Question

Find the power series solution of each of the initial-value problems in Exercises. $$ y^{\prime \prime}+x y^{\prime}-2 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1 $$

Short answer

The power series solution for the given initial-value problem is: $$ y(x) = x + \sum_{n=2}^{\infty} a_n x^n $$ where the coefficients \(a_n\) are recursively calculated using the recurrence relation: $$ (n+2)(n+1) a_{n+2} + (n+1) a_{n+1} - 2a_n = 0 $$ and the initial conditions \(a_0 = 0\) and \(a_1 = 1\).

Step-by-step solution

1. Assume the general power series form for y(x)

First, let's assume a general power series solution for y(x) as: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$

2. Substitute y(x) and its derivatives into the given differential equation

We need to find the first and second derivatives of the assumed power series solution for y(x): $$ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}, \quad y^{\prime \prime}(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} $$ Now, substitute y(x), y'(x), and y''(x) into the given differential equation: $$ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x \sum_{n=1}^{\infty} n a_n x^{n-1} - 2\sum_{n=0}^{\infty} a_n x^n = 0 $$

3. Compare coefficients of like powers to form a recurrence relation

To compare coefficients of like powers, we shift the indices so that all series start from n = 0: $$ \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n + \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n - 2\sum_{n=0}^{\infty} a_n x^n = 0 $$ Now, we can group the three sums into one: $$ \sum_{n=0}^{\infty} [(n+2)(n+1) a_{n+2} + (n+1) a_{n+1} - 2a_n]x^n = 0 $$ For the equation to be true, the coefficient of each power of x should be zero. Therefore, we have the recurrence relation: $$ (n+2)(n+1) a_{n+2} + (n+1) a_{n+1} - 2a_n = 0 $$

4. Find the solution using the given initial conditions

We are given the initial conditions: y(0) = 0 and y'(0) = 1. From the power series representation of y(x) and y'(x), we get: $$ y(0) = a_0 = 0 \\ y^{\prime}(0) = a_1 = 1 $$

5. Write the final power series solution

Using the recurrence relation and initial conditions, we can recursively calculate the coefficients \(a_n\) for n = 2,3,... and write the power series solution for y(x): $$ y(x) = x + \sum_{n=2}^{\infty} a_n x^n $$ This power series representation is the solution to the initial-value problem.

Key concepts

Power Series Solution

The power series solution is a method to solve differential equations. This approach is especially useful when it is difficult or impossible to find an explicit solution using other methods.

A power series is a series of the form \(y(x) = \sum_{n=0}^{\infty} a_n x^n\), where \(a_n\) are constants that need to be determined. The strategy is to assume the solution can be expressed with this series, then find the coefficients by substituting into the given differential equation.

By substituting back into the differential equation, we then compare like powers of \(x\) to determine equations for the coefficients \(a_n\), leading us to a formula or pattern that these coefficients follow.

Initial Value Problems

Initial value problems (IVPs) are differential equations accompanied by specific values, or conditions, that the solution must satisfy at a particular point. These conditions usually specify the value of the function, and possibly its derivatives, at a particular point.

For instance, in this exercise, our differential equation \(y'' + xy' - 2y = 0\) is supplemented with initial conditions \(y(0) = 0\) and \(y'(0) = 1\). These conditions help us uniquely determine the coefficients of the power series.

The initial conditions are substituted into the power series expansion, allowing us to solve for certain coefficients directly, or to begin a recursive process for calculating all necessary coefficients.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence. Each term is a function of preceding terms. In the context of power series solutions, the recurrence relation is derived by matching coefficients of like powers of \(x\) after substituting the power series into the differential equation.

In our problem, the recurrence relation is given by:
\((n+2)(n+1) a_{n+2} + (n+1) a_{n+1} - 2a_n = 0\).

This formula relates the coefficients of the power series at different steps \(n\), allowing us to calculate unknown coefficients based on the known initial conditions. The recurrence relation forms the backbone of many power-series solutions, acting as a bridge between the differential equation and the actual sequence of numbers (or coefficients) defining the function.

General Power Series

A general power series takes the form \(\sum_{n=0}^{\infty} a_n x^n\). It is a way to represent functions as infinite sums of powers centered around x = 0 (or another point).

This is a versatile representation since many functions can be expressed this way, including solutions to differential equations. By adjusting the coefficients \(a_n\), we can sculpt the series to match initial conditions and derive the specific functions that satisfy our differential equations.

The beauty of power series lies in its ability to approximate functions effectively over certain intervals, providing a useful tool in both theoretical math and applied sciences.

Ordinary Differential Equations

An ordinary differential equation (ODE) involves functions of one independent variable and their derivatives. These equations are fundamental in describing various phenomena, such as motion, growth, and decay.

The differential equation we solved, \(y'' + xy' - 2y = 0\), is a second-order linear ODE. Solving an ODE involves finding a function or a set of functions that satisfy the equation over some range of inputs.

ODEs can be tricky to solve, especially without simple closed-form solutions. This is where methods like power series come into play, allowing us to achieve solutions through infinite series rather than finite combinations of known functions. This approach broadens our ability to tackle various classes of differential equations that otherwise resist easy solution methods.