The indicated function 
y1(x)
 is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2,
y2 = y1(x) 
e−∫P(x) dx
y
2
1
(x)

dx
        (5)
as instructed, to find a second solution 
y2(x).
x2y'' − 7xy' + 16y = 0;    y1 = x4

We are given the second-order linear homogeneous differential equation:

\[
x^2 y'' - 7x y' + 16y = 0
\]
with a known solution \( y_1(x) = x^4 \).

To find a second linearly independent solution \( y_2(x) \), we can use the method of *reduction of order*. This method assumes the second solution has the form:

\[
y_2(x) = v(x) y_1(x)
\]

where \( v(x) \) is an unknown function to be determined.

### Step 1: Substitute \( y_2(x) = v(x) y_1(x) \) into the differential equation.

We know that \( y_1(x) = x^4 \), so we assume \( y_2(x) = v(x) x^4 \).

Now, we need to compute \( y_2'(x) \) and \( y_2''(x) \):

- First derivative:
\[
y_2'(x) = v'(x) x^4 + v(x) \cdot 4x^3
\]

- Second derivative:
\[
y_2''(x) = v''(x) x^4 + 2v'(x) \cdot 4x^3 + v(x) \cdot 12x^2
\]

### Step 2: Substitute \( y_2(x), y_2'(x), y_2''(x) \) into the original differential equation.

The original equation is:

\[
x^2 y'' - 7x y' + 16y = 0
\]

Substitute the expressions for \( y_2''(x), y_2'(x), \) and \( y_2(x) \):

\[
x^2 \left(v''(x) x^4 + 8v'(x) x^3 + 12v(x) x^2 \right) - 7x \left(v'(x) x^4 + 4v(x) x^3 \right) + 16 \cdot v(x) x^4 = 0
\]

### Step 3: Simplify the equation.

Simplify the powers of \( x \) in each term:

\[
x^6 v''(x) + 8x^5 v'(x) + 12x^4 v(x) - 7x^5 v'(x) - 28x^4 v(x) + 16x^4 v(x) = 0
\]

Combine like terms:

\[
x^6 v''(x) + (8x^5 - 7x^5) v'(x) + (12x^4 - 28x^4 + 16x^4) v(x) = 0
\]

This simplifies to:

\[
x^6 v''(x) + x^5 v'(x) = 0
\]

### Step 4: Solve the resulting equation for \( v(x) \).

We can factor out \( x^5 \) from the equation:

\[
x^5 \left(x v''(x) + v'(x)\right) = 0
\]

Since \( x^5 \neq 0 \) for \( x \neq 0 \), we are left with:

\[
x v''(x) + v'(x) = 0
\]

This is a first-order differential equation in \( v'(x) \). To solve this, we can rewrite it as:

\[
\frac{d}{dx}(v'(x)) + \frac{1}{x} v'(x) = 0
\]

### Step 5: Solve for \( v'(x) \).

This equation can be solved by separation of variables. Rewrite it as:

\[
\frac{dv'(x)}{v'(x)} = -\frac{dx}{x}
\]

Integrate both sides:

\[
\ln|v'(x)| = -\ln|x| + C
\]

Exponentiate both sides:

\[
v'(x) = \frac{C_1}{x}
\]

### Step 6: Solve for \( v(x) \).

Integrate again to find \( v(x) \):

\[
v(x) = C_1 \ln|x| + C_2
\]

### Step 7: Find \( y_2(x) \).

Now that we have \( v(x) \), we can substitute it back into the expression for \( y_2(x) \):

\[
y_2(x) = v(x) y_1(x) = (C_1 \ln|x| + C_2) x^4
\]

Since we are looking for a second solution independent from \( y_1(x) = x^4 \), we can discard the constant term \( C_2 x^4 \) (as it would simply multiply the known solution). Thus, the second linearly independent solution is:

\[
y_2(x) = C_1 x^4 \ln|x|
\]

Therefore, the second solution is:

\[
y_2(x) = x^4 \ln|x|
\]

### Final Answer:
The second solution to the differential equation is \( y_2(x) = x^4 \ln|x| \).

---

Would you like further clarification on any of these steps?

Here are 5 related questions to explore further:
1. How do you determine linear independence between two solutions of a differential equation?
2. What is the general solution to a second-order linear homogeneous differential equation?
3. How does the method of reduction of order work for non-constant coefficients?
4. What is the importance of the Wronskian in determining independent solutions?
5. Can you apply the reduction of order method to a non-homogeneous differential equation?

**Tip**: When solving second-order differential equations, always check if one solution is already given. It can greatly simplify finding the second solution through reduction of order!

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order to find a second solution y2(x). x^2y'' − 7xy' + 16y = 0; y1 = x^4.

Explore the step-by-step method to find the second solution to the differential equation x^2y'' − 7xy' + 16y = 0 using reduction of order. The known solution y1 = x^4 leads to the second solution y2 = x^4 ln|x|. This solution involves advanced concepts in differential equations, suitable for undergraduate math students.

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