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The given differential equation is:

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

We are tasked with finding \( y(x) \), and the hint suggests using the substitution:

\[
z(x) = \frac{y'(x)}{y(x)}
\]

Let's work through the problem step by step.

### Step 1: Express derivatives using \( z(x) \)

From the substitution \( z(x) = \frac{y'(x)}{y(x)} \), we can express \( y'(x) \) as:
\[
y'(x) = z(x) y(x)
\]

Now, let's compute \( y''(x) \) using the product rule:
\[
y''(x) = \frac{d}{dx}(z(x) y(x)) = z'(x) y(x) + z(x) y'(x)
\]
Since \( y'(x) = z(x) y(x) \), this simplifies to:
\[
y''(x) = z'(x) y(x) + z^2(x) y(x)
\]

### Step 2: Substitute into the original equation

Now, substitute \( y'(x) = z(x) y(x) \) and \( y''(x) = z'(x) y(x) + z^2(x) y(x) \) into the original equation:
\[
x y (z'(x) y(x) + z^2(x) y(x)) - x (z(x) y(x))^2 = y(x) z(x) y(x)
\]

This simplifies to:
\[
x y^2 (z'(x) + z^2(x)) - x z^2(x) y^2 = y^2 z(x)
\]

### Step 3: Simplify the equation

Cancel \( y^2 \) from all terms (assuming \( y \neq 0 \)):
\[
x (z'(x) + z^2(x)) - x z^2(x) = z(x)
\]

Simplifying further:
\[
x z'(x) = z(x)
\]

### Step 4: Solve the differential equation

This is a separable differential equation. We can rewrite it as:
\[
\frac{z'(x)}{z(x)} = \frac{1}{x}
\]

Integrating both sides:
\[
\ln |z(x)| = \ln |x| + C
\]

Exponentiating both sides:
\[
z(x) = Cx
\]

### Step 5: Return to \( y(x) \)

Recall that \( z(x) = \frac{y'(x)}{y(x)} \), so:
\[
\frac{y'(x)}{y(x)} = Cx
\]

This is another separable differential equation. Solving it:
\[
\frac{y'(x)}{y(x)} = Cx \implies \ln |y(x)| = \frac{C}{2} x^2 + D
\]

Exponentiating both sides:
\[
y(x) = e^{D} e^{\frac{C}{2} x^2}
\]

Let \( A = e^D \), so the solution is:
\[
y(x) = A e^{\frac{C}{2} x^2}
\]

### Final Answer:
\[
y(x) = A e^{\frac{C}{2} x^2}
\]

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

### Related questions:
1. How do you solve a first-order separable differential equation?
2. What is the product rule in differentiation, and how does it apply here?
3. How can you verify the solution to a differential equation?
4. Why can we assume \( y(x) \neq 0 \) when dividing by \( y(x) \)?
5. What are the general techniques for solving second-order differential equations?

### Tip:
When working with substitution in differential equations, carefully track how each derivative transforms, especially if applying product or chain rules.

Find y(x) given the equation: xyy'' - x(y')^2 = yy', using the substitution z(x) = y'(x)/y(x).

Learn how to solve the second-order differential equation xyy'' - x(y')^2 = yy' using substitution. This method involves transforming the equation into a separable form and solving step-by-step.

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