Problem 16 - Differential Equations and Linear Algebra

Question

Determine terms up to and including \(x^{5}\) in two linearly independent power series solutions of the given differential equation. State the radius of convergence of the series solutions. $$y^{\prime \prime}+(\sin x) y^{\prime}+y=0.$$

Step-by-Step Solution

Answer

The two linearly independent power series solutions up to \(x^5\) are: \(y_1(x)=a_0\left(1-\frac{a_1^2}{2}x^2+\frac{a_1^4}{8}x^4\right)\) and \(y_2(x)=a_1\left(1-\frac{a_1^2}{6}x^3+\frac{a_1^4}{20}x^5\right)\). The radius of convergence for the series solutions is infinite.

Step 1: Assume power series solutions for y, y', and y''

Assume power series solutions for \(y\), \(y^\prime\), and \(y^{\prime\prime}\): \(y(x)=\sum_{n=0}^{\infty}a_n x^n\) \(y^\prime(x)=\sum_{n=1}^{\infty}n a_n x^{n-1}\) \(y^{\prime\prime}(x)=\sum_{n=2}^{\infty}n(n-1) a_n x^{n-2}\)

Step 2: Substitute these expressions into the differential equation

Replace \(y\), \(y^\prime\), and \(y^{\prime\prime}\) in the differential equation with their power series representation: \(\left(\sum_{n=2}^{\infty}n(n-1) a_n x^{n-2}\right)+(\sin x)\left(\sum_{n=1}^{\infty}n a_n x^{n-1}\right)+\left(\sum_{n=0}^{\infty}a_n x^n\right)=0\)

Step 3: Find the recursion relation for an

Find the equation between the coefficients of the power series solutions, the so-called recursion relation: \(n(n-1)a_n=-\sum_{k=1}^{n-2}k a_k a_{n-k}\)

Step 4: Calculate the coefficients of the power series up to x^5

Now, calculate the coefficients for the two linearly independent power series up to \(x^5\), using the recursion relation: For \(n=0\) and \(n=1\), the coefficients are arbitrary and make up the two linearly independent series: \(a_0=a_0\) \(a_1=a_1\) For \(n=2\): \(2(1) a_2 = -a_1^2\) \(a_2 = -\frac{a_1^2}{2}\) For \(n=3\): \(3(2) a_3 = -a_1 a_2\) \(a_3 = -\frac{a_1^3}{6}\) For \(n=4\): \(4(3) a_4 = -2a_2^2\) \(a_4 = \frac{a_1^4}{8}\) For \(n=5\): \(5(4) a_5 = -\sum_{k=1}^{3}k a_k a_{5-k}\) \(a_5 = -\frac{a_1^5}{20}\) So, the two linearly independent power series solutions up to \(x^5\) are: \(y_1(x)=a_0\left(1-\frac{a_1^2}{2}x^2+\frac{a_1^4}{8}x^4\right)\) \(y_2(x)=a_1\left(1-\frac{a_1^2}{6}x^3+\frac{a_1^4}{20}x^5\right)\)

Step 5: Determine the radius of convergence

Since the power series solutions arise from a differential equation with a regular singular point at the origin x=0, the radius of convergence is the distance to the nearest singular point. In this case, the nearest point where the ODE becomes undefined is at infinity. Therefore, the radius of convergence of the series solutions is infinite.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. In the context of this problem, we are dealing with a second-order linear homogeneous differential equation with variable coefficients. This particular equation involves the unknown function y, its second derivative y'', and its first derivative y', where the coefficients of these derivatives depend on another function, sin x.

The presence of the sine function as a coefficient makes this differential equation more complex and suggests that standard solution methods, such as the method of undetermined coefficients or variation of parameters, might not be directly applicable. As a result, we turn to power series solutions to find a suitable method for solving the equation in cases where the usual techniques do not suffice.

Series Solution

The approach of finding a series solution to a differential equation involves expressing the solution as an infinite sum of powers of the independent variable, usually denoted as x. The series takes the form y(x) = \( \sum_{n=0}^{\infty}a_n x^n \), where a_n are the coefficients to be determined, and n is a non-negative integer.

To find these coefficients, the derivatives of the series are substituted back into the original differential equation, leading to a new equation where the coefficients of each power of x must sum to zero to satisfy the equation. This condition typically gives rise to a set of equations related to the coefficients, known as a recursion relation, which allows us to calculate the coefficients systematically. So in essence, the method transforms a differential problem into an algebraic one.

Radius of Convergence

The radius of convergence is a crucial concept when dealing with series solutions. It is the distance from the expansion point (in this case, the origin) within which the power series converges to a finite value. For the radius of convergence to be infinite, as in the problem provided, the series must converge for all values of x. This is indicative that the solution to the differential equation is well-behaved throughout the entire x-domain.

Determining the radius of convergence usually involves techniques such as the ratio test or the root test, among others. In the context of differential equations, if there are no singular points other than at infinity, then the series solutions can be assumed to have an infinite radius of convergence. Understanding how to determine this radius is crucial for validating whether the power series is an actual solution to the differential equation over the interval of interest.

Recursion Relation

A recursion relation is an equation that expresses the coefficients of a power series in terms of each other, providing us a way to compute them systematically. In the given exercise, after substituting the series expansions of y(x), y'(x), and y''(x) into the differential equation, comparing coefficients allows us to derive such a recursion relation.

These recursion relations are vital because they outline the relationship between subsequent coefficients starting from an initial arbitrary choice. For example, once you choose the initial values a_0 and a_1, the other coefficients can then be calculated using the established pattern. It's important to comprehend this recursion pattern as it guides the step-by-step calculations required to find the solution terms up to the necessary degree, in this case, up to x^5. Recursive relations are the backbone of finding series solutions because they make it possible to 'build' the solution, coefficient by coefficient, to any desired level of precision.