Question

Solve the given differential equation by means of a power series about the given point \(x_{0} .\) Find the recurrence relation; also find the first four terms in each of two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution. \(y^{\prime \prime}-y=0, \quad x_{0}=0\)

Short answer

Question: Find the first four terms in each of the two linearly independent power series solutions for the following differential equation: $$y^{\prime\prime}(x) - y(x) = 0$$ Answer: The first four terms of the two linearly independent power series solutions are given by: $$ y_1(x) = 1 + 0\cdot x + \frac{1}{2}x^2 + 0\cdot x^3 + \frac{1}{24}x^4 + \cdots $$ $$ y_2(x) = 0 + x + 0\cdot x^2 + \frac{1}{6}x^3 + 0\cdot x^4 + \cdots $$

Step-by-step solution

Write down the assumed solution

Assume a solution of the form: $$ y(x) = \sum_{n=0}^{\infty}a_n x^n $$ where \(n\) is a non-negative integer and each \(a_n\) is a constant.

Calculate the first and second derivatives of y(x)

Compute the first derivative of the assumed solution with respect to x: $$ y^{\prime}(x) = \frac{d}{dx} \left(\sum_{n=0}^{\infty}a_n x^n\right) = \sum_{n=1}^{\infty}na_n x^{n-1} $$ Compute the second derivative, y''(x): $$ y^{\prime\prime}(x) = \frac{d^2}{dx^2} \left(\sum_{n=1}^{\infty}na_n x^{n-1}\right) = \sum_{n=2}^{\infty}n(n-1)a_n x^{n-2} $$

Substitute y(x), y'(x), and y''(x) into the given equation

Plug the expressions from steps 1 and 2 for y(x), y'(x), and y''(x) into the given differential equation: $$ \sum_{n=2}^{\infty}n(n-1)a_n x^{n-2} - \sum_{n=0}^{\infty}a_n x^n = 0 $$

Match the coefficients

Now, match the coefficients of the same powers of x on both sides of the equation.

Find the recurrence relation

Equating the coefficients of the same powers of x on both sides, we get: $$ n(n-1)a_n = a_{n-2} $$ This is the recurrence relation. Rearrange to find \(a_n\) as a function of \(a_{n-2}\): $$ a_n = \frac{a_{n-2}}{n(n-1)} $$

Find the first four terms in each of two linearly independent solutions

To find the first four terms of each linearly independent solution, we can set \(a_0\) and \(a_1\) as arbitrary constants and compute the other \(a_n\)s using the recurrence relation. For the first solution, let \(a_0 = 1\) and \(a_1 = 0\). Then we compute the terms \(a_2, a_3, a_4\) using the recurrence relation: $$ a_2 = \frac{a_0}{2(2-1)} = \frac{1}{2} $$ $$ a_3 = \frac{a_1}{3(3-1)} = 0 $$ $$ a_4 = \frac{a_2}{4(4-1)} = \frac{1}{2(4)(3)} = \frac{1}{24} $$ So the first solution has the form: $$ y_1(x) = 1 + 0\cdot x + \frac{1}{2}x^2 + 0\cdot x^3 + \frac{1}{24}x^4 + \cdots $$ For the second solution, let \(a_0 = 0\) and \(a_1 = 1\). Then we compute the terms \(a_2, a_3, a_4\) using the recurrence relation: $$ a_2 = \frac{a_0}{2(2-1)} = 0 $$ $$ a_3 = \frac{a_1}{3(3-1)} = \frac{1}{6} $$ $$ a_4 = \frac{a_2}{4(4-1)} = 0 $$ So the second solution has the form: $$ y_2(x) = 0 + x + 0\cdot x^2 + \frac{1}{6}x^3 + 0\cdot x^4 + \cdots $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They describe how a certain quantity changes in relation to another, playing a significant role in various scientific fields like physics, engineering, and biology. For our exercise, the differential equation is given as \( y'' - y = 0 \) . This specific form is called a **linear second-order differential equation**. Here, the second derivative of a function is equal to a function of the variables involved, which are denoted by **\(y\)** and its derivatives.When solving such equations, one technique is to use a **power series solution**. This approach assumes the solution can be expressed as an infinite series, which can simplify the problem by turning it into a series of algebraic expressions.

Recurrence Relation

A recurrence relation is a way to define a sequence of numbers, where each term is expressed as a function of its preceding terms. In the problem, finding the recurrence relation is a crucial step involving comparing the coefficients of the series solution.The recurrence relation for this exercise is derived from equating coefficients of the power series derived from \( y'' - y = 0 \) . We arrive at: \[ n(n-1)a_n = a_{n-2} \] This relation helps build terms of the series step-by-step, enabling us to compute coefficients like \( a_2, a_3, \) and others from initial ones such as \( a_0 \) and \( a_1 \), allowing us to construct a series solution.

Linearly Independent Solutions

Keeping linearly independent solutions is important for representing the general solution of a differential equation. The concept of linear independence means that no solution in a set can be written as a linear combination of another in that same set.In this exercise, after assuming initial values like \( a_0 = 1, a_1 = 0 \) and \( a_0 = 0, a_1 = 1 \), we obtained two solutions:

• \( y_1(x) = 1 + \frac{1}{2}x^2 + \frac{1}{24}x^4 + \cdots \)
• \( y_2(x) = x + \frac{1}{6}x^3 + \cdots \)

These solutions are linearly independent because neither is simply a scalar multiple of the other. Combining linearly independent solutions allows us to write the general solution to our differential equation.

General Term

The general term of a series solution provides a formula for any term in the series, usually dependent on its position in the sequence. This is vital as it unfolds the pattern of terms in the series and aids in computing higher-order terms easily.In our power series, once the recurrence relation is identified, we can find an expression for the general term \( a_n \). Using: \[ a_n = \frac{a_{n-2}}{n(n-1)} \] The general solution for our differential equation involves combining both linearly independent solutions derived from the same recurrence relation. Understanding and finding the general term helps in identifying how the series progresses, offering insight into the structure of the solution.