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:

$y = \frac{C e^x}{x^3}$

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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.