Series Solutions | Mathematics

6.1 Power Series Method

For an ODE $y"' + p(x)y' + q(x)y = 0$ where $p$ and $q$ are analytic near $x_0$ Substitute the Power series $y = \sum_{n=0}^{\infty} a_n (x - x_0)^n$ and solve for the coefficients.

6.2 Ordinary and Regular Singular Points

$x_0$ is an ordinary point if $p$ and $q$ are analytic at $x_0$ . It is a regular singular Point if $(x - x_0)p(x)$ and $(x - x_0)^2 q(x)$ are analytic at $x_0$ .

6.3 Frobenius Method

At a regular singular point $x_0 = 0$ Substitute $y = \sum_{n=0}^{\infty} a_n x^{n + r}$ . The indicial equation determines the possible values of $r$ .

Theorem 6.1. If the roots $r_1 \geq r_2$ of the indicial equation differ by a non-integer, there Are two linearly independent solutions of the form $x^{r_1}\sum a_n x^n$ and $x^{r_2}\sum b_n x^n$ .

6.4 Bessel’s Equation

Bessel’s equation of order $\nu$ :

$x^2 y'' + xy' + (x^2 - \nu^2)y = 0$

For $\nu \notin \mathbb{Z}$ The solutions are $J_\nu(x)$ and $J_{-\nu}(x)$ (Bessel functions of the First kind). For $\nu = n \in \mathbb{N}$ The second solution is the Weber function $Y_n(x)$ .

6.4b Worked Example: Higher-Order ODE

Problem. Solve $y''' - 6y'' + 11y' - 6y = 0$ .

Solution

Solution. Characteristic equation: $r^3 - 6r^2 + 11r - 6 = 0$ .

Trying $r = 1$ : $1 - 6 + 11 - 6 = 0$ . Factor: $(r - 1)(r^2 - 5r + 6) = (r - 1)(r - 2)(r - 3) = 0$ .

Roots: $r = 1, 2, 3$ (three distinct real roots).

$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$ . $\blacksquare$

### 6.5 Worked Example: Power Series Method

Problem. Solve $y'' - xy = 0$ (Airy’s equation) using power series about $x_0 = 0$ .

Solution

Solution. Since $p(x) = 0$ and $q(x) = -x$ are both analytic everywhere, $x_0 = 0$ is an ordinary Point. Substitute $y = \sum_{n=0}^{\infty} a_n x^n$ :

$y' = \sum_{n=1}^{\infty} na_n x^{n-1}$ , $y'' = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}$ .

$y'' - xy = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} - \sum_{n=0}^{\infty} a_n x^{n+1} = 0$ .

Shift indices: first sum $\sum_{m=0}^{\infty} (m+2)(m+1)a_{m+2} x^m$ Second sum $\sum_{m=1}^{\infty} a_{m-1} x^m$ .

For $m = 0$ : $2 \cdot 1 \cdot a_2 = 0 \implies a_2 = 0$ .

For $m \geq 1$ : $(m+2)(m+1)a_{m+2} - a_{m-1} = 0 \implies a_{m+2} = \frac{a_{m-1}}{(m+2)(m+1)}$ .

This gives: $a_3 = \frac{a_0}{6}$ , $a_4 = \frac{a_1}{12}$ , $a_5 = \frac{a_2}{20} = 0$ $a_6 = \frac{a_3}{30} = \frac{a_0}{180}$ Etc.

Since $a_2 = 0$ All $a_{3k+2} = 0$ .

$y(x) = a_0\left(1 + \frac{x^3}{6} + \frac{x^6}{180} + \cdots\right) + a_1\left(x + \frac{x^4}{12} + \frac{x^7}{504} + \cdots\right)$ .

These are the Airy functions $\mathrm{Ai}(x)$ and $\mathrm{Bi}(x)$ (up to normalization). $\blacksquare$

6.6 Worked Example: Frobenius Method

Problem. Solve $2xy'' + y' + xy = 0$ near $x = 0$ using the Frobenius method.

Solution

Solution. Rewrite in standard form: $y'' + \frac{1}{2x}y' + \frac{1}{2}y = 0$ .

$x = 0$ is a regular singular point since $xp(x) = 1/2$ and $x^2 q(x) = x^2/2$ are analytic at $0$ .

Substitute $y = \sum_{n=0}^{\infty} a_n x^{n+r}$ , $a_0 \neq 0$ :

$y' = \sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1}$

$y'' = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2}$

Substituting into $2xy'' + y' + xy = 0$ :

$\sum_{n=0}^{\infty} 2(n+r)(n+r-1)a_n x^{n+r-1} + \sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1} + \sum_{n=0}^{\infty} a_n x^{n+r+1} = 0$

For $n = 0$ : $[2r(r-1) + r]a_0 = 0$ . Since $a_0 \neq 0$ : $r(2r - 2 + 1) = 0 \implies r(2r - 1) = 0$ .

Indicial equation: $r = 0$ or $r = 1/2$ .

For general $n \geq 1$ : $[2(n+r)(n+r-1) + (n+r)]a_n + a_{n-2} = 0$

$(n+r)(2n + 2r - 1)a_n = -a_{n-2}$

$a_n = -\frac{a_{n-2}}{(n+r)(2n + 2r - 1)}$

For $r = 0$ : $a_n = -\frac{a_{n-2}}{n(2n-1)}$ . Odd coefficients vanish ( $a_1 = 0$ ). Even: $a_2 = -\frac{a_0}{6}$ $a_4 = \frac{a_0}{120}$ Etc.

For $r = 1/2$ : $a_n = -\frac{a_{n-2}}{(n+1/2)(2n)} = -\frac{a_{n-2}}{n(2n+1)}$ .

$y = C_1 \sum_{k=0}^{\infty} a_{2k}^{(0)} x^{2k} + C_2 x^{1/2} \sum_{k=0}^{\infty} a_{2k}^{(1/2)} x^{2k}$ . $\blacksquare$

6.7 Worked Example: Bessel Functions

Problem. Find the first three nonzero terms of $J_0(x)$ The Bessel function of the first kind Of order zero, which satisfies $x^2 y'' + xy' + x^2 y = 0$ .

Solution

Solution. Here $\nu = 0$ . The indicial equation gives $r^2 = 0$ (repeated root $r = 0$ ).

Substituting $y = \sum_{n=0}^{\infty} a_n x^{2n}$ (we can show only even powers appear):

$y' = \sum_{n=1}^{\infty} 2n a_n x^{2n-1}$ , $y'' = \sum_{n=1}^{\infty} 2n(2n-1) a_n x^{2n-2}$ .

$x^2 y'' + xy' + x^2 y = \sum_{n=1}^{\infty} 2n(2n-1)a_n x^{2n} + \sum_{n=1}^{\infty} 2n a_n x^{2n} + \sum_{n=0}^{\infty} a_n x^{2n+2} = 0$ .

For $n = 0$ : $a_0$ is free.

For the recurrence: $4n^2 a_n + a_{n-1} = 0 \implies a_n = -\frac{a_{n-1}}{4n^2}$ for $n \geq 1$ .

$a_1 = -\frac{a_0}{4}$ , $a_2 = \frac{a_0}{64}$ , $a_3 = -\frac{a_0}{2304}$ .

Setting $a_0 = 1$ : $J_0(x) = 1 - \frac{x^2}{4} + \frac{x^4}{64} - \frac{x^6}{2304} + \cdots$ . $\blacksquare$

6.7 Common Mistakes

flowchart TD
    A[6_Series Solutions] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Intuition

When an ODE has no elementary closed-form solution, power series methods let you build a solution term by term. At an ordinary point, you can expand the solution as a Taylor series and substitute into the ODE to find recurrence relations for the coefficients. At a regular singular point, the solution may have a factor of $x^r$ multiplied by a power series — the Frobenius method. The indicial equation determines the possible exponents $r$ . Bessel functions, Legendre polynomials, and other special functions all arise from this process. The radius of convergence is determined by the distance to the nearest singular point of the ODE, linking local analyticity to global solution behaviour.

6.7 Common Mistakes

Mistake 1: Confusing ordinary and regular singular points. An ordinary point is where $p(x)$ and $q(x)$ are analytic. A regular singular point is where $(x - x_0)p(x)$ and $(x - x_0)^2 q(x)$ are analytic. An irregular singular point is where neither condition holds. Misclassifying the type of singular point leads to incorrect solution methods.

Mistake 2: Forgetting to check the indicial equation roots. When using the Frobenius method, the indicial equation determines the possible values of $r$ . If the roots differ by an integer, the second solution may involve a logarithmic term. Forgetting to check the roots can lead to missing solutions.

Mistake 3: Assuming that power series solutions always converge. Power series solutions converge only within the radius of convergence, which is determined by the distance to the nearest singular point. Do not assume that a power series solution converges everywhere; always check the radius of convergence.

Mistake 4: Forgetting that the Frobenius method requires regular singular points. The Frobenius method is applicable only at regular singular points. At irregular singular points, the method fails and other techniques (such as asymptotic expansions) are needed. Always verify that the singular point is regular before applying the Frobenius method.

Mistake 5: Confusing Bessel functions of the first and second kind. Bessel functions of the first kind $J_\nu(x)$ are analytic at $x = 0$ , while Bessel functions of the second kind $Y_\nu(x)$ are singular at $x = 0$ . Do not confuse the two; they serve different purposes in solving Bessel’s equation.

Cross-References

Second-Order Linear ODEs: The power series method extends the characteristic equation approach to variable-coefficient ODEs.
Fourier Series: Bessel functions arise logically when solving PDEs in cylindrical coordinates via separation of variables.
Laplace Transforms: The Laplace transform provides an alternative to series methods for solving ODEs with initial conditions.
Complex Numbers Review: The Frobenius method involves analysing singularities in the complex plane to determine convergence of series solutions.
Classical Mechanics
Electromagnetism

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