Problem 17 - Advanced Engineering Mathematics

Question

Find two power series solutions of the given differential equation about the ordinary point \(x=0\). $$ y^{\prime \prime}-3 x y=0 $$

Step-by-Step Solution

Answer

Two solutions are leading terms: \( y_1(x) = a_0(1 + \frac{3}{2}x^2 + \cdots) \) and \( y_2(x) = a_1(x + \frac{1}{2}x^3 + \cdots) \).

Step 1: Assume a Power Series Solution

First, we assume the power series solution for \( y(x) \) is of the form \[ y(x) = \sum_{n=0}^{\infty} a_n x^n. \] We aim to find an expression for this series that satisfies the differential equation.

Step 2: Find the Derivatives

Differentiate the series to find the first and second derivatives:\[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}, \]\[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}. \]

Step 3: Substitute Into the Differential Equation

Substitute \( y(x) \), \( y'(x) \), and \( y''(x) \) into the differential equation:\[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 3x \sum_{n=0}^{\infty} a_n x^n = 0. \]

Step 4: Align Powers of x

Align the powers of \( x \) by shifting indices where necessary. Start by re-indexing the series:\[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2} x^m, \]\[ 3x \sum_{n=0}^{\infty} a_n x^n = \sum_{m=0}^{\infty} 3 a_{m} x^{m+1}. \]

Step 5: Combine Series into Single Sum

Rewriting the equation, we have:\[ \sum_{m=0}^{\infty} \left((m+2)(m+1) a_{m+2} - 3 a_{m-1}\right)x^m = 0. \]This holds for all \( m \geq 0 \), implying coefficients must satisfy \((m+2)(m+1) a_{m+2} - 3 a_{m-1} = 0.\)

Step 6: Solve Recurrence Relation

From the relation \((m+2)(m+1) a_{m+2} = 3 a_{m-1}\), find\[ a_{m+2} = \frac{3 a_{m-1}}{(m+2)(m+1)}. \]Using initial conditions, solve sequentially for coefficients starting from known terms.

Step 7: Determine the First Few Terms

If \( a_0 = A \) and \( a_1 = B \), solve recursively:- For even terms, starting with \( a_0 \): \[ a_2 = \frac{3}{2}a_0, a_4 = \frac{3}{2 \cdot 4 \cdot 3} 3a_1 = \frac{3}{8}a_0, \ldots \]- For odd terms, starting with \( a_1 \): \[ a_3 = \frac{3}{3 \cdot 2} a_1 = \frac{1}{2}a_1, a_5 = \frac{3}{5 \cdot 4} a_2= \frac{3}{20}a_0, \ldots \]

Step 8: Write the Power Series Solutions

The two linearly independent solutions are\[ y_1(x) = a_0 \left(1 + \frac{3}{2}x^2 + \frac{3}{8}x^4 + \cdots \right), \]\[ y_2(x) = a_1 \left(x + \frac{1}{2}x^3 + \frac{3}{20}x^5 + \cdots \right). \]

Key Concepts

Power Series Solutions

In mathematics, the power series solution method plays a critical role in finding solutions for differential equations, especially when solutions can't be expressed in terms of elementary functions. A power series is essentially an infinite sum that represents a function. It takes the form:

• \[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]

Here, \(a_n\) are the coefficients to be determined, and \(x^n\) represents the terms of the series.
By assuming that the solution to the differential equation can be expressed as a power series, this method effectively finds a function that satisfies the equation within a radius of convergence.
Power series solutions are particularly useful when dealing with linear differential equations with variable coefficients, as the coefficients can often be expressed recursively. Additionally, it helps in generating approximate solutions where exact solutions are not feasible, providing clarity into the behavior of solutions near particular points.

Ordinary Point

Differential equations often have points where their coefficients feature abrupt changes or complexities. An ordinary point is a specific type of point where the functions involved in a differential equation are analytic, meaning they can be expressed by a power series.
In the context of our original problem, the ordinary point is located at \(x=0\). Why does this matter? Because power series solutions are especially handy when starting at an ordinary point. Around these points, the differential equation can be transformed into a format where standard series manipulation techniques apply more smoothly.
For the differential equation:

• \[ y'' - 3xy =0 \]

recognizing \(x=0\) as an ordinary point allows us to leverage the power series expansion effectively, enabling the isolation of coefficients systematically. It's this simplification that makes the series solution an attractive method for such problems.

Recurrence Relation

The journey to find a power series solution often leads to establishing a recurrence relation. This is an equation that spells out how each coefficient in the power series relates to others.
In our problem, the recurrence relation derived is:

• \[ a_{m+2} = \frac{3 a_{m-1}}{(m+2)(m+1)} \]

This relation is essential because it transforms the challenge of solving a differential equation into a more straightforward algebraic problem. Starting from known initial terms (like \(a_0\) and \(a_1\)), one can calculate further terms of the series, step by step.
Once this relation is established, the task becomes methodically applying it to generate the series's terms. By systematically solving for terms sequentially, we can capture as many terms in the series as needed for precision, comprehensively describing the solution's behavior around the ordinary point \(x=0\).