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To solve this problem, we will apply the **method of reduction of order** to find the second solution \( y_2(x) \) of the differential equation:

\[
-40x^2y'' - 16x(x-5)y' + 16(x-5)y = 0, \quad x > 0
\]

### Step-by-step breakdown of the solution process:

1. **Given solution**: We are provided with the first solution \( y_1(x) = 5x \).

2. **Reduction of order**: For the second solution, we assume \( y_2(x) = v(x)y_1(x) = v(x)(5x) \), where \( v(x) \) is a function to be determined.

3. **Substitute into the differential equation**:
   - First, we calculate the derivatives of \( y_2(x) \):
     \[
     y_2'(x) = v'(x)5x + v(x)(5)
     \]
     \[
     y_2''(x) = v''(x)5x + v'(x)(5) + v'(x)(5)
     \]
     Simplifying the second derivative:
     \[
     y_2''(x) = v''(x)5x + 10v'(x)
     \]

4. **Substitute \( y_2(x) \), \( y_2'(x) \), and \( y_2''(x) \) into the original equation**:
   \[
   -40x^2(v''(x)5x + 10v'(x)) - 16x(x-5)(v'(x)5x + v(x)5) + 16(x-5)v(x)(5x) = 0
   \]
   Since \( y_1(x) = 5x \) is already a solution, simplifying this equation will help us find a simpler equation for \( v(x) \).

5. **Simplified equation for \( v(x) \)**:
   Using the fact that this is a linear equation in \( v(x) \), the equation simplifies to a first-order equation in terms of \( w(x) = v'(x) \), where \( w(x) \) satisfies the following:
   \[
   y_1 w' + (2y_1' + p(x)y_1)w = 0
   \]
   Substituting \( y_1 = 5x \) into this equation, we get the equation to solve for \( w(x) \).

6. **Solve the differential equation for \( w(x) \)**, given that \( w(1) = -4 \).

Let me calculate the simplified equation for \( w(x) \).The general solution for \( w(x) \) is:

\[
w(x) = \frac{C_1}{x^2}
\]

To determine the constant \( C_1 \), we use the initial condition \( w(1) = -4 \):

\[
w(1) = \frac{C_1}{1^2} = C_1 = -4
\]

Thus, the specific solution for \( w(x) \) is:

\[
w(x) = \frac{-4}{x^2}
\]

Now that we have \( w(x) \), we can proceed to find \( v(x) \) by integrating \( w(x) \):

\[
v'(x) = w(x) = \frac{-4}{x^2}
\]

Let me know if you'd like the detailed calculation of \( v(x) \) or if you have any questions!

Here are 5 related questions:
1. How do we use the reduction of order method in general for second-order ODEs?
2. What is the purpose of the initial condition \( w(1) = -4 \)?
3. How do we interpret the roles of \( y_1(x) \) and \( y_2(x) \) in forming the fundamental solution set?
4. Can you derive \( v(x) \) from \( w(x) = \frac{-4}{x^2} \)?
5. How does this method differ from solving a homogeneous second-order equation directly?

**Tip**: Remember, the second solution \( y_2(x) \) is often constructed by multiplying the first solution by a function, so the key is finding that function using reduction of order.

Consider the differential equation: -40x^2y'' - 16x(x - 5)y' + 16(x - 5)y = 0, x > 0. Given that y1(x) = 5x is a solution, apply the reduction of order method to find another solution y2 for which y1 and y2 form a fundamental solution set. Starting with y1, solve for w in y1 w' + (2y1' + p(x)y1)w = 0 so that w(1) = -4.

This problem involves solving a second-order differential equation -40x^2y'' - 16x(x - 5)y' + 16(x - 5)y = 0 using the method of reduction of order. The first solution y1(x) = 5x is given, and we apply the reduction method to find the second solution y2(x) to form a fundamental solution set.

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