Question

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

Short answer

The power series solution of the given initial-value problem \(y''(x) - xy'(x) - y(x) = 0\), with \(y(0) = 1\) and \(y'(0) = 0\) is found by expressing \(y(x)\) as a power series, finding its first and second derivatives, and substituting them into the given differential equation. Using the initial conditions and a recursion relation, we find the coefficients of the power series solution. The power series solution is: \[y(x) = \sum_{n=0}^{\infty} a_n x^n\] where \(a_0 = 1\), \(a_1 = 0\), and the subsequent coefficients \(a_n\) are determined using the recursion relation obtained from the differential equation.

Step-by-step solution

Express y as a power series

Let's start by expressing \(y\) as a power series: \[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]

Find the first and second derivatives

Next, we need to find the first and second derivatives of \(y(x)\). For the first derivative \(y'(x)\), we have: \[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \] For the second derivative \(y''(x)\), we have: \[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \]

Substitute the derivatives into the differential equation

Now, let's substitute the power series for \(y(x)\), \(y'(x)\), and \(y''(x)\) into the given differential equation: \[ y''(x) - x y'(x) - y(x) = 0 \] We then get: \[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} -x\sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{n=0}^{\infty} a_n x^n = 0\]

Rewrite the series to have matching exponents

In order to combine the series, we should rewrite each series with the same exponents in the power of x. For the first and second series, we can replace \(n\) with \(n+2\) and \(n+1\), respectively, and adjust the summation bounds accordingly. \[\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n -x\sum_{n=0}^{\infty} (n+1) a_{n+1} x^n - \sum_{n=0}^{\infty} a_n x^n = 0\]

Combine the series

Now we can combine the series: \[\sum_{n=0}^{\infty} [(n+2)(n+1) a_{n+2} -x(n+1) a_{n+1} - a_n] x^n = 0\]

Equate coefficients to zero

Since the combined series must be zero, this means that its coefficients must be zero: \[(n+2)(n+1) a_{n+2} - x(n+1) a_{n+1} - a_n = 0\] This is a recursion relation which will help us find the coefficients of \(y(x)\) in terms of \(a_0\) and \(a_1\).

Use the initial conditions

Now we can use the initial conditions \(y(0)=1\) and \(y'(0)=0\) to determine some of the coefficients: \[ y(0) = \sum_{n=0}^{\infty} a_n \cdot 0^n = a_0 = 1\] \[ y'(0) = \sum_{n=1}^{\infty} n a_n \cdot 0^{n-1} = a_1 = 0\]

Find the power series solution

With initial conditions \(a_0 = 1\) and \(a_1 = 0\), we can now use the recursion relation to find the power series solution of the given initial-value problem. Plug in the values of \(a_0\) and \(a_1\) and calculate subsequent coefficients \(a_n\). Then, substitute the coefficients back into the power series representation of \(y(x)\) to get the power series solution.

Key concepts

Power Series Solution

Power series solutions are a powerful technique used in solving differential equations, such as initial-value problems. Essentially, a power series represents a function as an infinite sum of terms, each composed of a coefficient and a power of the variable. This allows complex functions to be expressed in a more manageable form. It is especially helpful when exact solutions are challenging to find. The general form of a power series is:

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

Here, \(a_n\) are the series coefficients and \(n\) counts up from zero to infinity. In the exercise, the function \(y(x)\) is approximated as a power series, enabling the investigation of its properties through its terms. Each term in the series represents a simpler polynomial function, making analysis or computation possible.

Initial-Value Problem

An initial-value problem (IVP) in differential equations involves finding a solution that satisfies a given ordinary differential equation (ODE) and specific conditions at a starting point. These conditions specify the values of the function and its derivatives at a particular point, usually given as \(y(a) = b\) and sometimes \(y'(a) = c\). For the problem at hand:

• \(y(0)=1\)
• \(y'(0)=0\)

The initial conditions help us determine the specific power series solution from the general family of solutions to the differential equation. They essentially "anchor" the solution curve, tailoring the general solution to meet the specific criteria at the point \(x=0\).

Recursion Relation

A recursion relation is an equation that defines a sequence of numbers where each term is a function of its preceding terms. In solving differential equations using power series, recursion relations connect the coefficients of the power series. Once known, these relations allow for the systematic computation of successive coefficients based on initial ones. In this exercise, the recursion relation is established by equating the series' coefficients to zero:

• \((n+2)(n+1) a_{n+2} - (n+1) a_{n+1} - a_n = 0\)

This equation allows us to solve for \(a_{n+2}\) using previous values \(a_n\), essentially constructing the sequence of coefficients necessary to form the power series solution.

Coefficients

In a power series, each term is associated with a coefficient \(a_n\), just as a normal polynomial has coefficients. These coefficients scale the contribution of each power of \(x\) in the series to the overall function. In the solution of differential equations, finding the correct coefficients is tantamount to constructing the function that fits the differential equation and its initial conditions. The coefficients often start with some determined based on the initial conditions, such as \(a_0\) and \(a_1\) in our problem:

• \(a_0 = 1\)
• \(a_1 = 0\)

Subsequent coefficients are determined by applying the recursion relation, leading to a complete series representation of the solution.

Differential Equations Theory

Differential equations theory encompasses various methodologies to find solutions to differential equations - the mathematical equations that describe how a quantity changes with respect to another. It lays the groundwork for understanding phenomena in physics, engineering, biology, and more. The theory aims to provide tools to analyze the behavior of systems described by these equations. This particular exercise falls under this theory as it addresses linear second-order differential equations using power series. Understanding these concepts allows us to solve more complex equations analytically or approximatively, facilitating deeper insights into dynamic systems. Approaches like power series offer a robust framework for attaining solutions that may not be easily achievable through other means, highlighting the breadth and efficacy of differential equations theory in practical applications.