Math Problem Statement

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

Solution

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

x2y+xy3y=0,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=0x = 0, which is a regular singular point since the coefficients p(x)=1xp(x) = \frac{1}{x} and q(x)=3q(x) = -3 have singularities at x=0x = 0.

Step 2: Assume a Frobenius Series Solution

We assume a solution of the form:

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

where rr is the indicial exponent to be determined, and ana_n are the coefficients.

Step 3: Compute Derivatives

We compute the derivatives of y(x)y(x):

y=n=0an(n+r)xn+r1,y' = \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1},

y=n=0an(n+r)(n+r1)xn+r2.y'' = \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2}.

Step 4: Substitute into the ODE

Substituting yy, yy', and yy'' into the original ODE:

x2(n=0an(n+r)(n+r1)xn+r2)+x(n=0an(n+r)xn+r1)3(n=0anxn+r)=0.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:

n=0an(n+r)(n+r1)xn+r+n=0an(n+r)xn+r3n=0anxn+r=0.\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:

n=0an((n+r)(n+r1)+(n+r)3)xn+r=0.\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 nn:

(n+r)(n+r1)+(n+r)3=0.(n+r)(n+r-1) + (n+r) - 3 = 0.

This simplifies to:

(n+r)23=0.(n+r)^2 - 3 = 0.

Step 7: Find the Indicial Roots

Setting n=0n=0:

(r)23=0    r=±3.(r)^2 - 3 = 0 \implies r = \pm \sqrt{3}.

Step 8: Recurrence Relation

For n0n \geq 0, we have:

an=3(n+r)(n+r1)(n+r)(n+r1)an1.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)=n=0anxn+r,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 a0a_0 or a1a_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 ana_n for r=3r = \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 nn to observe patterns before generalizing the recurrence relation.

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Math Problem Analysis

Mathematical Concepts

Ordinary Differential Equations
Series Solutions
Frobenius Method

Formulas

y(x) = ∑(n=0 to ∞) a_n x^(n+r)
(n+r)(n+r-1) + (n+r) - 3 = 0
a_n = (3 - (n+r)(n+r-1)) / ((n+r)(n+r-1)) a_{n-1}

Theorems

Existence and Uniqueness Theorem for ODEs
Frobenius Theorem

Suitable Grade Level

Graduate Level