Problem 5 - Differential Equations and Linear Algebra

Question

Determine two linearly independent power series solutions to the given differential equation centered at \(x=0 .\) Also determine the radius of convergence of the series solutions. $$y^{\prime \prime}+x y=0.$$

Step-by-Step Solution

Answer

The two linearly independent power series solutions to the differential equation \(y''+xy=0\) are: \[y_1(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}\] \[y_2(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}\] The radius of convergence for both series solutions is infinity.

Step 1: Assume a power series solution for y

Let's assume that the function y has a power series representation: \[y(x)=\sum_{n=0}^{\infty} a_n x^n\]

Step 2: Differentiate the power series

Now, we find the first and second derivatives of y with respect to x: \[y'(x) = \sum_{n=1}^{\infty} na_n x^{n-1}\] \[y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\]

Step 3: Substitute the derivatives into the differential equation

Now, substitute the expressions for \(y''\) and y into the given differential equation: \[\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x\sum_{n=0}^{\infty} a_n x^n=0\]

Step 4: Find the recurrence relation for the coefficients

To find the recurrence relation, let's first rewrite the summations with a common index and adjust the powers of x: \[\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n + \sum_{n=0}^{\infty} a_n x^{n+1}=0\] \[\sum_{n=0}^{\infty} ((n+2)(n+1) a_{n+2} + a_n x)x^n=0\] Since this equation holds for all values of x, we have the recurrence relation: \[(n+2)(n+1) a_{n+2} + a_n x=0\]

Step 5: Write the general series solution and determine the coefficients

Using the recurrence relation, we can write the general series solution as: \[y(x) = a_0\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} + a_1\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}\] where \(a_0\) and \(a_1\) are arbitrary constants.

Step 6: Obtain two linearly independent solutions

The two linearly independent solutions can be obtained by taking \(a_0=1\) and \(a_1=0\) for one solution and \(a_0=0\) and \(a_1=1\) for the other solution. Putting these values into the general series solution, we get the two linearly independent series solutions: \[y_1(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}\] \[y_2(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}\]

Step 7: Determine the radius of convergence

We will examine the series convergence using the ratio test: \[\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\frac{(-1)^{n+1}}{(2(n+1))!}}{\frac{(-1)^n}{(2n)!}} = \lim_{n \to \infty} \frac{(2n)!}{(2(n+1))!} = \lim_{n \to \infty} \frac{(2n)!}{(2n+2)(2n+1)(2n)!} = 0\] Since the limit of the ratio test for both \(y_1(x)\) and \(y_2(x)\) is zero, the radius of convergence for both series solutions is infinity.

Key Concepts

Linear Independence

In the context of power series solutions to differential equations, linear independence refers to solutions that cannot be expressed as a linear combination of one another. This is crucial because having two linearly independent solutions allows us to form the general solution to a second-order linear differential equation.

For the differential equation given in the exercise, both solutions derived from the power series method must be linearly independent. Consider the two forms
\[y_1(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}\]
and
\[y_2(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}.\]
These two series are linearly independent because one contains only even powers of \(x\) and the other contains only odd powers. Using these two distinct series, we can construct a general solution to our differential equation by a linear combination of \(y_1\) and \(y_2\), like

• \(y(x) = C_1 y_1(x) + C_2 y_2(x)\),

where \(C_1\) and \(C_2\) are constants determined by initial or boundary conditions.

Radius of Convergence

The radius of convergence is a measure of the interval within which a power series converges to a function. For power series solutions of differential equations, it's essential to determine this radius to understand where the solution is valid.

In our exercise, the series solutions were actually found to have an infinite radius of convergence. This conclusion allows the power series to be valid everywhere on the real line, a significant result when studying the behavior of solutions.

• The ratio test is a common method used to determine the radius of convergence for power series.
• For the series provided in the solution, applying this test led us to the conclusion that the limit of the resulting ratio was zero as \(n\) approaches infinity.

A ratio of zero suggests that the series converges for all real values of \(x\). Thus, the solutions can be evaluated at any point within \([-\infty, \infty]\), providing a comprehensive understanding of the solutions across the entire domain of \(x\).

Recurrence Relation

A recurrence relation is an equation that expresses each term of a sequence as a function of its preceding terms. Such relations are indispensable when breaking down power series solutions for differential equations, as they reveal patterns in the coefficients.

• In this exercise, starting from the differential equation \(y'' + x y = 0\), we derived a recurrence relation after substituting our power series expressions for \(y\) and its derivatives.
• The recurrence relation was \((n+2)(n+1) a_{n+2} + a_n = 0\), critical for computing the coefficients \(a_n\).

Given an initial condition or an initial set of coefficients, we used this relation to express higher coefficients in terms of earlier ones, systematically finding all coefficients. This method made it possible to develop two separate and complete power series (one for even powers and another for odd powers). This allowed us to propose two linearly independent solutions for the differential equation.

Differential Equation Solutions

Solving differential equations using power series involves finding solutions in the form of an infinite series. The technique is particularly helpful for handling ordinary differential equations around ordinary points.

• The solutions are found by assuming a power series and determining the coefficients of this series using techniques like recurrence relations.
• In our problem, once the series form of the solution was assumed, substitutions into the differential equation allowed us to systematically solve for all coefficients.

By manipulating these series and identifying patterns (e.g., even and odd powers), we successfully uncovered two distinct solutions:
1. \(y_1 = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}\)
2. \(y_2 = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}\)
Combining these two forms yields the general solution. This approach showcases the power of series solutions, providing a deeper understanding of the behavior of the system described by the differential equation over its entire domain.