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I have completed everything but the last question. Can you explain how to find the homogeneous equation?

To find the homogeneous equation for this second-order differential equation:

\[
x^2 \frac{d^2y}{dx^2} - 7x \frac{dy}{dx} + 16y = 0,
\]

we follow a process similar to solving linear differential equations with constant coefficients, except we account for the fact that the equation involves \(x^2\) and \(x\), which suggest a solution of the form \(y = x^r\).

### Steps to find the homogeneous equation:

1. **Assume a solution**: 
   We assume a solution of the form \(y = x^r\), where \(r\) is some unknown constant. The reason we choose this form is because when dealing with equations of this type (Euler-Cauchy equations), solutions of the form \(x^r\) often fit naturally.

2. **Find derivatives**: 
   We compute the first and second derivatives of \(y = x^r\):
   \[
   \frac{dy}{dx} = r x^{r-1},
   \]
   \[
   \frac{d^2y}{dx^2} = r(r-1) x^{r-2}.
   \]

3. **Substitute into the original equation**: 
   Substituting \(y = x^r\), \(\frac{dy}{dx}\), and \(\frac{d^2y}{dx^2}\) into the differential equation:
   \[
   x^2 \frac{d^2y}{dx^2} - 7x \frac{dy}{dx} + 16y = 0,
   \]
   becomes
   \[
   x^2 \left[ r(r-1) x^{r-2} \right] - 7x \left[ r x^{r-1} \right] + 16 x^r = 0.
   \]

4. **Simplify**: 
   Simplifying each term:
   \[
   x^2 \cdot r(r-1) x^{r-2} = r(r-1) x^r,
   \]
   \[
   -7x \cdot r x^{r-1} = -7r x^r,
   \]
   \[
   16 x^r = 16 x^r.
   \]

   Putting them together:
   \[
   r(r-1) x^r - 7r x^r + 16 x^r = 0.
   \]

5. **Factor out \(x^r\)**:
   Since \(x^r \neq 0\), we can factor it out:
   \[
   x^r \left[ r(r-1) - 7r + 16 \right] = 0,
   \]
   \[
   r(r-1) - 7r + 16 = 0.
   \]

6. **Solve the characteristic equation**:
   The equation becomes:
   \[
   r^2 - r - 7r + 16 = 0,
   \]
   \[
   r^2 - 8r + 16 = 0.
   \]
   This is a quadratic equation. Solving it using the quadratic formula:
   \[
   r = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(16)}}{2(1)},
   \]
   \[
   r = \frac{8 \pm \sqrt{64 - 64}}{2},
   \]
   \[
   r = \frac{8 \pm 0}{2} = 4.
   \]

   Hence, \(r = 4\).

### Conclusion:
The homogeneous equation has a double root \(r = 4\), which means the general solution to the homogeneous equation will have the form:
\[
y_h = (C_1 + C_2 \ln x) x^4,
\]
where \(C_1\) and \(C_2\) are constants.

Do you have any more questions or would you like further details on any step?

Here are 5 follow-up questions you might find helpful:
1. Why do we assume a solution of the form \(y = x^r\) for Euler-Cauchy equations?
2. How would the solution change if the roots were complex instead of real?
3. What happens if the characteristic equation has distinct real roots?
4. Can the integrating factor method be applied directly to solve this differential equation?
5. How does the presence of logarithmic terms arise when solving for repeated roots?

**Tip**: When solving homogeneous equations, always simplify carefully and check for common factors to avoid algebraic mistakes.

Explore the solution to a second-order homogeneous differential equation: x^2 y'' - 7x y' + 16y = 0. This detailed guide covers key steps, including the Euler-Cauchy equation method, derivative calculations, and solving the characteristic equation. Ideal for students studying calculus and differential equations.

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