Given a second order ODE x^2y'' + xy' - 3y = 0. How to find its forbenius series solution and recurrence relation

To find the Frobenius series solution for the given second-order ordinary differential equation (ODE):

\[
x^2 y'' + x y' - 3y = 0,
\]

we follow these steps:

### Step 1: Identify the Regular Singular Point
The equation is in the form of a Frobenius series around \(x = 0\), which is a regular singular point since the coefficients \(p(x) = \frac{1}{x}\) and \(q(x) = -3\) have singularities at \(x = 0\).

### Step 2: Assume a Frobenius Series Solution
We assume a solution of the form:

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

where \(r\) is the indicial exponent to be determined, and \(a_n\) are the coefficients.

### Step 3: Compute Derivatives
We compute the derivatives of \(y(x)\):

\[
y' = \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1},
\]

\[
y'' = \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2}.
\]

### Step 4: Substitute into the ODE
Substituting \(y\), \(y'\), and \(y''\) into the original ODE:

\[
x^2 \left( \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} \right) + x \left( \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1} \right) - 3 \left( \sum_{n=0}^{\infty} a_n x^{n+r} \right) = 0.
\]

This simplifies to:

\[
\sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r} + \sum_{n=0}^{\infty} a_n (n+r) x^{n+r} - 3 \sum_{n=0}^{\infty} a_n x^{n+r} = 0.
\]

### Step 5: Combine Terms
Combining these series gives:

\[
\sum_{n=0}^{\infty} a_n \left( (n+r)(n+r-1) + (n+r) - 3 \right) x^{n+r} = 0.
\]

### Step 6: Set the Coefficient to Zero
The expression in parentheses must equal zero for all \(n\):

\[
(n+r)(n+r-1) + (n+r) - 3 = 0.
\]

This simplifies to:

\[
(n+r)^2 - 3 = 0.
\]

### Step 7: Find the Indicial Roots
Setting \(n=0\):

\[
(r)^2 - 3 = 0 \implies r = \pm \sqrt{3}.
\]

### Step 8: Recurrence Relation
For \(n \geq 0\), we have:

\[
a_n = \frac{3 - (n+r)(n+r-1)}{(n+r)(n+r-1)} a_{n-1}.
\]

This can be rearranged to get a clearer recurrence relation.

### Summary
The solution can be expressed as a power series:

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

with the recurrence relation determined by the equation derived.

### Final Notes
To solve for specific coefficients, you would typically choose \(a_0\) or \(a_1\) and compute the rest recursively. 

Would you like to explore specific coefficients, or do you have any questions about this process? Here are some related topics you might consider:

1. What are the specific values of \(a_n\) for \(r = \sqrt{3}\)?
2. How do you determine the radius of convergence for this series?
3. Can you show how to apply this method to a different ODE?
4. What are other methods to solve second-order ODEs?
5. How does the choice of initial conditions affect the series solution?

**Tip:** When calculating coefficients, start with small values of \(n\) to observe patterns before generalizing the recurrence relation.

Given a second order ODE x^2y'' + xy' - 3y = 0. How to find its Frobenius series solution and recurrence relation?

Explore the detailed Frobenius series solution for the second order ODE x^2y'' + xy' - 3y = 0. This comprehensive guide covers the identification of singular points, the construction of the series solution, and the derivation of the recurrence relation, suitable for graduate-level mathematics students.

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