
\mu(y) \cdot 12y^2 = 4y \cdot \mu(y),
\]

which implies \( \mu(y) \) must cancel out any inconsistencies. We solve the simplified equation (or test for the most probable integrating factor). A reasonable guess is \( \mu(y) = \frac{1}{y^2} \) to make the equation exact.

Let's multiply the whole equation by \( \mu(y) = \frac{1}{y^2} \):

\[
\frac{1}{y^2} \cdot (2x + 4y^3) \, dx + \frac{1}{y^2} \cdot 4xy \, dy = 0.
\]

This simplifies to:

\[
\left(\frac{2x}{y^2} + 4y\right) \, dx + \frac{4x}{y} \, dy = 0.
\]

Now, compute the partial derivatives again to check if the equation is exact.

\[
\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}\left(\frac{2x}{y^2} + 4y\right) = -\frac{4x}{y^3} + 4,
\]

\[
\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}\left(\frac{4x}{y}\right) = \frac{4}{y}.
\]

Since the equation is still not exact with this form, the integrating factor must involve both \(x\) and \(y\). However, considering the complex interaction of the terms, the exact solution strategy involves deeper methods or testing likely candidates manually until an exact match is found.

Let's try finding the solution using a general solution approach with a hypothetical integrating factor or manual inspection.

Given the complexity, I'll analyze further or adjust based on deeper or simpler methods upon request.

### Next Steps
Would you like to continue exploring this approach or need detailed analysis of possible methods, or do you have questions?

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**Relative Questions:**

1. What is the process to determine the integrating factor when it depends on both \(x\) and \(y\)?
2. How do we verify if an integrating factor indeed makes a DE exact?
3. Can the method of integrating factors be generalized to non-linear DEs?
4. What other common methods are there to solve non-exact differential equations?
5. How does the presence of a specific variable in the integrating factor simplify solving DEs?

**Tip:** Always start by checking exactness before applying any integrating factors. This ensures you use the most straightforward method available.

Find an integrating factor for the DE (2x + 4y^3) dx + 4xy dy = 0 to be exact and hence find the general solution of the differential equation.

Learn how to find an integrating factor for the differential equation (2x + 4y^3) dx + 4xy dy = 0 and determine the general solution. This problem explores exact differential equations and integrating factors, suitable for college or university students.

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