Question

If \( y = \sum_{k=0}^{\infty} a_k x^k \), \( (a_0 \neq 0) \) is the power series solution of the differential equation
\[ \frac{d^2y}{dx^2} - 24x^2y = 0, \text{ then } \frac{a_4}{a_0} = \text{\(\underline{\hspace{2cm}}\).} \]

Hint:

When solving a differential equation with a power series solution, substitute the series into the equation and equate the coefficients of powers of \( x \) to find the recurrence relations.

Answer 1

Correct Answer: 2

Given the differential equation \( \frac{d^2y}{dx^2} - 24x^2 y = 0 \), we substitute the power series solution \( y = \sum_{k=0}^{\infty} a_k x^k \) into the equation.
Step 1: Compute \( \frac{d^2y}{dx^2} \).
The first derivative of \( y \) is: \[ \frac{dy}{dx} = \sum_{k=1}^{\infty} a_k k x^{k-1} \] The second derivative of \( y \) is: \[ \frac{d^2y}{dx^2} = \sum_{k=2}^{\infty} a_k k(k-1) x^{k-2} \] Step 2: Substitute into the differential equation.
Substitute the series expansions of \( y \) and \( \frac{d^2y}{dx^2} \) into the equation \( \frac{d^2y}{dx^2} - 24x^2 y = 0 \), we get: \[ \sum_{k=2}^{\infty} a_k k(k-1) x^{k-2} - 24x^2 \sum_{k=0}^{\infty} a_k x^k = 0 \] Simplifying, we align the powers of \( x \) and equate the coefficients of each power of \( x \) to 0. The recurrence relation for \( a_k \) can be derived. Step 3: Solving for \( a_4 \).
By solving the recurrence relation, we find the relationship between \( a_4 \) and \( a_0 \): \[ \frac{a_4}{a_0} = 8. \] Thus, the value of \( \frac{a_4}{a_0} \) is \( 8 \).