Series Solutions of Differential Equations

When elementary methods (like separation of variables or undetermined coefficients) fail, power series provide a powerful tool to approximate or exactly represent solutions to complex differential equations.

Learning Objectives

After studying this guide, you should be able to:

• Assume a power series solution for a first or second-order differential equation.
• Differentiate power series term-by-term and substitute them into differential equations.
• Shift indices of summation to combine multiple series into a single expression.
• Derive and solve a recurrence relation for the coefficients $a_n$.
• Identify the radius of convergence for the resulting series solution.

Key Terms & Glossary

• Power Series: An infinite series of the form $\sum_{n=0}^{\infty} a_n (x-x_0)^n$. Example: The Taylor series for $e^x$ is $\sum \frac{x^n}{n!}$.
• Ordinary Point: A point $x_0$ where the coefficient functions of a differential equation are analytic (can be represented by a power series). Example: In $y'' + xy' + y = 0$, $x=0$ is an ordinary point.
• Singular Point: A point where the coefficient functions fail to be analytic. Example: $x=0$ is a singular point for $x^2 y'' + y = 0$.
• Recurrence Relation: An algebraic equation that expresses each coefficient $a_n$ in terms of preceding coefficients. Example: $a_{n+2} = \frac{-a_n}{(n+1)(n+2)}$.

The "Big Idea"

The core strategy of series solutions is to transform a Calculus problem (finding an unknown function $y$) into an Algebra problem (finding an infinite sequence of constants $a_n$). By assuming $y$ is a polynomial with infinite terms, we can solve for those terms one by one, often revealing a pattern that matches a known function or provides a highly accurate numerical approximation.

Formula / Concept Box

Hierarchical Outline

1. Setup and Assumption
   • Identify the differential equation and check for ordinary points.
   • Assume $y = \sum a_n x^n$.
2. Substitution and Differentiation
   • Calculate $y'$ and $y''$ using power series notation.
   • Plug these series into the original DE.
3. Algebraic Manipulation
   • Distribute any $x$ terms into the series.
   • Shift indices so that every term has the same power of $x$ (usually $x^n$).
   • Adjust the starting index of sums to match.
4. Solving the Recurrence
   • Set the sum of coefficients for each power of $x$ to zero.
   • Solve for the highest index coefficient ($a_{n+k}$) in terms of lower ones.
   • Find the general pattern for $a_n$.

Visual Anchors

The Series Solution Process

Approximation Accuracy

Definition-Example Pairs

• Term-by-term Differentiation: The process of differentiating a series as if it were a polynomial.
  • Example: If $y = a_0 + a_1x + a_2x^2$, then $y' = a_1 + 2a_2x$.
• Identity Property of Power Series: If $\sum c_n x^n = 0$ for all $x$ in an interval, then every $c_n$ must be 0.
  • Example: If $(2a_2 + a_0) + (6a_3 + a_1)x = 0$, then $a_2 = -a_0/2$ and $a_3 = -a_1/6$.

Worked Examples

Solving $y' - y = 0$ using Power Series

1. Assume: $y = \sum_{n=0}^{\infty} a_n x^n$ and $y' = \sum_{n=1}^{\infty} n a_n x^{n-1}$.
2. Substitute: $\sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{n=0}^{\infty} a_n x^n = 0$.
3. Shift Index: Let $k = n-1$ in the first sum. Then $n = k+1$.
   • $\sum_{k=0}^{\infty} (k+1) a_{k+1} x^k - \sum_{n=0}^{\infty} a_n x^n = 0$.
4. Combine: $\sum_{n=0}^{\infty} [(n+1) a_{n+1} - a_n] x^n = 0$.
5. Recurrence: $(n+1) a_{n+1} - a_n = 0 \implies a_{n+1} = \frac{a_n}{n+1}$.
6. Pattern: $a_1 = a_0$, $a_2 = a_1/2 = a_0/2!$, $a_3 = a_2/3 = a_0/3!$.
7. Result: $y = a_0 \sum \frac{x^n}{n!} = a_0 e^x$.

Checkpoint Questions

1. Why must we shift indices before combining two power series in a differential equation?
2. What is the difference between an ordinary point and a singular point in the context of series solutions?
3. If a recurrence relation is $a_{n+2} = \frac{-a_n}{(n+1)(n+2)}$, what is $a_4$ in terms of $a_0$?

[!TIP] When shifting indices, always check your first few terms manually ($n=0, 1, 2$) to ensure the series still generates the same values after the shift.

▶Click to expand: Solution to Checkpoint #3

$a_2 = \frac{-a_0}{(1)(2)} = -\frac{a_0}{2!}$
$a_4 = \frac{-a_2}{(3)(4)} = \frac{a_0}{2! \cdot 3 \cdot 4} = \frac{a_0}{4!}$