Question

In each exercise, \(t=0\) is an ordinary point of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\). (a) Find the recurrence relation that defines the coefficients of the power series solution \(y(t)=\sum_{n=0}^{\infty} a_{n} t^{n} .\) (b) As in equation (7), find the first three nonzero terms in each of two linearly independent solutions. (c) State the interval \(-R

Short answer

**Short Answer:** The first three non-zero terms of the two linearly independent solutions are: \(y_1(t) = 1 - t^2 + \frac{t^4}{3!}\) and \(y_2(t) = t - \frac{t^3}{2\cdot3} + \frac{t^5}{5!}\). The solutions are even and odd: \(y_1(t)\) is an even function, and \(y_2(t)\) is an odd function.

Step-by-step solution

1. Setup the power series solution and differentiate

The given power series solution is \(y(t) = \sum_{n=0}^{\infty} a_{n} t^{n}\). To find the recurrence relation, we first need to differentiate \(y(t)\) and \(y'(t)\) with respect to \(t\). First, differentiate \(y(t)\) to get \(y'(t)\): $$y'(t) = \sum_{n=1}^{\infty} na_{n}t^{n-1}$$ Now, differentiate \(y'(t)\) to obtain \(y''(t)\): $$y''(t) = \sum_{n=2}^{\infty} n(n-1)a_{n}t^{n-2}$$

2. Substitute the derivatives into the given differential equation

Now, we substitute the power series of \(y(t)\), \(y'(t)\), and \(y''(t)\) into the given differential equation \((1+t)y''(t) + y(t) = 0\): $$ (1+t)\sum_{n=2}^{\infty} n(n-1)a_{n}t^{n-2} + \sum_{n=0}^{\infty} a_{n}t^n = 0. $$

3. Rewrite equation for easier manipulation

Our goal is to find a recurrence relation for the coefficients \(a_n\). First, let's rewrite the above equation in a more convenient form, to compare the coefficient of each power of \(t\): $$ \sum_{n=2}^{\infty} n(n-1)a_{n}t^{n-1} + \sum_{n=2}^{\infty} n(n-1)a_{n}t^{n} + \sum_{n=0}^{\infty} a_{n}t^n = 0. $$

4. Finding the recurrence relation

To find the recurrence relation, we equate the coefficients of the same powers of \(t\) on both sides: For \(n = 0\), we have: $$a_0 = 0$$ For \(n \geq 1\), we have: $$n(n-1)a_n + a_n = 0$$ This implies: $$a_{n} = \frac{a_{n-1}}{n(n-1)}, \quad \text{for } n \geq 1.$$ This is the required recurrence relation.

5. Finding linearly independent solutions

Using the recurrence relation, we can find the coefficients for two linearly independent solutions: Recursion for even indexes: $$a_{2n} = \frac{a_{2n-2}}{(2n)(2n-1)}, \quad \text{for } n \geq 1 \text{ and } a_0 = 1.$$ Recursion for odd indexes: $$a_{2n+1} = \frac{a_{2n-1}}{(2n+1)(2n)}, \quad \text{for } n \geq 1 \text{ and } a_1 = 1.$$ Starting with \(a_0 = 1\), and using the even recursion, we obtain the nonzero terms for the first solution: $$y_1(t) = 1-t^2+\frac{t^4}{3!}-\cdots$$ Starting with \(a_1 = 1\), and using the odd recursion, we obtain the nonzero terms for the second solution: $$y_2(t) = t-\frac{t^3}{2\cdot3}+\frac{t^5}{5!}-\cdots$$ The first three non-zero terms of the two linearly independent solutions are: \(y_1(t) = 1 - t^2 + \frac{t^4}{3!}\) and \(y_2(t) = t - \frac{t^3}{2\cdot3} + \frac{t^5}{5!}\).

6. Determine the interval of convergence

For a power series solution, the interval of convergence will be \(-R < t < R\). Since our differential equation contains the term \((1+t)y''(t)\), the power series converges for: $$ \Rvert 1+t \Lvert < R \Rightarrow -R < t < R $$ Since there is no constraint on \(R\) imposed by the differential equation, the interval of convergence is: $$ -R < t < R, \quad \text{for all }R. $$

7. Check if the solutions are even and odd

From Theorem 8.2, we need to check if the coefficients of the power series solutions have a consistent pattern of alternating non-zero and zero coefficients. If so, the solutions are even and odd functions. From the first few terms, we can notice that \(y_1(t)\) has nonzero coefficients only for even powers of \(t\), and \(y_2(t)\) has nonzero coefficients only for odd powers of \(t\). Thus, we conclude that the solutions are even and odd: $$ y_1(t) \text{ is an even function and } y_2(t) \text{ is an odd function.} $$

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving derivatives of a function with respect to one independent variable. An ODE expresses the rate of change of a dependent variable, typically denoted as \( y \), corresponding to the change in the independent variable, often denoted as \( t \). For example, the ODE \( y'' + p(t)y' + q(t)y = 0 \) contains a function \( y(t) \) and its derivatives \( y' \) and \( y'' \), with \( p(t) \) and \( q(t) \) representing given functions of \( t \). Solving an ODE means finding a function \( y(t) \) that satisfies the equation throughout its domain.

In the exercise provided, \( t=0 \) represents an ordinary point, which means the functions \( p(t) \) and \( q(t) \) are analytical at \( t=0 \), allowing us to seek a power series solution in the form of \( y(t) = \sum_{n=0}^{\infty} a_{n} t^{n} \). This series solution method is particularly useful when explicit solutions are not readily obtainable by other means.

Recurrence Relation

A recurrence relation in the context of differential equations is a way of defining the coefficients of a power series solution based on a pattern or recursive formula. It helps in determining all the coefficients of the series from a small number of initial terms. In the step-by-step solution for the given differential equation, after substituting the differentiated power series into the ODE and equating coefficients of like powers of \( t \), a pattern emerges that allows the expression of each coefficient in terms of preceding ones.

For the exercise, we derive a recurrence relation by equating the coefficients of the same power of \( t \) to zero because the entire series must sum to zero for all values of \( t \). This leads to formulas like \( a_{n} = \frac{a_{n-1}}{n(n-1)} \) for \( n \geq 1 \), which indicates each coefficient can be calculated by a simple division involving the preceding coefficient and the square of a natural number.

Convergence of Power Series

The convergence of a power series is essential to ensure that the series solution to an ODE represents a valid function. A power series \( \sum_{n=0}^{\infty} a_{n}t^{n} \) converges within a certain interval around the center of the series, denoted as \( -R < t < R \), called the radius of convergence. The convergence can be conditional or absolute, depending on the values of \( t \) within this interval. In the case of our differential equation, the theorem stated guarantees convergence, but the explicit radius \( R \) is not provided since it might depend on the analytical properties of \( p(t) \) and \( q(t) \) which are not specified in the exercise.

The power series for the solutions of the differential equation converges when the value of \( t \) is such that the series sums to a finite number, ensuring that the function defined by the series is meaningful and can represent the solution of the differential equation over that interval.

Linearly Independent Solutions

For second-order ODEs, like the one in our exercise, two solutions are needed to form the general solution, and these solutions must be linearly independent. Linear independence in this context means that no solution is simply a constant multiple of another. To test for linear independence, one could use the Wronskian determinant, but in the case of power series solutions, we look for differing patterns in the coefficients.

In the exercise, the step-by-step solution finds two series, one with non-zero coefficients for even powers of \( t \) and another with non-zero coefficients for odd powers of \( t \). These correspond to even and odd functions, respectively, and are inherently linearly independent. Knowing the first few terms of these solutions, we are able to understand the nature of the full solutions and to verify their linear independence, an important step in constructing the general solution to the ODE.