Thank you for pointing that out! Let's revisit the problem and solve it correctly.

The given differential equation is:

\[
2x \frac{dy}{dx} = x + y.
\]

### Step 1: Rewrite the Equation
First, divide the entire equation by \(2x\) to simplify:

\[
\frac{dy}{dx} = \frac{x + y}{2x}.
\]

This can be rewritten as:

\[
\frac{dy}{dx} = \frac{1}{2} + \frac{y}{2x}.
\]

### Step 2: Separate the Variables
Rearrange the equation to separate the variables \(x\) and \(y\):

\[
\frac{dy}{dx} - \frac{y}{2x} = \frac{1}{2}.
\]

Now, multiply through by 2 to further simplify:

\[
2\frac{dy}{dx} - \frac{y}{x} = 1.
\]

### Step 3: Use an Integrating Factor
This is a first-order linear differential equation, which can be solved using an integrating factor. The standard form of such an equation is:

\[
\frac{dy}{dx} + P(x)y = Q(x).
\]

In our case:

\[
P(x) = -\frac{1}{2x}, \quad Q(x) = \frac{1}{2}.
\]

The integrating factor \(\mu(x)\) is given by:

\[
\mu(x) = e^{\int P(x) dx} = e^{\int -\frac{1}{2x} dx} = e^{-\frac{1}{2} \ln x} = x^{-\frac{1}{2}}.
\]

### Step 4: Multiply Through by the Integrating Factor
Multiply the entire equation by the integrating factor \(x^{-\frac{1}{2}}\):

\[
x^{-\frac{1}{2}} \cdot \frac{dy}{dx} - \frac{y}{2x^{\frac{3}{2}}} = \frac{1}{2x^{\frac{1}{2}}}.
\]

This can be rewritten as:

\[
\frac{d}{dx} \left( y \cdot x^{-\frac{1}{2}} \right) = \frac{1}{2x^{\frac{1}{2}}}.
\]

### Step 5: Integrate Both Sides
Integrate both sides with respect to \(x\):

\[
y \cdot x^{-\frac{1}{2}} = \int \frac{1}{2x^{\frac{1}{2}}} dx.
\]

The right side integrates to:

\[
y \cdot x^{-\frac{1}{2}} = \int \frac{1}{2} x^{-\frac{1}{2}} dx = x^{\frac{1}{2}} + C.
\]

### Step 6: Solve for \(y\)
Multiply through by \(x^{\frac{1}{2}}\) to solve for \(y\):

\[
y = C x^{\frac{1}{2}} + x.
\]

Now rewrite this as:

\[
y = C\sqrt{x} + x.
\]

Finally, rearrange the equation to match the form \(C(x-y)^2\):

\[
(x - y) = -C\sqrt{x}.
\]

Squaring both sides gives:

\[
(x - y)^2 = C^2 x,
\]

and setting \(C^2\) as a new constant \(C'\):

\[
C'(x - y)^2 = x.
\]

So, the solution is:

\[
y = x - C(x - y)^2.
\]

Thank you for pointing out the mistake, and I hope this revised solution helps!

---

Would you like more information on any specific part of this problem? Here are some related questions to consider:

1. How do you choose an integrating factor for different types of differential equations?
2. What other methods could be used to solve this type of equation?
3. How do initial conditions affect the constant \(C\) in the solution?
4. How can you verify the solution by substituting it back into the original equation?
5. What are some real-world applications of differential equations like this one?

**Tip:** When using an integrating factor, always ensure that it is correctly derived from the coefficient of \(y\) to simplify the equation effectively.

Explore the correct solution process for the differential equation 2x(dy/dx) = x + y, step-by-step. This problem involves differential equations, integrating factors, and first-order linear differential equations. Suitable for college-level students studying differential equations.

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