Question

First Order Equations. The series methods discussed in this section are directly applicable to the first order linear differential equation \(P(x) y^{\prime}+Q(x) y=0\) at a point \(x_{0}\), if the function \(p=Q / P\) has a Taylor series expansion about that point. Such a point is called an ordinary point, and further, the radius of convergence of the series \(y=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\) is at least as large as the radius of convergence of the series for \(Q / P .\) In each of Problems 16 through 21 solve the given differential equation by a series in powers of \(x\) and verify that \(a_{0}\) is arbitrary in each case. Problems 20 and 21 involve nonhomogeneous differential equations to which series methods can be easily extended. Where possible, compare the series solution with the solution obtained by using the methods of Chapter 2 . $$ y^{\prime}-x y=0 $$

Short answer

Question: Find the series solution for the given first-order linear differential equation \(y^{\prime} - xy = 0\) in the form of \(y = \sum_{n=0}^{\infty} a_n (x - x_0)^n\). Answer: The series solution for the given differential equation is \(y= a_{0} \sum_{n=0}^{\infty} \frac{(x - x_0)^{2n}}{(2n)!}\).

Step-by-step solution

Rewrite the differential equation in the desired form

We can rewrite the given differential equation \(y^{\prime} - xy = 0\) in the first-order linear form \(P(x)y^{\prime}+Q(x)y=0\) by defining \(P(x) = 1\) and \(Q(x) = -x\).

Find the Taylor series expansion for Q/P

Next, we will find the Taylor series expansion for \(p=Q/P\). Here \(p=-x\). Since \(p\) is a polynomial of degree 1, it is already in its Taylor series expansion form.

Formulate the series solution

Now, we will assume the series solution in terms of \(a_n\). Let \(y = \sum_{n=0}^{\infty} a_n (x - x_0)^n\). Differentiating this series solution with respect to \(x\), we have \(y^{\prime} = \sum_{n=1}^{\infty} n a_n (x - x_0)^{n-1}\).

Substitute the series in the differential equation

Substitute the series expression for \(y\) and \(y^{\prime}\) in the given differential equation \((y^{\prime} - x y = 0)\). We have: $$ \sum_{n=1}^{\infty} n a_n (x - x_0)^{n-1} - x\sum_{n=0}^{\infty} a_n (x-x_{0})^{n} = 0 $$

Equate coefficients to determine the series

As the series representation needs to be true for all \(x\) within the radius of convergence, the coefficients of each power \((x-x_0)^n\) must be equal to 0. Shifting indices for both summations to align the powers, we get: $$ \sum_{n=0}^{\infty}(n+1)a_{n+1}(x-x_0)^{n} - \sum_{n=1}^{\infty} a_{n-1}(x-x_0)^{n}=0 $$ Comparing coefficients of each power, we have: $$ (n+1)a_{n+1} = a_{n-1} \Rightarrow a_{n+1} = \frac{a_{n-1}}{n+1} $$

Find the general expression for the coefficients

Substituting \(n = 0, 1, 2, \ldots\) into the recurrence relation, we obtain $$ a_{1} = \frac{a_{-1}}{1}, \quad a_{2} = \frac{a_{0}}{2},\quad a_{3} = \frac{a_{1}}{3}, \quad a_{4} = \frac{a_{2}}{4}, \ldots $$ Since \(a_{-1}\) is undefined, we have \(a_1 = 0\). Also, noting that odd indices have even index coefficients (\(a_3 = a_1 = 0\), \(a_5 = a_3 = 0\), etc.), we can separate the sum into even and odd terms. Since all odd terms are zero, we only consider even terms: $$ y = a_{0} + a_{2}(x - x_0)^{2} + a_{4}(x - x_0)^{4} + \ldots $$

Substitute the expression for coefficients into the series solution

Substituting \(a_{2} = \frac{a_{0}}{2}\), \(a_{4} = \frac{a_{2}}{4} = \frac{a_{0}}{8}\), and so on, we have: $$ y = a_{0}\left(1 + \frac{(x - x_0)^{2}}{2} + \frac{(x - x_0)^{4}}{8} + \ldots\right) = a_{0} \sum_{n=0}^{\infty} \frac{(x - x_0)^{2n}}{(2n)!} $$

Confirm that a_0 is arbitrary

As seen in the final solution, \(a_0\) is a factor in the series solution. This confirms that \(a_0\) is arbitrary, allowing for different solutions based on different initial conditions. The series solution for the given differential equation \(y^{\prime} - xy = 0\) is: $$ y= a_{0} \sum_{n=0}^{\infty} \frac{(x - x_0)^{2n}}{(2n)!} $$

Key concepts

First Order Linear Differential Equation

A first order linear differential equation is fundamental in the study of differential equations and is expressed in the form \(P(x) y' + Q(x) y = 0\). These equations contain only the first derivative \(y'\) of the unknown function \(y\) and no higher derivatives.

When solving such an equation, the first step is often to identify the coefficients \(P(x)\) and \(Q(x)\). These functions play a crucial role in determining the behavior of the solution to the differential equation. In the given exercise, identifying these coefficients allowed the problem to be rewritten in a form that was amenable to finding a series solution.

Taylor Series Expansion

The Taylor series expansion is a powerful tool in mathematics that represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. More formally, if \(f(x)\) is infinitely differentiable at a point \(x_0\), then the Taylor series of \(f\) at \(x_0\) is given by:
\[f(x) = \f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \ldots\]Each term of the series involves a higher derivative of \(f\) at \(x_0\) and provides a polynomial approximation that becomes increasingly accurate as more terms are included.

The Taylor series is crucial in solving differential equations where we seek to express the solution as a series. It's especially useful when \(Q/P\) from the linear differential equation is not already a simple function that can be easily integrated or has a known series expansion.

Ordinary Point

In the context of differential equations, an ordinary point is a crucial concept. It refers to a point \(x_0\) where the functions \(P(x)\) and \(Q(x)\) in a differential equation \(P(x)y'+Q(x)y=0\) are well-behaved, which typically means they are analytic and thus can be expanded into a Taylor series.

When a first order differential equation has an ordinary point, it allows the solution to be represented as a power series centered on that point. This approach leads to a series solution for \(y\), where the solution is applicable in a particular interval around \(x_0\), defined by its radius of convergence. If a point is not ordinary, special methods are applied to tackle the resulting singular point.

Radius of Convergence

The radius of convergence is a measure of the interval in which a Taylor series is valid and converges to the function it represents. In mathematical terms, if we have a power series \(\sum_{n=0}^{\infty}a_{n}(x-x_0)^n\), its radius of convergence is the distance from \(x_0\) within which this series converges.

For differential equations, the radius of convergence of the series solution is at least as large as the radius of convergence of the Taylor series for the coefficient function \(Q/P\). Thus, understanding and determining the radius of convergence is essential, as it informs us where our series approximation for the solution of a differential equation is valid and reliable.

Recurrence Relation

Lastly, recurrence relations are equations that define a sequence of numbers using recurrence: each term of the sequence is defined in terms of previous terms. In the case of series solutions to differential equations, a recurrence relation helps us calculate the coefficients \(a_n\) of the series in a step-by-step manner.

For the exercise in question, the recurrence relation \(a_{n+1} = \frac{a_{n-1}}{n+1}\) was utilized to connect the coefficients of the series. It simplified the process of finding the series representation of the solution by providing a systematic way to determine all coefficients from a given initial set. This approach underscores the iterative nature of solving differential equations via series solutions.