Question

Find the coefficients \(a_{0}, \ldots, a_{N}\) for \(N\) at least 7 in the series solution \(y=\sum_{n=0}^{\infty} a_{n} x^{n}\) of the initial value problem. $$ (1+3 x) y^{\prime \prime}+x y^{\prime}+2 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-3 $$

Short answer

In this problem, we were asked to find a power series solution for an initial value problem with a non-homogeneous differential equation with variable coefficients. We used the power series method and first wrote the general power series form, then found the recurrence relation for the coefficients. After using the initial conditions to find the first few coefficients, we found a general formula for all \(a_n\). We then computed additional coefficients up to \(a_7\) and obtained the power series solution up to \(x^7\): $$ y(x) = 2 - 3x + 18x^2 - \frac{324}{5}x^3 + \frac{5832}{25}x^4 - \frac{262260}{175}x^5 + \frac{2034244}{1435}x^6 - \frac{60963732}{4305}x^7 + \dots $$ This solution will provide a good approximation near \(x=0\), and additional coefficients can be computed as needed to improve the degree of accuracy.

Step-by-step solution

Write the general power series form

Let's start by writing the general power series solution for \(y(x)\): $$ y(x) = \sum_{n=0}^{\infty} a_n x^n. $$ We will need to find derivatives for \(y(x)\) and plug them into the differential equation, so let's find \(y'(x)\) and \(y''(x)\): $$ y'(x) = \sum_{n=1}^{\infty} na_n x^{n-1} $$ $$ y''(x) = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} $$

Substitute the series into the differential equation

Now, substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the given differential equation: $$ (1+3x)\left(\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}\right) + x\left(\sum_{n=1}^{\infty} na_n x^{n-1}\right) + 2\left(\sum_{n=0}^{\infty} a_n x^n\right) = 0. $$

Simplify the equation and find the recurrence relation

To simplify the equation, we can distribute each term: $$ \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + 3\sum_{n=2}^{\infty} n(n-1)a_n x^{n-1} + \sum_{n=1}^{\infty} na_n x^n + 2\sum_{n=0}^{\infty} a_n x^n = 0. $$ Now, to find the recurrence relation, equate the coefficients of the powers of \(x\) on both sides. Since the equation holds for all values of \(x\), the coefficients of each power of \(x\) on both sides must be equal. Setting the coefficients of \(x^n\) for \(n \geq 0\) equal to each other, we get the following recurrence relation: $$ n(n-1)a_n + 3(n-1)(n-2)a_{n-1} + na_{n-1} + 2a_n = 0 $$

Simplify the recurrence relation

Solve for \(a_n\): $$ a_n = -\frac{3(n-1)(n-2)a_{n-1} + na_{n-1}}{n(n-1) + 2}. $$ Now, express \(a_n\) as a function of \(a_{n-1}\): $$ a_n = -\frac{(3n^2 - 9n + 6 + n^2 - n)a_{n-1}}{(n^2 - n + 2)} $$ $$ a_n = -\frac{4n^2 - 10n + 6}{n^2 - n + 2} a_{n-1}. $$

Use the initial conditions to find coefficients

We have the initial conditions \(y(0) = 2\) and \(y'(0) = -3\). Use these to find the first few coefficients: $$ y(0) = 2 = a_0 $$ $$ y'(0) = -3 = a_1. $$ Plug in these values and compute the next coefficients using the recurrence relation: $$ a_2 = -\frac{4(2)^2 - 10(2) + 6}{(2)^2 - (2) + 2} a_1 = -\frac{12}{2}(-3) = 18 $$ $$ a_3 = -\frac{4(3)^2 - 10(3) + 6}{(3)^2 - (3) + 2} a_2 = -\frac{18}{5}(18) = -\frac{324}{5} $$ Now, six more coefficients should be computed deriving the general formula for \(a_n\).

Compute the next coefficients

Compute \(a_4, \ldots, a_7\) using the recurrence relation: $$ a_4 = -\frac{4(4)^2 - 10(4) + 6}{(4)^2 - (4) + 2} a_3 = -\frac{54}{15}(-\frac{324}{5}) = \frac{5832}{25} $$ $$ a_5 = -\frac{4(5)^2 - 10(5) + 6}{(5)^2 - (5) + 2} a_4 = -\frac{90}{28}( \frac{5832}{25}) = -\frac{262260}{175} $$ $$ a_6 = -\frac{4(6)^2 - 10(6) + 6}{(6)^2 - (6) + 2} a_5 = -\frac{126}{41}(-\frac{262260}{175}) = \frac{2034244}{1435} $$ $$ a_7 = -\frac{4(7)^2 - 10(7) + 6}{(7)^2 - (7) + 2} a_6 = -\frac{162}{54}( \frac{2034244}{1435}) = -\frac{60963732}{4305} $$

Write the final solution

Now that we have the coefficients \(a_0, \ldots, a_7\), let's write down the power series solution for this initial value problem up to \(x^7\): $$ y(x) = 2 - 3x + 18x^2 - \frac{324}{5}x^3 + \frac{5832}{25}x^4 - \frac{262260}{175}x^5 + \frac{2034244}{1435}x^6 - \frac{60963732}{4305}x^7 + \dots $$ Keep in mind that this solution will be a good approximation near \(x=0\). Further coefficients can be computed using the recurrence relation if a higher degree of accuracy is needed.

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In the context of our problem, we are dealing with a second-order linear differential equation. This type of equation is characterized by the presence of both the second derivative, denoted as \(y''\), and the first derivative, \(y'\), of the unknown function \(y\). The equation may also contain the function \(y\) itself and is set equal to zero or some function of \(x\).

The given differential equation in our example, \((1+3x)y'' + xy' + 2y = 0\), shows how the function \(y\) is connected to its second and first derivatives. Differential equations are fundamental in physics, engineering, economics, and various other fields because they describe many phenomena such as motion, heat, and growth rates. To solve such an equation, we look for a function \(y(x)\) that satisfies the equation for all values of \(x\). Power series solutions are one way to find such a function when the solution cannot be expressed in terms of elementary functions.

In our step-by-step solution, we began by expressing \(y(x)\) as an infinite power series with coefficients \(a_n\) that we need to find. By plugging the power series for \(y\), \(y'\), and \(y''\) into the differential equation and manipulating the terms, we can derive a system of equations to solve for these coefficients.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation along with specified values, known as initial conditions, which the solution must satisfy at the beginning of the interval for which the solution is sought. In our example, the initial conditions given are \(y(0) = 2\) and \(y'(0) = -3\), which means that when \(x = 0\), the function \(y\) and its first derivative \(y'\) take these values, respectively.

Solving an initial value problem involves finding a function that not only satisfies the differential equation but also passes through the given initial points. This is a crucial step in ensuring that the solution is unique to the problem at hand. To apply the initial conditions in a power series solution like the one we are dealing with, one typically plugs \(x = 0\) into the series expressions for \(y(x)\) and \(y'(x)\) to relate the coefficients directly to these initial values. This gives us the base cases in a recursive sequence of equations, which then enables us to calculate subsequent coefficients in the series.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence, each term of the sequence is defined as a function of the preceding terms. In the case of power series solutions to differential equations, the recurrence relation provides a method to calculate the coefficients of the series. After substituting the power series expressions into the differential equation, we derive a relationship between the coefficients \(a_n\) and the terms that come before it, such as \(a_{n-1}\), \(a_{n-2}\), and so on.

The structure of the recurrence relation is typically influenced by the form of the differential equation. For our problem, the recurrence relation was found to be \(a_n = -\frac{4n^2 - 10n + 6}{n^2 - n + 2} a_{n-1}\), which allows us to compute newer coefficients based on the previous ones, starting from the initial conditions tied to \(a_0\) and \(a_1\). It is a step-by-step approach, where each coefficient is dependent on the previous one, thus forming a chain that builds the whole solution. For students, understanding how to develop and use a recurrence relation is vital because it simplifies the process of solving complex problems by breaking them down into smaller, more manageable pieces.