The given equation is:

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

This is a **differential equation**. Let's check if it is exact or can be solved using an integrating factor.

---

### Step 1: Rewrite the equation in the standard form
We compare it to the standard form of a first-order differential equation:

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

where:
- \( M(x, y) = x^2 + 4xy \)
- \( N(x, y) = x \)

---

### Step 2: Check for exactness
A differential equation \( M(x, y) \, dx + N(x, y) \, dy = 0 \) is **exact** if:

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

Let's compute the partial derivatives:

\[
\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^2 + 4xy) = 4x
\]
\[
\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x) = 1
\]

Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is **not exact**.

---

### Step 3: Check for an integrating factor
Since the equation is not exact, we need an integrating factor. An integrating factor \( \mu(x, y) \) is a function that, when multiplied with the given equation, makes it exact. 

For simplicity, we’ll assume an integrating factor depends only on \(x\) or \(y\). If it’s in terms of \(x\), the form \( \mu(x) = \frac{1}{x} \) is a useful candidate here.

---

### Step 4: Multiply by \( \frac{1}{x} \)
Multiplying the original equation by \( \frac{1}{x} \):

\[
\frac{1}{x} \left( (x^2 + 4xy) \, dx + x \, dy \right) = 0
\]

This simplifies to:

\[
(x + 4y) \, dx + dy = 0
\]

---

### Step 5: Solve the new exact equation
Now the new equation is:

\[
(x + 4y) \, dx + dy = 0
\]

This is an **exact differential equation**, as:

\[
M(x, y) = x + 4y, \quad N(x, y) = 1
\]

\[
\frac{\partial M}{\partial y} = 4, \quad \frac{\partial N}{\partial x} = 4
\]

Both partial derivatives are equal, confirming the equation is exact.

---

### Step 6: Find the solution
Since the equation is exact, there exists a function \( F(x, y) \) such that:

\[
\frac{\partial F}{\partial x} = M(x, y) = x + 4y, \quad \frac{\partial F}{\partial y} = N(x, y) = 1
\]

To find \( F(x, y) \), integrate \( M(x, y) = x + 4y \) with respect to \(x\):

\[
F(x, y) = \int (x + 4y) \, dx = \frac{x^2}{2} + 4xy + g(y)
\]

Here, \( g(y) \) is an arbitrary function of \(y\). 

Now, differentiate \( F(x, y) \) with respect to \(y\):

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

Since \( \frac{\partial F}{\partial y} = N(x, y) = 1 \), we have:

\[
4x + g'(y) = 1 \quad \Rightarrow \quad g'(y) = 1 \quad \Rightarrow \quad g(y) = y
\]

Thus, the function \( F(x, y) \) becomes:

\[
F(x, y) = \frac{x^2}{2} + 4xy + y = C
\]

---

### Step 7: General solution
The general solution to the differential equation is:

\[
\frac{x^2}{2} + 4xy + y = C
\]

where \(C\) is an arbitrary constant.

---

Do you need further clarification or steps explained in more detail?  

---

Here are 5 related questions:  
1. How do you find the general solution for a non-exact differential equation?  
2. What other methods exist to solve non-exact equations if integrating factors are hard to find?  
3. How do we determine if a differential equation is linear?  
4. Can integrating factors depend on both \(x\) and \(y\)?  
5. How is the method of solving exact equations related to potential functions in physics?

**Tip:** Integrating factors are often easier to find when they depend on just one variable (like \(x\) or \(y\)).

Learn how to solve the non-exact differential equation (x^2 + 4xy)dx + xdy = 0 using the integrating factor method. This step-by-step guide explains the exactness condition, partial derivatives, and finding the general solution. Ideal for students in Grades 11-12 studying differential equations.

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