Consider The Following. (2 + X^2)y'' - Xy' + 4y = 0, X_0 = 0 Seek Power Series Solutions Of The Given

Consider The Following. (2 + X^2)y'' - Xy' + 4y = 0, X_0 = 0 Seek Power Series Solutions Of The Given

Understanding and solving differential equations is a fundamental aspect of advanced mathematics, physics, and engineering. Among various methods, the power series approach is particularly powerful for solving linear differential equations with variable coefficients. In this article, we explore the process of finding power series solutions to the differential equation:

(2 + X²) y'' - X y' + 4 y = 0, with the initial point X₀ = 0

This equation presents an interesting case due to its variable coefficients, making direct methods less straightforward. We will systematically develop the power series solution, analyze the recurrence relations, and discuss the convergence and implications of the solutions obtained.

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Understanding the Differential Equation

Equation Overview

The given differential equation is:

(2 + X²) y'' - X y' + 4 y = 0

This is a second-order linear ordinary differential equation (ODE) with variable coefficients, where the coefficients depend explicitly on X. The initial point X₀ = 0 suggests that we are interested in solutions centered around X = 0, making it suitable for a power series expansion about this point.

Significance of Variable Coefficients

Variable coefficients often prevent straightforward solutions using elementary functions. Instead, power series methods allow us to construct solutions locally around a point of interest, providing insights into the behavior of solutions near that point.

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Methodology for Power Series Solutions

Assuming a Power Series Solution

We seek a solution in the form of a power series centered at X=0:

y(X) = Σ{n=0}^∞ an X^n

where a_n are coefficients to be determined.

Similarly, derivatives are expressed as:

y'(X) = Σ{n=1}^∞ n an X^{n-1}

y''(X) = Σ{n=2}^∞ n(n-1) an X^{n-2}

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Substituting into the Differential Equation

Substituting the power series into the differential equation involves expressing (2 + X²) y'', X y', and 4 y, all as power series:

1. (2 + X²) y'' = 2 y'' + X² y''
2. -X y' remains as is
3. 4 y remains as is

Each term is expanded as a power series, and then all are combined to match coefficients of like powers of X.

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Deriving the Recurrence Relation

Expressing Each Term

Let's write each term explicitly:

• Term 1: 2 y'' = 2 Σ{n=2}^∞ n(n-1) an X^{n-2} = Σ{n=2}^∞ 2 n(n-1) an X^{n-2}
• Term 2: X² y'' = X² Σ{n=2}^∞ n(n-1) an X^{n-2} = Σ{n=2}^∞ n(n-1) an X^{n}
• Term 3: -X y' = -X Σ{n=1}^∞ n an X^{n-1} = - Σ{n=1}^∞ n an X^{n}
• Term 4: 4 y = 4 Σ{n=0}^∞ an X^n

Reindexing Series for Consistency

To combine these series, we need all sums expressed with the same powers of X:

• For the first term, set m = n - 2:

When n=2, m=0; as n→∞, m→∞

So, 2 y'' = Σ{m=0}^∞ 2 (m+2)(m+1) a{m+2} X^{m}

• For the second term, n remains as is:

Σ{n=2}^∞ n(n-1) an X^{n} = Σ{n=2}^∞ n(n-1) an X^{n}

• For the third term, n remains as is:


• Σ{n=1}^∞ n an X^{n}


• For the fourth term:

4 Σ{n=0}^∞ an X^{n}

Now, the entire differential equation becomes:

Σ{m=0}^∞ 2 (m+2)(m+1) a{m+2} X^{m} + Σ{n=2}^∞ n(n-1) an X^{n} - Σ{n=1}^∞ n an X^{n} + 4 Σ{n=0}^∞ an X^{n} = 0

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Aligning the Series

Express all sums over X^{n} or X^{m} with the same index:

• For the second sum, start from n=0:

When n=2, sum begins; for n=0, term is zero because n(n-1)=0

• For the third sum, n=1 to ∞
• For the last sum, n=0 to ∞

Expressed uniformly:

• First sum: m=0 to ∞
• Second sum: n=2 to ∞
• Third sum: n=1 to ∞
• Fourth sum: n=0 to ∞

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Formulating the Recurrence Relations

Equating Coefficients

Since the sum equals zero for all X, each coefficient of X^n must be zero:

For each power n ≥ 0,

Coefficient sum:

[2 (n+2)(n+1) a{n+2}] + [n(n-1) an if n ≥ 2] - [n an if n ≥ 1] + [4 an] = 0

Note that for n=0:

• The second sum does not contribute (starts from n=2)


• The third sum: n=1, so for n=0, no contribution


• The fourth sum: n=0

Thus, for n=0:

2 (2)(1) a2 + 4 a0 = 0

which simplifies to:

2 2 1 a2 + 4 a0 = 0

4 a2 + 4 a0 = 0

=> a2 = - a0

Similarly, for n ≥ 1, the general recurrence relation is:

2 (n+2)(n+1) a{n+2} + [n(n-1) - n + 4] an = 0

which simplifies to:

2 (n+2)(n+1) a{n+2} + (n^2 - n + 4) an = 0

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Final Recurrence Formula

Thus, the recurrence relation for n ≥ 0 is:

a{n+2} = - [(n^2 - n + 4) / (2 (n+2)(n+1))] an

with initial coefficients a0 and a1 remaining arbitrary (as per the nature of second-order linear ODEs).

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Constructing the Power Series Solution

Explicit Expressions for Coefficients

Using the recurrence:

a{n+2} = - [(n^2 - n + 4) / (2 (n+2)(n+1))] an

and initial values a0, a1, we can compute subsequent coefficients iteratively:

• a2 = - a0 (from earlier)


• a3 = - [(1^2 - 1 + 4) / (2 3 2)] a1 = - [(1 -1 +4)/(12)] a1 = - (4/12) a1 = - (1/3) a_1


• a4 = - [(2^2 - 2 + 4) / (2 4 3)] a2 = - [(4 - 2 + 4)/24] a2 = - (6/24) a2 = - (1/4) a_2

But since a2 = - a0, we substitute back:

a4 = - (1/4)(- a0) = (1/4) a_0

Similarly, higher coefficients are obtained recursively.

Series Solution Structure

The general solution is a linear combination:

y(X) = a0 Σ{k=0}^∞ c{2k} X^{2k} + a1 Σ{k=0}^∞ c{2k+1} X^{2k+1}

where the coefficients c{n} are computed via the recurrence relation, and a0, a_1 are arbitrary constants.

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Analysis of