Great! Let's solve the given differential equation using the **integrating factor method**.

The given equation is:

\[
(x + 2) y' = 5 - 8y - 4xy
\]

First, we rewrite it in the standard linear form:

\[
y' + P(x)y = Q(x)
\]

We need to divide the whole equation by \(x + 2\):

\[
y' + \frac{4x + 8}{x + 2} y = \frac{5}{x + 2}
\]

This simplifies to:

\[
y' + 4y = \frac{5}{x + 2}
\]

Now, this is a linear differential equation of the form \(y' + P(x)y = Q(x)\), where \(P(x) = 4\) and \(Q(x) = \frac{5}{x + 2}\).

### Step 1: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[
\mu(x) = e^{\int P(x) dx} = e^{\int 4 \, dx} = e^{4x}
\]

### Step 2: Multiply the equation by the integrating factor
Now, multiply both sides of the differential equation by \(e^{4x}\):

\[
e^{4x} y' + 4e^{4x} y = e^{4x} \cdot \frac{5}{x + 2}
\]

This simplifies the left-hand side to the derivative of \(e^{4x}y\):

\[
\frac{d}{dx}(e^{4x} y) = e^{4x} \cdot \frac{5}{x + 2}
\]

### Step 3: Integrate both sides
Now, integrate both sides with respect to \(x\):

\[
e^{4x} y = \int e^{4x} \cdot \frac{5}{x + 2} \, dx
\]

This integral is non-trivial, and I'll solve it by substitution or other methods.

Let me calculate the integral for the right-hand side.The integral \( \int e^{4x} \cdot \frac{5}{x + 2} \, dx \) does not simplify directly using standard methods, so it remains in the form:

\[
\int e^{4x} \cdot \frac{5}{x + 2} \, dx = 5 \cdot \int \frac{e^{4x}}{x + 2} \, dx
\]

This is a more complex integral, typically requiring advanced techniques or approximation methods, such as integration by parts or using special functions.

Would you like to proceed with numerical methods to approximate the solution, or attempt a series expansion to handle the integral more easily?

Learn how to solve a differential equation using the integrating factor method step-by-step. This problem involves transforming the differential equation using an integrating factor and integrating to find the solution. Suitable for advanced undergraduate students and those studying differential equations.

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