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The given equation is:

\[
y(x + y + 1)dx + (x + 2y)dy = 0
\]

This is a first-order differential equation. Let's solve it using an appropriate method, likely an exact equation approach.

### Step 1: Rewriting the equation
We have:

\[
M(x, y)dx + N(x, y)dy = 0
\]
where:
- \( M(x, y) = y(x + y + 1) \)
- \( N(x, y) = x + 2y \)

### Step 2: Check if the equation is exact
To verify if the equation is exact, we compute the partial derivatives:

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

\[
\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}[x + 2y] = 1
\]

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

### Step 3: Find the integrating factor
An integrating factor is often a function of \( x \) or \( y \). Here, we will try \( \mu(y) \), meaning the integrating factor depends only on \( y \).

We need the following condition for \( \mu(y) \) to hold:

\[
\frac{1}{\mu(y)} \frac{d\mu(y)}{dy} = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}
\]

Substituting the values:

\[
\frac{1}{\mu(y)} \frac{d\mu(y)}{dy} = \frac{(x + 2y + 1) - 1}{x + 2y} = \frac{x + 2y}{x + 2y} = 1
\]

Solving this differential equation for \( \mu(y) \):

\[
\frac{d\mu(y)}{dy} = \mu(y)
\]

The solution to this is:

\[
\mu(y) = e^y
\]

### Step 4: Multiply the original equation by the integrating factor
Multiplying the entire equation by \( e^y \), we get:

\[
e^y y(x + y + 1)dx + e^y(x + 2y)dy = 0
\]

This new equation should now be exact.

### Step 5: Solve the exact equation
Now that the equation is exact, we can find a potential function \( \psi(x, y) \) such that:

\[
\frac{\partial \psi}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial \psi}{\partial y} = N(x, y)
\]

I'll solve this next.The potential function \( \psi(x, y) \) for this exact differential equation is:

\[
\psi(x, y) = \frac{x^2 y e^y}{2} + x(y^2 e^y + y e^y) + \frac{(-x^2 y - 2xy^2 - 2xy + 2x + 4y - 4)e^y}{2}
\]

### Step 6: General solution
The general solution to the equation is found by setting \( \psi(x, y) = C \), where \( C \) is a constant. Therefore, the solution to the given differential equation is:

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

This represents the implicit solution of the differential equation.

Let me know if you'd like further details on any step!

---

### Related Questions:
1. How do you determine if a differential equation is exact?
2. What are some common methods to find an integrating factor?
3. Can an integrating factor be a function of both \( x \) and \( y \)?
4. How does multiplying by an integrating factor make a non-exact equation exact?
5. What are other types of first-order differential equations, and how are they solved?

#### Tip:
For exact equations, always verify that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) before solving directly!

Learn how to solve the first-order differential equation y(x + y + 1)dx + (x + 2y)dy = 0 using the method of exact equations and an integrating factor. This problem covers key concepts such as exactness conditions and solving with integrating factors. Suitable for undergraduate-level math students.

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