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The problem asks us to solve a second-order differential equation using the reduction of order method. Here’s the differential equation and instructions from the image:

### Given:
\[
x^2 y'' + 2xy' - 2y = 0, \quad x > 0
\]
with the known solution \( y_1(x) = x \).

### (a) Use the reduction of order method to find the general solution.

#### Solution Process:

1. **Step 1: Form of the Solution**  
Since we are given that \( y_1(x) = x \) is a solution, we assume the second solution is of the form:
\[
y(x) = v(x) y_1(x) = v(x) x
\]
where \( v(x) \) is a function to be determined.

2. **Step 2: Derivatives of \( y(x) \)**  
Now, calculate the first and second derivatives of \( y(x) \):
\[
y' = v'(x) x + v(x) 
\]
\[
y'' = v''(x) x + 2 v'(x)
\]

3. **Step 3: Substitute into the Differential Equation**  
Substitute \( y = v(x) x \), \( y' = v'(x) x + v(x) \), and \( y'' = v''(x) x + 2 v'(x) \) into the original differential equation:
\[
x^2 (v''(x) x + 2 v'(x)) + 2x(v'(x) x + v(x)) - 2(v(x) x) = 0
\]
Simplify each term:
\[
x^3 v''(x) + 2x^2 v'(x) + 2x^2 v'(x) + 2x v(x) - 2x v(x) = 0
\]
\[
x^3 v''(x) + 4x^2 v'(x) = 0
\]

4. **Step 4: Simplify the Equation**  
Divide the whole equation by \( x^2 \) (since \( x > 0 \)):
\[
x v''(x) + 4 v'(x) = 0
\]

5. **Step 5: Solve the First-Order Equation**  
This is now a first-order differential equation in \( v'(x) \). Let \( u(x) = v'(x) \), so the equation becomes:
\[
x u'(x) + 4u(x) = 0
\]
This is a separable differential equation. Divide both sides by \( u(x) \) and \( x \):
\[
\frac{u'(x)}{u(x)} = -\frac{4}{x}
\]
Integrating both sides:
\[
\ln |u(x)| = -4 \ln |x| + C_1
\]
\[
u(x) = \frac{C_1}{x^4}
\]

6. **Step 6: Find \( v(x) \)**  
Now, integrate \( u(x) = v'(x) = \frac{C_1}{x^4} \) to find \( v(x) \):
\[
v(x) = \int \frac{C_1}{x^4} dx = -\frac{C_1}{3x^3} + C_2
\]

7. **Step 7: General Solution**  
The general solution is:
\[
y(x) = v(x) x = \left(-\frac{C_1}{3x^3} + C_2\right)x = -\frac{C_1}{3x^2} + C_2 x
\]
Thus, the general solution to the differential equation is:
\[
y(x) = C_1 x + \frac{C_2}{x^2}
\]

Let me know if you need further clarification or have any questions!

### 5 Related Questions:
1. How does the reduction of order method work in general for second-order linear differential equations?
2. Why is it necessary to use \( y_1(x) = x \) as part of the solution?
3. What are the conditions for applying the reduction of order method?
4. How would the solution change if the initial conditions were provided?
5. Can this method be extended to higher-order differential equations?

### Tip:  
When solving second-order differential equations, always verify if one solution is already known. This can simplify the process significantly using methods like reduction of order.

Consider the differential equation: x^2 y'' + 2xy' - 2y = 0, with y_1(x) = x being a solution. Use the reduction of order method to find the general solution.

This problem involves solving the second-order differential equation x^2 y'' + 2xy' - 2y = 0 using the reduction of order method. The known solution y_1(x) = x simplifies the process, and the solution method includes finding a second independent solution.

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