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. $$ \left(1+x+2 x^{2}\right) y^{\prime \prime}+(2+8 x) y^{\prime}+4 y=0, \quad y(0)=-1, \quad y^{\prime}(0)=2 $$

Short answer

The first 8 coefficients are \(a_0 = -1\), \(a_1 = 2\), \(a_2 = -2\), \(a_3 = \frac{4}{3}\), \(a_4 = 1\), \(a_5 = -\frac{4}{3}\), \(a_6 = \frac{8}{7}\), and \(a_7 = -\frac{2}{3}\).

Step-by-step solution

Series representation of the differential equation

Represent the function \(y\) as a power series: $$ y(x) = \sum_{n=0}^{\infty} a_{n} x^n $$

Differentiating the series representation

Differentiate the power series for \(y(x)\) and \(y'(x)\): $$ y'(x)=\sum_{n=1}^{\infty} na_{n}x^{n-1} $$ and $$ y''(x)=\sum_{n=2}^{\infty} n(n-1)a_{n}x^{n-2}. $$

Substitute the series expressions into the equation

Substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the given differential equation \((1+x+2x^2)y'' + (2+8x)y' + 4y = 0\): $$ (1+x+2x^2) \sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}+(2+8x)\sum_{n=1}^{\infty}na_{n}x^{n-1}+4\sum_{n=0}^{\infty}a_{n}x^{n}=0 $$

Determine the recurrence relation between coefficients

Now, we need to find a recurrence relation between the coefficients \(a_n\). First, we equate the coefficients of the same powers of \(x\): $$ \begin{cases} a_2+2a_1+4a_0=0 \\ 3a_3+8a_2+4a_1=0 \\ (n+2)(n+1)a_{n+2}+2(n+1)na_n+4na_{n-1}=0, \quad n \ge 2 \end{cases} $$

Find the coefficients using the initial conditions

Using the initial conditions \(y(0) = -1\) and \(y'(0) = 2\), we have: $$ \begin{cases} a_0=-1 \\ a_1=2 \end{cases} $$

Compute the coefficients \(a_{0}, \ldots, a_{N}\) for \(N \ge 7\)

Now, we can compute the coefficients \(a_2, a_3, \ldots, a_N\) for \(N \ge 7\) using the recurrence relation from step 4: From the first equation: $$ a_2=-\frac{2a_1+4a_0}{1}=-2 $$ From the second equation: $$ a_3=-\frac{8a_2+4a_1}{3}=\frac{4}{3} $$ And for \(n\ge 2\): \begin{align*} a_{n+2}&=-\frac{2(n+1)na_n+4na_{n-1}}{(n+2)(n+1)} \\ a_{4}&=-\frac{2(3 \cdot 2)a_2+4 \cdot 2 \cdot a_1}{4\cdot 3}=1 \\ a_{5}&=-\frac{2(4 \cdot 5)a_3+4 \cdot 5 \cdot a_2}{5 \cdot 6}=-\frac{4}{3} \\ a_{6}&=-\frac{2(6 \cdot 7)a_4+4 \cdot 7 \cdot a_3}{7 \cdot 8}=\frac{8}{7} \\ a_{7}&=-\frac{2(8 \cdot 9)a_5+4 \cdot 9 \cdot a_4}{9 \cdot 10}=-\frac{2}{3} \\ \cdots \end{align*} Now, we have found the coefficients \(a_{0}, \ldots, a_{N}\) for \(N \ge 7\): $$ a_0 = -1, \quad a_1 = 2, \quad a_2 = -2, \quad a_3 = \frac{4}{3}, \quad a_4 = 1, \quad a_5 = -\frac{4}{3}, \quad a_6 = \frac{8}{7}, \quad a_7 = -\frac{2}{3}, \ldots $$

Key concepts

Power Series

A power series is a way of expressing a function as an infinite sum of terms, where each term is a power of the variable with a specific coefficient. In this type of solution, which is common in differential equations, the function is broken down into simpler, polynomial-like expressions.
The general form of a power series is:

• \(y(x) = \sum_{n=0}^{\infty} a_{n} x^n\)

This is particularly useful in differential equations because it allows us to manipulate and differentiate the function term by term. Each coefficient \(a_n\) plays a critical role as it defines the specific behavior and solution of the function.
The concept of power series is essential for finding solutions to differential equations when simple analytical methods are not applicable, especially around points where the functions may not behave nicely or have clearly defined behaviors.

Recurrence Relations

Recurrence relations are equations that recursively define sequences, meaning each term of the sequence is expressed as a function of its preceding terms. In solving differential equations using power series, a recurrence relation is used to determine the coefficients \(a_n\) of the series.
In our differential equation problem, after substituting the power series, we equated terms to find relationships between coefficients:

• For example, \(a_2 = -\frac{2a_1 + 4a_0}{1}\)
• The general form for \(n \ge 2\) was found as:
  \( (n+2)(n+1)a_{n+2} + 2(n+1)na_n + 4na_{n-1} = 0 \)

This approach enables us to calculate unknown coefficients step-by-step, starting from known initial values. This is often key to building a solution that satisfies both the differential equation and any given initial conditions.

Initial Value Problem

An initial value problem (IVP) is a differential equation paired with specified values, known as the initial conditions, at a particular point. The role of initial conditions is to ensure a unique solution to the differential equation.
In the example provided, the conditions given were \(y(0) = -1\) and \(y'(0) = 2\). Specifically:

• \(a_0 = -1\)
• \(a_1 = 2\)

These initial conditions are crucial because they allow us to solve for the first few coefficients directly and anchor the solution, ensuring that the series solution fits within the broader mathematical context provided by the differential equation.

Higher-Order Differential Equations

Higher-order differential equations involve derivatives greater than the first order. The differential equation provided:\[(1+x+2x^2)y'' + (2+8x)y' + 4y = 0\] is a second-order equation because it includes the second derivative \(y''\).
Solving such equations typically involves more complex methods, as demonstrated through series solutions and recurrence relations in this example.
Higher-order equations often model more complex systems and can have more diverse behaviors compared to first-order equations. They, therefore, require solutions that incorporate more comprehensive methods, like the power series approach, which elegantly handles the intricacies these systems present.