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The given initial value problem (IVP) is:

\[
y' = \frac{x}{2y} + \frac{y}{x}, \quad y(1) = 3
\]

The hint states that this is a Bernoulli equation. Let's break this down and solve it step-by-step.

### Step 1: Recognize the Bernoulli form
The Bernoulli differential equation has the general form:

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

To solve it, we first try to manipulate the given equation into this form.

The equation is:

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

Multiply both sides by \( y \) to remove the denominator on the right-hand side:

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

Now, rearrange the terms to:

\[
y y' - \frac{y^2}{x} = \frac{x}{2}
\]

### Step 2: Set up a substitution
This is not quite in the Bernoulli form yet. We make a substitution to simplify it further. Let's use:

\[
v = y^2 \quad \text{so that} \quad y' = \frac{1}{2} \frac{dv}{dx}
\]

Thus, the equation becomes:

\[
\frac{1}{2} v' - \frac{v}{x} = \frac{x}{2}
\]

### Step 3: Solve the linear first-order equation
Now, multiply the entire equation by 2 to eliminate the fraction:

\[
v' - \frac{2v}{x} = x
\]

This is now a linear first-order differential equation. To solve it, we can use an integrating factor.

The integrating factor is:

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

### Step 4: Multiply through by the integrating factor
Now, multiply the whole equation by \(\frac{1}{x^2}\):

\[
\frac{1}{x^2} v' - \frac{2v}{x^3} = \frac{x}{x^2}
\]

This simplifies to:

\[
\left( \frac{v}{x^2} \right)' = \frac{1}{x}
\]

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

\[
\frac{v}{x^2} = \ln|x| + C
\]

Thus,

\[
v = x^2 \ln|x| + C x^2
\]

### Step 6: Back-substitute \( v = y^2 \)
Now, substitute back \( v = y^2 \):

\[
y^2 = x^2 \ln|x| + C x^2
\]

Taking the square root of both sides:

\[
y = \pm x \sqrt{\ln|x| + C}
\]

### Step 7: Use the initial condition
We are given the initial condition \( y(1) = 3 \). Substituting \( x = 1 \) and \( y = 3 \) into the solution:

\[
3 = \pm 1 \cdot \sqrt{\ln(1) + C}
\]

Since \( \ln(1) = 0 \), this simplifies to:

\[
3 = \sqrt{C}
\]

Thus, \( C = 9 \).

### Final Solution:
The solution to the initial value problem is:

\[
y = x \sqrt{\ln|x| + 9}
\]

Would you like further clarifications or details on any step? Here are a few questions to extend your understanding:

1. What is the method for solving Bernoulli differential equations in general?
2. How can you verify that the final solution satisfies the original differential equation?
3. Why is an integrating factor useful for solving linear first-order differential equations?
4. What are some other types of substitutions commonly used to simplify differential equations?
5. Can Bernoulli equations always be solved using this method?

**Tip:** Always check if an equation is linear or can be transformed into a recognizable form like Bernoulli or separable to simplify the solving process.

Solve the IVP y' = x/2y + y/x, y(1) = 3. (Hint: the equation is Bernoulli.)

Learn how to solve the Bernoulli initial value problem y' = x/2y + y/x with y(1) = 3. This solution involves using a substitution and solving a linear first-order differential equation using an integrating factor. Suitable for students in calculus or differential equations courses.

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