Math Problem Statement

Find a Frobenius type solution for the differential equation xy" + y' - y = 0

Solution

To find a Frobenius series solution for the differential equation

xy+yy=0,xy'' + y' - y = 0,

we will attempt to find a solution in the form of a power series around x=0x = 0:

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 series coefficients.

Step 1: Compute derivatives of the series

First, we calculate the first and second derivatives of the proposed solution.

  1. The first derivative y(x)y'(x) is:

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

  1. The second derivative y(x)y''(x) is:

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

Step 2: Substitute into the differential equation

We now substitute y(x)y(x), y(x)y'(x), and y(x)y''(x) into the original differential equation:

xy+yy=0.x y'' + y' - y = 0.

Substitute each term:

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

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Frobenius Series
Power Series Solutions

Formulas

y(x) = ∑ a_n x^{n+r}
y'(x) = ∑ a_n (n+r) x^{n+r-1}
y''(x) = ∑ a_n (n+r)(n+r-1) x^{n+r-2}

Theorems

Frobenius Method
Power Series Expansion

Suitable Grade Level

College-Level Mathematics