Comprehensive Power Series Solutions for the Differential Equation y'' - xy = 0 at x=0

Exploring Two Independent Power Series Solutions Around the Ordinary Point x = 0

Key Takeaways

• Methodology: Utilize the power series approach to assume and derive solutions for the differential equation.
• Recurrence Relation: Establish a relation that connects the coefficients of the power series, enabling the calculation of higher-order terms.
• Independent Solutions: Derive two linearly independent solutions that form the general solution to the differential equation.

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Introduction

Differential equations are fundamental in modeling various physical phenomena, ranging from mechanics to electrical circuits. Solving differential equations often requires sophisticated techniques, especially when dealing with non-constant coefficients or non-linear terms. One such powerful method for solving linear differential equations is the power series method. This approach involves expressing the solution as an infinite sum of terms, each multiplied by a power of the independent variable. In this comprehensive guide, we delve into finding two power series solutions for the differential equation:

$$ y'' - x y = 0 $$

centered around the ordinary point \( x = 0 \). We will systematically explore the steps involved, derive the necessary recurrence relations, and construct the independent solutions that form the general solution to the equation.

Understanding the Differential Equation

The given differential equation is:

$$ y'' - x y = 0 $$

where:

• y'' denotes the second derivative of \( y \) with respect to \( x \).
• x y represents the product of \( x \) and \( y \).

This is a linear, second-order differential equation with variable coefficients, specifically due to the presence of \( x \) multiplying \( y \).

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Step 1: Assuming a Power Series Solution

Formulating the Series Representation

To utilize the power series method, we begin by expressing the solution \( y(x) \) as an infinite power series centered at \( x = 0 \), also known as a Maclaurin series:

$$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$

Here, \( a_n \) represents the coefficients of the series that we aim to determine. This form assumes that the solution can be expressed as a sum of terms with increasing powers of \( x \), each scaled by a coefficient \( a_n \).

Calculating Derivatives

To substitute \( y(x) \) into the differential equation, we need its first and second derivatives:

$$ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} $$

$$ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} $$

These expressions are obtained by differentiating the power series term-by-term, a procedure valid within the interval of convergence of the series.

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Step 2: Substituting into the Differential Equation

Substituting \( y(x) \), \( y'(x) \), and \( y''(x) \) into the original differential equation yields:

$$ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - x \sum_{n=0}^{\infty} a_n x^n = 0 $$

Simplifying, the equation becomes:

$$ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \sum_{n=0}^{\infty} a_n x^{n+1} = 0 $$

The goal now is to align the powers of \( x \) in both summations to facilitate the combination of like terms.

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Step 3: Aligning the Powers of x

Adjusting the Indices

To combine the two series, we need them to have the same exponent on \( x \). This involves shifting the index of one of the series.

For the first summation, let \( k = n - 2 \). This implies \( n = k + 2 \) and \( n - 2 = k \). Substituting, we obtain:

$$ \sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2} x^k $$

For the second summation, multiply both sides by \( x \):

$$ x \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+1} $$

Let \( k = n + 1 \), which implies \( n = k - 1 \) and \( n + 1 = k \). Substituting, we get:

$$ \sum_{k=1}^{\infty} a_{k-1} x^k $$

Therefore, the differential equation now becomes:

$$ \sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2} x^k - \sum_{k=1}^{\infty} a_{k-1} x^k = 0 $$

Combining the Summations

To combine the two series into a single summation, we separate the \( k = 0 \) term from the first summation:

$$ (2 \cdot 1) a_2 + \sum_{k=1}^{\infty} \left[ (k+2)(k+1) a_{k+2} - a_{k-1} \right] x^k = 0 $$

This equation must hold for all \( x \), which implies that the coefficients of each power of \( x \) must individually equate to zero.

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Step 4: Establishing the Recurrence Relation

Deriving the Relation Between Coefficients

From the combined summation:

$$ 2a_2 + \sum_{k=1}^{\infty} \left[ (k+2)(k+1) a_{k+2} - a_{k-1} \right] x^k = 0 $$

Setting the coefficients of like powers of \( x \) to zero gives:

For \( k = 0 \):

$$ 2a_2 = 0 \quad \Rightarrow \quad a_2 = 0 $$

For \( k \geq 1 \):

$$ (k+2)(k+1) a_{k+2} - a_{k-1} = 0 $$

Rearranging provides the recurrence relation:

$$ a_{k+2} = \frac{a_{k-1}}{(k+2)(k+1)} $$

This relation allows us to express higher-order coefficients in terms of lower-order ones, paving the way to determine the entire series.

Solving the Recurrence Relation

The recurrence relation enables us to compute successive coefficients based on initial values. To fully determine the series solution, we need two initial conditions corresponding to the two independent solutions.

We consider two cases:

• Case 1: \( a_0 = 1 \) and \( a_1 = 0 \)
• Case 2: \( a_0 = 0 \) and \( a_1 = 1 \)

These initial conditions ensure the linear independence of the solutions, allowing us to construct the general solution as a linear combination of the two.

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Step 5: Generating the Power Series Solutions

First Independent Solution (\( y_1(x) \))

For the first solution, we set:

• Initial Conditions: \( a_0 = 1 \), \( a_1 = 0 \)

Using the recurrence relation:

$$ a_{k+2} = \frac{a_{k-1}}{(k+2)(k+1)} $$

We compute the coefficients as follows:

• For \( k = 1 \): \( a_3 = \frac{a_0}{3 \cdot 2} = \frac{1}{6} \)
• For \( k = 2 \): \( a_4 = \frac{a_1}{4 \cdot 3} = 0 \)
• For \( k = 3 \): \( a_5 = \frac{a_2}{5 \cdot 4} = 0 \)
• For \( k = 4 \): \( a_6 = \frac{a_3}{6 \cdot 5} = \frac{1}{180} \)
• For \( k = 5 \): \( a_7 = \frac{a_4}{7 \cdot 6} = 0 \)

Substituting these coefficients back into the power series:

$$ y_1(x) = 1 - \frac{x^3}{6} + \frac{x^6}{180} - \cdots $$

This series represents one of the two linearly independent solutions to the differential equation.

Second Independent Solution (\( y_2(x) \))

For the second solution, we set:

• Initial Conditions: \( a_0 = 0 \), \( a_1 = 1 \)

Applying the recurrence relation:

$$ a_{k+2} = \frac{a_{k-1}}{(k+2)(k+1)} $$

We compute the coefficients as follows:

• For \( k = 1 \): \( a_3 = \frac{a_0}{3 \cdot 2} = 0 \)
• For \( k = 2 \): \( a_4 = \frac{a_1}{4 \cdot 3} = \frac{1}{12} \)
• For \( k = 3 \): \( a_5 = \frac{a_2}{5 \cdot 4} = 0 \)
• For \( k = 4 \): \( a_6 = \frac{a_3}{6 \cdot 5} = 0 \)
• For \( k = 5 \):strong> \( a_7 = \frac{a_4}{7 \cdot 6} = \frac{1}{840} \)

Substituting these coefficients back into the power series:

$$ y_2(x) = x + \frac{x^4}{12} + \frac{x^7}{840} + \cdots $$

This series constitutes the second linearly independent solution to the differential equation.

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Step 6: Constructing the General Solution

The general solution to a linear homogeneous differential equation is a linear combination of its independent solutions. Therefore, combining \( y_1(x) \) and \( y_2(x) \), we obtain:

$$ y(x) = C_1 y_1(x) + C_2 y_2(x) $$

where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial or boundary conditions.

Substituting the explicit forms of \( y_1(x) \) and \( y_2(x) \), the general solution becomes:

$$ y(x) = C_1 \left(1 - \frac{x^3}{6} + \frac{x^6}{180} - \cdots \right) + C_2 \left(x + \frac{x^4}{12} + \frac{x^7}{840} + \cdots \right) $$

This expression encapsulates all possible solutions to the differential equation in the vicinity of \( x = 0 \).

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Convergence of the Power Series

To ensure the validity of the power series solution, it is essential to consider the radius of convergence. The given differential equation has its coefficients as polynomials in \( x \), meaning that \( x = 0 \) is an ordinary point. Hence, the power series solutions will converge in some neighborhood around \( x = 0 \).

The specific radius of convergence can be determined using the ratio test on the coefficients \( a_n \). However, given the factorial growth in the denominators of the coefficients, the series are likely to have an infinite radius of convergence, making them valid for all finite \( x \). Nevertheless, computing the exact radius requires a detailed analysis, which is beyond the scope of this discussion.

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Practical Applications

Solving differential equations using power series is not merely an academic exercise; it has profound applications in physics, engineering, and other sciences. For instance:

• Quantum Mechanics: Power series solutions are employed to solve the Schrödinger equation for systems like the quantum harmonic oscillator.
• Fluid Dynamics: In solving Navier-Stokes equations under specific conditions.
• Electromagnetism: Analyzing electromagnetic fields in various media.

The ability to express solutions as power series allows for approximate calculations and insights into the behavior of systems near specific points, such as equilibrium positions or points of symmetry.

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Summary

In this comprehensive exploration, we addressed the process of finding two power series solutions to the differential equation \( y'' - x y = 0 \) around the ordinary point \( x = 0 \). By assuming a solution in the form of a power series, calculating derivatives, substituting back into the equation, and establishing a recurrence relation, we systematically derived two independent solutions. These solutions form the foundation for the general solution, encompassing all possible behaviors of the system described by the differential equation within the vicinity of \( x = 0 \).

Power series methods are invaluable tools in the mathematician's arsenal, providing a structured approach to solving complex differential equations that may not be amenable to elementary techniques. The elegance of this method lies in its generality and the profound insights it offers into the nature of differential equations and their solutions.

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References