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(5 pts) Find recurrence relationship and first four non-zero term of each of the fundamental solutions of y ′′ +3x^2

y′+5xy=0

User Duyuanchao
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1 Answer

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The given second-order linear differential equation is \(y'' + 3x^2 y' + 5xy = 0\).

To find the recurrence relationship and the first four non-zero terms of each fundamental solution, we can use the Frobenius method, assuming that the solutions can be written in the form of a power series \(y = \sum_{n=0}^\infty a_nx^{n+r}\), where \(r\) is a root of the indicial equation.

First, we find the indicial equation by substituting \(y = x^r \sum_{n=0}^\infty a_nx^n\) into the differential equation and equating coefficients of \(x^{n+r}\) to zero:

\[
r(r-1) + 3rx + 5x = 0
\]

Solving for \(r\), we get the roots \(r = 0\) and \(r = -2\).

For the root \(r = 0\):
The recurrence relation is \(a_{n+2} = -\frac{5a_n}{(n+2)(n+1)}\).
The first four non-zero terms are:
\(a_0, a_2, a_4, a_6\).

For the root \(r = -2\):
The recurrence relation is \(a_{n+2} = -\frac{3a_n}{(n+2)(n+1)}\).
The first four non-zero terms are:
\(a_{-2}, a_0, a_2, a_4\).

Please note that in the case of negative indices, the term \(a_{-2}\) represents the coefficient corresponding to the term \(x^{-2}\). The recurrence relations provided above allow you to generate the terms of each fundamental solution.
User Sawo Cliff
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