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!

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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!