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.