Power Series Solutions of Differential Equations

Why Use Power Series Methods?

Power series methods are indispensable tools for solving ordinary differential equations (ODEs), particularly when other methods fall short. Here are key reasons for their use:

Variable Coefficients: They are effective for linear ODEs with variable coefficients.
Non-Elementary Solutions: For ODEs (e.g., Bessel's, Legendre's) whose solutions aren't elementary functions.
Solutions Near Singular Points: The Frobenius method handles regular singular points.
Approximation: Truncated series provide polynomial approximations.
Defining Special Functions: Many special functions are defined via ODEs solved by power series.

The idea is to assume $y(x) = \sum_{n=0}^{\infty} a_n (x-x_0)^n$.

$$ y(x) = a_0 + a_1(x-x_0) + a_2(x-x_0)^2 + \dots $$

Higher-Order Differential Coefficients as a Series

If $y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + \dots$

First Derivative:

$$ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = a_1 + 2a_2 x + \dots $$

Second Derivative:

$$ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = 2a_2 + 6a_3 x + \dots $$

k-th Derivative:

$$ y^{(k)}(x) = \sum_{n=k}^{\infty} \frac{n!}{(n-k)!} a_n x^{n-k} $$

Leibniz's Theorem for the n-th Derivative of a Product

If $y = uv$, its $n$-th derivative is:

$$ y^{(n)} = (uv)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(n-k)} v^{(k)} $$

Leibniz-Maclaurin Method

Finds series solution $y(x) = \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}x^n$ by repeatedly differentiating the ODE and evaluating at $x_0=0$.

Interactive Solver for $y' = y + x$

Initial value $y(0)$:

Number of terms in series (after $y(0)$, max 10):

Frobenius Method

For linear ODEs around a regular singular point $x_0$. Assume $y(x) = x^r \sum_{n=0}^{\infty} a_n x^n$.

Steps:

Verify regular singular point.
Substitute series into ODE.
Find Indicial Equation for $r$.
Find Recurrence Relation for $a_n$.
Determine solutions based on roots $r_1, r_2$.

Bessel's Equation

Bessel's equation of order $\nu$: $x^2 y'' + xy' + (x^2 - \nu^2)y = 0$.

Solution $J_\nu(x)$ (Bessel function of the first kind):

$$ J_\nu(x) = \sum_{m=0}^{\infty} \frac{(-1)^m}{m! \Gamma(m+\nu+1)} \left(\frac{x}{2}\right)^{2m+\nu} $$

Calculate Terms of $J_\nu(x)$

Order $\nu$ (non-negative):

Value of $x$:

Number of terms in sum ($m=0$ to $N-1$, max 15):

Bessel's Equation and Gamma Functions

Gamma function $\Gamma(z) = \int_0^\infty t^{z-1}e^{-t} dt$. Properties: $\Gamma(z+1) = z\Gamma(z)$, $\Gamma(n) = (n-1)!$. Crucial for $J_\nu(x)$ for non-integer $\nu$.

$$ J_\nu(x) = x^\nu \sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+\nu} m! \Gamma(m+\nu+1)} $$

Legendre's Equation

Legendre's equation: $(1-x^2)y'' - 2xy' + n(n+1)y = 0$.

$x=0$ is ordinary point. Solution $y(x) = \sum_{k=0}^{\infty} a_k x^k$. Recurrence:

$$ a_{k+2} = \frac{(k-n)(k+n+1)}{(k+1)(k+2)} a_k $$

Legendre Polynomials

When $n$ is a non-negative integer, one solution is $P_n(x)$, normalized by $P_n(1)=1$.

$P_0(x)=1, P_1(x)=x, P_2(x)=\frac{1}{2}(3x^2-1)$, etc.

Generate and Evaluate $P_n(x)$ (using Series)

Order $n$ (integer, 0-10):

Value of $x$ to evaluate $P_n(x)$ (optional):

Legendre Polynomials using Rodrigue's Formula

Rodrigue's formula:

$$ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2-1)^n] $$

Interactive Calculator for $P_n(x)$ via Rodrigue's Formula

Order $n$ (integer, 0-7):