Question

Solve the given differential equation by means of a power series about the given point \(x_{0} .\) Find the recurrence relation; also find the first four terms in each of two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution. \((1-x) y^{\prime \prime}+x y^{\prime}-y=0, \quad x_{0}=0\)

Short answer

Question: Solve the following differential equation by using a power series method: \((1-x) y^{\prime \prime} + x y^{\prime} - y = 0\) Answer: The general solution of the given differential equation is \(y(x) = C_1 y_1(x) + C_2 y_2(x)\), where \(y_1(x) = 1 + \frac{1}{2}x^2+\frac{1}{24}x^4\) and \(y_2(x) = x+\frac{1}{6}x^3+\frac{1}{120}x^5\), and \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Substituting power series into the differential equation

First, we'll substitute the power series \(y=\sum_{n=0}^{\infty} a_nx^n\) into the given differential equation \((1-x) y^{\prime \prime} + x y^{\prime} - y = 0\). But before that, we need to find the first and second derivatives of \(y\), which are \(y^{\prime}\) and \(y^{\prime\prime}\): \(y^\prime = \sum_{n=1}^{\infty} na_nx^{n-1}\) \(y^{\prime\prime} = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\). Now substitute \(y\), \(y^\prime\), and \(y^{\prime\prime}\) into the given differential equation: $(1-x)\left(\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\right) +x\left(\sum_{n=1}^{\infty} na_nx^{n-1}\right) -\left(\sum_{n=0}^{\infty} a_nx^n\right)=0$

Simplifying and combining terms

To simplify and combine the terms, we need to adjust the summation indices and distribute the coefficients. Multiply the first summation by \((1-x)\), the second summation by \(x\), and leave the third one as it is: $\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}-\sum_{n=2}^{\infty} n(n-1)a_nx^{n-1} +\sum_{n=1}^{\infty} na_nx^n -\sum_{n=0}^{\infty} a_nx^n=0$ Now, rearrange the series to have the same base for each term: \(\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2}-(n+2)(n+1)a_{n+2}x+(n+1)a_nx^n -a_nx^n]x^{n}=0\)

Equating coefficients and finding the recurrence relation

Since the powers of \(x\) are linearly independent, the coefficients must be equal for the equation to hold. Set the coefficients equal to zero: \((n+2)(n+1)a_{n+2}-(n+2)(n+1)a_{n+2}x+(n+1)a_nx^n-a_nx^n=0\) Now, we can find the recurrence relation for \(a_n\). First, divide both sides by \((n+1)(n+2)\): \(a_{n+2}=\frac{a_n}{(n+2)(n+1)}\)

Finding the first four terms of the two linearly independent solutions

We will find the first four terms for two linearly independent solutions using the recurrence relation. For the first solution, let \(a_0 = 1\) and \(a_1 = 0\): \(a_2 = \frac{a_0}{2\cdot 1} = \frac{1}{2}\), \(a_3 = \frac{a_1}{3\cdot 2} = 0\), \(a_4 = \frac{a_2}{4\cdot 3} = \frac{1}{2\cdot 4\cdot 3} = \frac{1}{24}\), \(a_5 = \frac{a_3}{5\cdot 4} = 0\) Therefore, the first solution is given by \(y_1(x) = a_0 + a_2 x^2 + a_4 x^4 = 1 + \frac{1}{2}x^2+\frac{1}{24}x^4\) For the second solution, let \(a_0 = 0\) and \(a_1 = 1\): \(a_2 = \frac{a_0}{2\cdot 1} = 0\), \(a_3 = \frac{a_1}{3\cdot 2} = \frac{1}{6}\), \(a_4 = \frac{a_2}{4\cdot 3} = 0\), \(a_5 = \frac{a_3}{5\cdot 4} = \frac{1}{6\cdot 5\cdot 4} = \frac{1}{120}\) The second solution is given by \(y_2(x) = a_1 x + a_3 x^3 + a_5 x^5 = x+\frac{1}{6}x^3+\frac{1}{120}x^5\)

Finding the general term of each solution (if possible)

For the first solution: \(y_1(x) = \sum_{k=0}^{\infty} a_{2k} x^{2k}\) where \(a_{2k+1}=0\) and \(a_{2k} = \frac{a_{2(k-1)}}{(2k)(2k-1)}.\) For the second solution: \(y_2(x) = \sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}\) where \(a_{2k}=0\) and \(a_{2k+1} = \frac{a_{2k-1}}{(2k+1)(2k)}.\) The general solution of the given differential equation is: \(y(x) = C_1 y_1(x) + C_2 y_2(x),\) where \(C_1\) and \(C_2\) are constants.

Key concepts

Recurrence Relation

A recurrence relation is a fundamental piece of the puzzle when working with power series solutions of differential equations. It provides a means to determine the coefficients of the series systematically. In simpler terms, it's like a recipe that tells us how to find each term of the solution based on previous terms.

In our exercise, after substituting a power series into the differential equation and differentiating, we look for a pattern in the coefficients. This is achieved by equating the coefficients of like powers of \(x\) because terms with different powers of \(x\) cannot add up to zero unless their coefficients are zero – as \(x\) to different powers are linearly independent. The relation that connects these coefficients is called the recurrence relation. The relation we derived, \(a_{n+2} = \frac{a_n}{(n+2)(n+1)}\), is our key to unlocking all the terms of the series. With this relation, we can generate as many terms of the power series as needed, simply by using the known initial values.

Understanding and correctly applying the recurrence relation ensures that students can not only find the first few terms of a series solution but also the general term, if the pattern allows for it. This method is powerful because it transforms the complex problem of solving a differential equation into a more manageable task of algebraic computations.

Series Solution of Differential Equations

Solving differential equations using series solutions is a versatile technique that you’ll often encounter in mathematical physics and engineering. The idea is to assume that the solution to the differential equation can be represented as an infinite sum of powers of \(x\), known as a power series.

Let's dive into why this method is useful. Some differential equations are stubborn and refuse to be solved using standard methods. This is where the series solution comes in as a savior. By expressing the solution as a power series and plugging it back into the original equation, as demonstrated in our exercise, we can unravel the behavior of the equation term by term.

In step 1 of our provided solution, we start by identifying a power series for \(y\), its first, and second derivatives. We then substitute these series into the original differential equation. By gathering like terms and simplifying the equation, we lay the groundwork to extract the necessary coefficients that will build up our series solution, as seen in the following steps. The ability to represent the solution this way gives us a powerful tool to approximate solutions numerically and understand the qualitative behavior of the solutions around the point of expansion.

Exercise improvement advice

When presenting these steps to students, it is critical to emphasize the importance of accurate algebraic manipulation. Discussing each substitution carefully and illustrating how terms are combined can help avoid confusion. Moreover, providing real-world applications of where series solutions come into play could spark interest and a better grasp of the concept.

Linearly Independent Solutions

The notion of linear independence is crucial when discussing solutions to differential equations. Two solutions to a differential equation are termed linearly independent if there is no constant \(c\) (other than zero) for which \(c\cdot y_1(x) = y_2(x)\). In other words, you cannot express one solution as a simple scalar multiple of the other. This concept is the cornerstone for formulating the general solution to a second-order differential equation.

In the context of our exercise, we were asked to find two linearly independent solutions to the given differential equation. To do so, we began by choosing different initial conditions for the series coefficients. This guarantees that the resulting series solutions won't simply be scalar multiples of each other. For instance, setting \(a_0 = 1\) and \(a_1 = 0\) leads us to one solution, and starting with \(a_0 = 0\) and \(a_1 = 1\) to another. Following our recurrence relation, we computed the first few terms, leading to two distinct solutions, \(y_1(x)\) and \(y_2(x)\).

The general solution of the differential equation is a linear combination of these independent solutions. The coefficients in this combination, labeled \(C_1\) and \(C_2\), are determined by the boundary conditions or initial values specific to a particular problem. It's like having two different building blocks, and the general solution is a structure you build according to your design specifications (the initial conditions).

Exercise improvement advice

To help students appreciate this concept in practice, exercises that involve graphing the independent solutions can be valuable. Seeing how the solutions don't just scale or shift each other on a graph reinforces their independence. Moreover, discussing cases where linear independence is not immediately obvious, such as solutions involving trigonometric functions, can deepen understanding.