Question

Consider each of the following problems as illustrations showing that, in a power series solution, we must be cautious about using the general recursion relation between the coefficients for the first few terms of the series. Solve \(y^{\prime \prime}=-y\) by the Frobenius method. You should find that the roots of the indicial equation are \(s=0\) and \(s=1 .\) The value \(s=0\) leads to the solutions \(\cos x\) and \(\sin x\) as you would expect. For \(s=1,\) call the series \(y=\sum_{n=0}^{\infty} b_{n} x^{n+1},\) and find the relation $$b_{n+2}=-\frac{b_{n}}{(n+3)(n+2)}.$$ Show that the \(b_{0}\) series obtained from this relation is just \(\sin x,\) but that the \(b_{1}\) series is not a solution of the differential equation. What is wrong?

Short answer

The series for \(s=1\) with \(b_0\) produces \(\sin x\), but the series with \(b_1\) does not form a solution for the differential equation.

Step-by-step solution

Write down the differential equation

The differential equation given is: \[ y'' = -y \]

Expand the solution as a power series

Assume a Frobenius series solution of the form \[ y = x^s \sum_{n=0}^{fty} a_n x^n \]

Find the indicial equation and roots

By substituting the power series into the differential equation, we obtain the indicial equation. The roots of this equation are \(s = 0\) and \(s = 1\).

Consider the solution for s=0

For \(s=0\), the solution reduces to a standard power series: \[ y(x) = \sum_{n=0}^{fty} a_n x^n \] Solving the recursion relations leads to the solutions \(\cos x\) and \(\sin x\), as expected.

Consider the solution for s=1

For the case \(s=1\), assume the solution of the form \[ y = \sum_{n=0}^{\infty} b_n x^{n+1} \]The recursion relation for the coefficients \(b_n\) can be found to be \[b_{n+2} = -\frac{b_n}{(n+3)(n+2)}\]

Analyze the series for b_0

By using \(b_0\), the series obtained from this relation is \[ y = b_0 x \sum_{n=0}^{\infty} \-\frac{1}{(n+3)(n+2)} x^n,\] which corresponds to the series expansion for \(\sin x\).

Analyze the series for b_1

Using \(b_1\) instead of \(b_0\), the series does not satisfy the same differential equation. This indicates that the series for \(b_1\) does not form a solution. The issue lies in the initial conditions and the nature of the Frobenius method around singular points.

Final Step: Summarize why s=1 has issues

When \(s=1\), the series must match the boundary conditions for it to be valid. While \(b_0\) corresponds to a valid solution in the form of \(\sin x\), the series with \(b_1\) leads to inconsistencies, and thus is not a valid solution.

Key concepts

Power Series Solution

In differential equations, a power series solution is a technique where the solution is expressed as an infinite sum of terms involving powers of the independent variable. This approach is highly useful, especially when solving equations near singular points.
To start, we assume the solution can be written in the form:
\[ y = \sum_{n=0}^{\infty} a_n x^n \]
Here, \(a_n\) are the coefficients to be determined. The goal is to substitute this series into the differential equation and then identify a recursion relation for the coefficients \(a_n\). This helps in finding a pattern or rule that the coefficients follow.
Using a power series solution allows us to approximate complex functions with a polynomial, making it easier to analyze and solve differential equations.

Indicial Equation

An indicial equation arises when solving differential equations using the Frobenius method. It helps in determining the exponents at which the series solution is valid.
For a differential equation in the form:
\[ y'' + p(x)y' + q(x)y = 0 \]
we assume a solution of the form:
\[ y = x^s \sum_{n=0}^{\infty} a_n x^n \]
By substituting this into the given differential equation, we equate coefficients of the same powers of \(x\), leading to an equation whose roots give us the possible values of \(s\).
These roots are crucial because they influence the form of the series solution. In the original exercise, the roots are found to be \(s=0\) and \(s=1\). Each of these roots leads to a different form of the solution, helping to capture the behavior of the function at the point of interest.

Recursion Relation

A recursion relation is a formula that connects each term in a series to its predecessors. In the context of power series solutions to differential equations, it defines how each coefficient \(a_n\) or \(b_n\) can be computed from the previous ones.
For the given problem, we solve for the coefficients based on an assumed form of the solution. After substituting the assumed solution into the differential equation and equating terms of the same power in \(x\), we derived the recursion relation:
\[ b_{n+2} = -\frac{b_n}{(n+3)(n+2)} \]
This recursion relation allows us to compute all coefficients from initial ones, \(b_0\) and \(b_1\), hence building the series term by term. While \(b_0\) series provided a meaningful solution (\(\sin x\)), the \(b_1\) series did not satisfy the differential equation, showing the importance of correct initial conditions.

Differential Equations

Differential equations involve functions and their derivatives and describe how these functions change. They are fundamental in modeling various physical phenomena.
The exercise involves solving the second-order differential equation:
\[ y'' = -y \]
To solve such equations, multiple methods can be used, including separation of variables, integrating factors, and, as seen here, the Frobenius method. The Frobenius method is particularly useful for handling equations near singular points and involves expanding the solution in a power series.
This equation is a classic example and its general solution is well-known to involve trigonometric functions like \(\cos x\) and \(\sin x\). The series solutions derived show different approaches to a known problem, emphasizing the robustness of series expansions and the subtle nuances in solving such fundamental equations.