Series of ODE

3 min read Updated Fri Apr 24 2026 07:36:29 GMT+0000 (Coordinated Universal Time)

Consider:

P0(x)y+P1(x)y+P2(x)y=0P_0(x)y'' + P_1(x)y' + P_2(x)y = 0

Here P0,P1,P2P_0, P_1, P_2 are analytic functions of xx.

These equations rarely have elementary solutions; power series gives a general workable form. Special functions (Bessel, Legendre, Laguerre, Hermite, Chebyshev) arise naturally.

Terminology

Ordinary Point

x=ax=a is ordinary if P0(a)0P_0(a) \neq 0.

Singular Point

When x=ax=a is not ordinary. When P0(a)=0P_0(a)=0.

Regular Singular Point

A singular point x=ax=a is regular when the ODE is rewritten as:

y+Q1(x)xay+Q2(x)(xa)2y=0y'' + \frac{Q_1(x)}{x-a} y' + \frac{Q_2(x)}{(x-a)^2} y = 0

And Q1(x),Q2(x)Q_1(x), Q_2(x) are analytic at x=ax=a.

Irregular Singular Point

A singular point that not is not regular.

Series Solutions

A second-order ODE has two independent series solutions y=ay1+by2y = a y_1 + b y_2 where a,ba,b are constants.

Solution About Ordinary Points

Suppose x=ax=a is an ordinary point. The solution is of the form:

y=n=0an(xa)ny = \sum_{n=0}^{\infty} a_n (x-a)^n

Procedure:

  1. Compute y,yy',y''
  2. Substitute in ODE
  3. Collect powers of xx
  4. Set coefficient of each xnx^n to zero
    → gives recurrence relation
  5. Use recurrence to express all ana_n in terms of a0,a1a_0, a_1

Solution About Singular Points

Suppose x=ax=a is a regular singular point. The solution is of the form:

y=(xa)mn=0an(xa)ny = (x-a)^m \sum_{n=0}^{\infty} a_n (x-a)^n

Here a00a_0 \neq 0.

Frobenius Method

Used to find series solutions about regular singular point x=ax=a.

y=(xa)mn=0an(xa)ny = (x-a)^m \sum_{n=0}^{\infty} a_n (x-a)^n

Procedure:

  1. Compute derivatives y,yy', y''
  2. Substitute in ODE
  3. Set coefficient of the lowest power term to 00
    Which gives an indicial equation. The equatiion is quadratic in mm.
  4. Solve for m1,m2m_1, m_2
  5. Based on nature of roots, construct solutions

Case 1

Distinct roots, not differing by integer.

Two independent Frobenius series:

y=c1ym1+c2ym2y = c_1 y_{m_1} + c_2 y_{m_2}

Case 2

Equal roots.

One Frobenius solution; second solution involves

y=c1(y1)m1+c2(ym)m1y = c_1 (y_1)_{m_1} + c_2 \left(\frac{\partial y}{\partial m}\right)_{m_1}

Case 3

Roots differ by integer. Larger root (corresponding to m2m_2) always gives a valid solution. Smaller root may or may not.

Subcase 3a

If the recurrence relation produces finite coefficients when m=m1m=m_1, then another Frobenius solution exists.

y=c1(y1)m2+c2(y2)m1y = c_1 (y_1)_{m_2} + c_2(y_2)_{m_1}

Subcase 3b

If recurrence breaks, the complete solution is:

y=c1(ym)m1+c2y(y)m2y = c_1 \left( \frac{\partial y}{\partial m}\right)_{m_1} + c_2 y \left( y \right)_{m_2}

Here c1c_1, c2c_2 are constants.

Bessel’s Equation

The equation of the form:

x2y+xy+(x2n2)y=0x^2 y'' + x y' + (x^2 - n^2) y = 0

Bessel Functions

Solutions of Bessel’s equation.

The solutions are Bessel functions of order nn:

Jn(x)=k=0(1)kk!(n+k)!(x2)n+2kJ_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (n+k)!} \left( \frac{x}{2} \right)^{n + 2k}