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. $$ (4+x) y^{\prime \prime}+(2+x) y^{\prime}+2 y=0, \quad y(0)=2, \quad y^{\prime}(0)=5 $$

Short answer

Question: Determine the coefficients up to \(a_7\) of the power series solution for the initial value problem $$(4+x)y''+(2+x)y'+2y=0, \quad y(0)=2, \quad y'(0)=5$$. Answer: The coefficients up to \(a_7\) are \(a_0=2\), \(a_1=-5\), \(a_2=-\frac{1}{2}\), \(a_3=-\frac{8}{81}\), \(a_4=-\frac{5}{126}\), \(a_5=-\frac{12}{741}\), \(a_6=-\frac{7}{726}\), and \(a_7=-\frac{16}{3945}\).

Step-by-step solution

Differentiate

Differentiate the power series for y twice to obtain the first and second derivatives: $$y' = \sum_{n=1}^{\infty} na_nx^{n-1}$$ and $$y'' = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}.$$

Apply initial conditions

Plug in the initial values \(y(0)=2\) and \(y'(0)=5\) to find \(a_0\) and \(a_1\): $$y(0) = a_0 = 2$$ and $$y'(0) = a_1 = 5$$

Substitute derivatives back into the original equation

Substitute the power series for \(y\), \(y'\), and \(y''\) back into the original equation and use \(a_0 = 2\) and \(a_1 = 5\): $$ (4+x)\left(\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\right)+(2+x)\left(\sum_{n=1}^{\infty} na_nx^{n-1}\right)+2\left(\sum_{n=0}^{\infty} a_nx^{n}\right)=0 $$

Make expressions equal coefficients

Group terms with equal powers of x and factor out the constant term to make sure expressions equal the coefficients: $$ \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} + \sum_{n=1}^{\infty}na_nx^{n-1} + \sum_{n=0}^{\infty}(2+2n)a_nx^{n} = 0 $$

Find a recurrence relation

Group the expressions with equal powers of x and factor out the common terms to find a recurrence relation. For \(n \ge 2\): $$ a_{n-2}\left[n(n-1)+(n-1)(n+1)+2(n+1)\right]=0 $$ This leads to the recurrence relation: $$ a_n = -\frac{2(n+1)}{n(n^2-2)} $$

Calculate coefficients using the recurrence relation

We can now calculate the coefficients up to \(a_7\): $$ a_2 = -\frac{2(2+1)}{2(2^2-2)} = -\frac{6}{12}=-\frac{1}{2} $$ $$ a_3 = -\frac{2(3+1)}{3(3^2-2)} = -\frac{8}{81} $$ $$ a_4 = -\frac{2(4+1)}{4(4^2-2)} = -\frac{10}{252}=-\frac{5}{126} $$ $$ a_5 = -\frac{2(5+1)}{5(5^2-2)} = -\frac{12}{741} $$ $$ a_6 = -\frac{2(6+1)}{6(6^2-2)} = -\frac{14}{1452}=-\frac{7}{726} $$ $$ a_7 = -\frac{2(7+1)}{7(7^2-2)} = -\frac{16}{3945} $$

Put solution in terms of the power series and found coefficients

The power series solution for the initial value problem is: $$ y(x) = \sum_{n=0}^{\infty} a_nx^n = 2 - 5x - \frac{1}{2}x^2 - \frac{8}{81}x^3 - \frac{5}{126}x^4 - \frac{12}{741}x^5 - \frac{7}{726}x^6 - \frac{16}{3945}x^7 + \cdots $$

Key concepts

Initial Value Problem

When it comes to differential equations, an initial value problem is a fundamental concept that involves finding a function that satisfies not only the differential equation itself but also meets specific conditions at the starting point, termed 'initial values'.

Let's put this into context with our exercise. Here, we have a second-order differential equation and the initial conditions provided are for the function's value and first derivative at zero: explicitly, we are given that at the point where our variable, let's call it x, is zero, the function's value, which we denote as y(0), is 2, and its first derivative, y'(0), is 5.

These initial conditions are critical; they serve as a starting point that has unique significance in the solution process. By applying these initial points, we effectively anchor the general solution to a particular solution that exclusively corresponds to our initial conditions. This step is an integral part of the process that ensures we resolve the problem for a specific case rather than dealing with a broad set of potential functions.

Recurrence Relation

The term recurrence relation may seem daunting at first, but it's a straightforward yet powerful tool in mathematics, especially in solving series solutions of differential equations. In essence, it's a formula that relates each term of a sequence to some of its predecessors.

In our scenario, the recurrence relation arises when we attempt to simplify the complicated-looking differential equation in terms of just the coefficients in the series expansion of our solution. By assertively organizing terms and examining powers of x, we discover a pattern that tells us exactly how to find any coefficient provided we know one or more of the starting coefficients, which we get from the initial conditions.

What this means is, as opposed to finding each coefficient from scratch, we employ this recurrence formula as a stepping stone, hopping from one coefficient to the next in a systematic manner. The beauty of this lies in its efficiency; by repeating a single action, we can unravel an entire sequence of coefficients—one piece after another—like a mathematician's version of a domino cascade.

Differential Equations Series Solution

To talk about differential equations series solution, it's necessary to understand that many differential equations can't be solved with simple functions. Instead, we sum an entire series of functions into one 'power series', hoping this infinite sum will behave nicely and satisfy our original equation.

In practice, this approach involves expressing the unknown function y as a power series, just like in our textbook example. We then differentiate this series, term by term (thanks to the properties of power series and their convergence), to find the derivatives of y. Next, we plug these into the given differential equation.

But here’s where the clever part comes in: since the equation must hold true for all x within the interval of convergence, we intelligently equate the coefficients of equivalent powers of x from both sides of the equation to zero, effectively setting up a system of equations that need to be satisfied. This process gives us a recipe for determining the coefficients of the series, namely the recurrence relation that we discussed earlier. While the calculations can be intricate, the logic behind them is incredibly orderly and linear. This method's real delight lies in witnessing the evolution of a complicated, perhaps mysterious, differential equation into an orderly sequence of coefficients that complies, term-by-term, with the initial problem.