The given differential equation is:

\[
xy' + 3y - xy = 0
\]

We can rewrite it in a more standard form by rearranging and factoring terms:

\[
xy' - xy + 3y = 0
\]

\[
xy' + y(-x + 3) = 0
\]

To solve this, we first note that this is a **first-order linear differential equation** of the form \( y' + p(x)y = q(x) \), and we can solve it using an **integrating factor**.

### Step 1: Rewrite the equation
We want to rewrite the equation in the form:

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

Dividing through by \(x\), we have:

\[
y' + \left(\frac{3 - x}{x}\right)y = 0
\]

Here, \( P(x) = \frac{3 - x}{x} \) and \( Q(x) = 0 \).

### Step 2: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[
\mu(x) = e^{\int P(x) dx} = e^{\int \frac{3 - x}{x} dx}
\]

We can split this integral:

\[
\mu(x) = e^{\int \left( \frac{3}{x} - 1 \right) dx} = e^{\int \frac{3}{x} dx - \int 1 dx}
\]

\[
\mu(x) = e^{3 \ln|x| - x} = e^{\ln|x^3| - x} = x^3 e^{-x}
\]

### Step 3: Multiply through by the integrating factor
Now, we multiply the entire differential equation by the integrating factor \( \mu(x) = x^3 e^{-x} \):

\[
x^3 e^{-x} \cdot y' + x^3 e^{-x} \cdot \frac{(3 - x)}{x} \cdot y = 0
\]

This simplifies to:

\[
\frac{d}{dx} \left( x^3 e^{-x} y \right) = 0
\]

### Step 4: Integrate both sides
Now integrate both sides:

\[
x^3 e^{-x} y = C
\]

where \( C \) is the constant of integration.

### Step 5: Solve for \( y \)
Finally, solving for \( y \), we get:

\[
y = \frac{C}{x^3 e^{-x}} = \frac{C e^x}{x^3}
\]

Thus, the general solution to the differential equation is:

\[
y = \frac{C e^x}{x^3}
\]

---

Would you like more details on any step or have questions?

Here are some additional problems to explore:
1. Solve \( y' - \frac{2}{x}y = x^2 \).
2. What is the integrating factor for \( y' + y = x^2 \)?
3. Solve the equation \( y' + 2y = 0 \).
4. Solve the equation \( xy' - 4y = 0 \).
5. Can you derive the solution to \( y' + \frac{y}{x} = 0 \)?

**Tip:** Integrating factors are a powerful method for solving first-order linear differential equations, making them essential for tackling a broad class of problems.

Learn how to solve the first-order linear differential equation xy' + 3y - xy = 0 using the integrating factor method. This solution provides a step-by-step approach, covering key concepts in differential equations, such as integrating factors. Suitable for college-level calculus students.

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