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The problem in the image asks to solve the given second-order differential equation using a power series method. The equation and initial conditions are as follows:

\[
(x^2 - 4)y'' + 8xy' + 6y = 0
\]
with initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \).

The given answer structure is:

\[
y = \sum_{n=0}^{\infty} (n+1) x^{2n} + \sum_{n=0}^{\infty} 0 \cdot x^{2n+1}
\]

This indicates that the solution is expressed as a power series, with terms in powers of \(x^{2n}\) (even powers of \(x\)) and the odd powers have zero coefficients.

### Breakdown of the steps for power series solution:

1. **Assume a power series solution for \( y \):**

\[
y(x) = \sum_{n=0}^{\infty} a_n x^n
\]

2. **Compute derivatives for the power series:**

\[
y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}, \quad y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}
\]

3. **Substitute into the differential equation:** 
Substitute the series for \( y \), \( y' \), and \( y'' \) into the equation \( (x^2 - 4)y'' + 8xy' + 6y = 0 \).

4. **Equate the coefficients of powers of \( x \) to zero:** 
This step involves solving for the coefficients \( a_n \) in the power series expansion by setting the resulting terms equal to zero.

5. **Use initial conditions:** 
The initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \) are used to determine specific values for the coefficients \( a_n \).

The form of the answer suggests that only the even powers of \(x\) contribute to the solution, while the odd powers have coefficients of zero. Thus, the solution takes the form of:

\[
y(x) = \sum_{n=0}^{\infty} (n+1) x^{2n}
\]

This corresponds to the solution involving even powers of \( x \).

---

Would you like a detailed explanation of how to compute the coefficients, or do you have any other specific questions?

Here are 5 follow-up questions to expand on this:

1. How do you find the coefficients \( a_n \) from the power series?
2. Why do the odd powers of \(x\) have zero coefficients in the solution?
3. How are the initial conditions used to determine specific terms in the series?
4. Can this power series solution be written in a closed form?
5. What is the radius of convergence for this power series?

**Tip:** Power series methods are especially useful for solving differential equations near regular points, where the coefficients of the equation do not introduce singularities.

Use power series to solve the initial-value problem (x^2 - 4)y'' + 8xy' + 6y = 0, y(0) = 1, y'(0) = 0.

Learn how to solve the differential equation (x^2 - 4)y'' + 8xy' + 6y = 0 using power series with initial conditions y(0) = 1, y'(0) = 0. This solution involves computing coefficients in a power series expansion and analyzing even and odd powers of x.

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