We are given the differential equation:

\[
4y \frac{dy}{dx} = x
\]

We are also provided with the initial condition \( y(4) = 9 \). 

### Step 1: Separate Variables
We can rewrite the differential equation to separate variables:

\[
4y \frac{dy}{dx} = x \quad \Rightarrow \quad y \, dy = \frac{x}{4} \, dx
\]

### Step 2: Integrate Both Sides
Now, integrate both sides:

\[
\int y \, dy = \int \frac{x}{4} \, dx
\]

The left-hand side gives:

\[
\frac{y^2}{2}
\]

The right-hand side gives:

\[
\frac{x^2}{8}
\]

So, we have:

\[
\frac{y^2}{2} = \frac{x^2}{8} + C
\]

### Step 3: Solve for \( y \)
Multiply the entire equation by 2 to simplify:

\[
y^2 = \frac{x^2}{4} + 2C
\]

Let \( 2C \) be a new constant, say \( C_1 \):

\[
y^2 = \frac{x^2}{4} + C_1
\]

Now, take the square root of both sides:

\[
y = \pm \sqrt{\frac{x^2}{4} + C_1}
\]

### Step 4: Apply the Initial Condition
We are given that \( y(4) = 9 \). Plugging this into the equation:

\[
9 = \sqrt{\frac{4^2}{4} + C_1} \quad \Rightarrow \quad 9 = \sqrt{4 + C_1}
\]

Square both sides:

\[
81 = 4 + C_1 \quad \Rightarrow \quad C_1 = 77
\]

### Final Answer:
Thus, the solution to the differential equation is:

\[
y = \sqrt{\frac{x^2}{4} + 77}
\]

Would you like a detailed explanation of any of the steps?

Here are 5 related questions to deepen your understanding:

1. How do you solve differential equations using separation of variables?
2. What role does the initial condition play in solving differential equations?
3. Can a differential equation have more than one solution without initial conditions?
4. How would the solution change if the initial condition were different?
5. What are other methods to solve first-order differential equations?

**Tip:** Always check that your solution satisfies both the differential equation and the initial condition to verify correctness.

Explore how to solve the differential equation 4yy' = x with the initial condition y(4) = 9 using separation of variables. This step-by-step solution covers key differential equation concepts and applies initial value problem-solving techniques, suitable for advanced high school and early college students.

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