Download original notes
PAGE 1

5.2 Series solutions near ordinary point

NOT on exam 3

\[ P(x)y'' + Q(x)y' + R(x)y = 0 \quad \text{2nd-order linear homogeneous} \]

\( x = x_0 \) where \( P(x_0) \neq 0 \) is called an ordinary point.

Goal: solve the equation with a power series solution

Try this example: \( y'' - y = 0 \)

Assume:

\[ y = a_0 + a_1x + a_2x^2 + a_3x^3 + \dots = \sum_{n=0}^{\infty} a_n x^n \quad \text{goal: } a_n = ? \]
\[ y' = a_1 + 2a_2x + 3a_3x^2 + \dots = \sum_{n=1}^{\infty} a_n \cdot n x^{n-1} \]

Where the summation starts at \( n=1 \) due to the loss of \( a_0 \).

\[ y'' = 2a_2 + 3 \cdot 2 a_3 x + 4 \cdot 3 a_4 x^2 + \dots = \sum_{n=2}^{\infty} a_n \cdot n(n-1) x^{n-2} \]

Sub into \( y'' - y = 0 \)

PAGE 2
\[ \underbrace{\sum_{n=2}^{\infty} a_n (n)(n-1) x^{n-2}}_{y''} - \underbrace{\sum_{n=0}^{\infty} a_n x^n}_{y} = 0 \]

Re-index such that all series have \( x^k \) as generic terms.

Let \( k = n - 2 \) (so \( n = k + 2 \))

\[ \sum_{k=0}^{\infty} a_{k+2} (k+2)(k+1) x^k - \sum_{k=0}^{\infty} a_k x^k = 0 \]

This must hold for all \( k \).

That means for every \( k \),

\( a_{k+2}(k+2)(k+1) - a_k = 0 \) recurrence relation

\[ a_{k+2} = \frac{a_k}{(k+2)(k+1)} \]
  • \( k=0: \quad a_2 = \frac{a_0}{2 \cdot 1} = \frac{a_0}{2!} \)
  • \( k=1: \quad a_3 = \frac{a_1}{3 \cdot 2} = \frac{a_1}{3!} \)
  • \( k=2: \quad a_4 = \frac{a_2}{4 \cdot 3} = \frac{a_0}{4 \cdot 3 \cdot 2 \cdot 1} = \frac{a_0}{4!} \)
PAGE 3
\[ k=3: \quad a_5 = \frac{a_3}{5 \cdot 4} = \frac{a_1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{a_1}{5!} \]

Generalize:

\[ a_n = \frac{a_0}{n!} \text{ if } n \text{ is even} \]\[ a_n = \frac{a_1}{n!} \text{ if } n \text{ is odd} \]

Series solution:

\( y = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots \)

or \( y = (a_0 + a_2 x^2 + a_4 x^4 + a_6 x^6 + \dots) + (a_1 x + a_3 x^3 + a_5 x^5 + \dots) \)

Where the first group contains even \( n \) and the second group contains odd \( n \).

\[ = (a_0 + \frac{a_0}{2!} x^2 + \frac{a_0}{4!} x^4 + \frac{a_0}{6!} x^6 + \dots) + (a_1 x + \frac{a_1}{3!} x^3 + \frac{a_1}{5!} x^5 + \dots) \]\[ = a_0 (1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots) + a_1 (x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \dots) \]

Where the first series is \( \cosh(x) \) and the second series is \( \sinh(x) \).

\( y = a_0 \cosh(x) + a_1 \sinh(x) \)

\( a_0, a_1 \) depend on initial conditions

PAGE 4

\( y'' - y = 0 \) solved the "old" way

characteristic eq: \( r^2 - 1 = 0 \quad r = 1, -1 \)

\( y = c_1 e^x + c_2 e^{-x} \)

\( \cosh(x) = \frac{1}{2} e^x + \frac{1}{2} e^{-x} \)

\( \sinh(x) = \frac{1}{2} e^x - \frac{1}{2} e^{-x} \)

can be made to agree with the series solution.

for this one it's an overkill

but, for this one it's not: \( y'' + xy = 0 \)

non constant coefficient "old" way does not work.

Series solution method works

\[ y = \sum_{n=0}^{\infty} a_n x^n \quad y' = \sum_{n=1}^{\infty} a_n \cdot n x^{n-1} \quad y'' = \sum_{n=2}^{\infty} a_n (n)(n-1) x^{n-2} \]\[ \sum_{n=2}^{\infty} a_n (n)(n-1) x^{n-2} + x \sum_{n=0}^{\infty} a_n x^n = 0 \]

Note: The \( x \) is distributed into the second summation.

PAGE 5

Re-indexing Power Series

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

Make them match by re-indexing series

1st series:

\( k = n - 2 \)

\( n = k + 2 \)

2nd series:

\( k = n + 1 \)

\( n = k - 1 \)

\[ \sum_{k=0}^{\infty} a_{k+2} (k+2)(k+1) x^k + \sum_{k=1}^{\infty} a_{k-1} x^k = 0 \]

They match now

when \( k = 0 \): \( a_2 (2)(1) = 0 \rightarrow a_2 = 0 \)

for \( k \ge 1 \): \( a_{k+2} (k+2)(k+1) + a_{k-1} = 0 \)

\[ a_{k+2} = \frac{-a_{k-1}}{(k+2)(k+1)} \]
PAGE 6

Calculating Coefficients and General Solution

\( k = 1 \): \( a_3 = \frac{-a_0}{3 \cdot 2} \)

\( k = 2 \): \( a_4 = \frac{-a_1}{4 \cdot 3} \)

\( k = 3 \): \( a_5 = -\frac{a_2}{5 \cdot 4} = 0 \) (since \( a_2 = 0 \))

\( k = 4 \): \( a_6 = -\frac{a_3}{6 \cdot 5} = \frac{a_0}{6 \cdot 5 \cdot 3 \cdot 2} \)

\( k = 5 \): \( a_7 = -\frac{a_4}{7 \cdot 6} = \frac{a_1}{7 \cdot 6 \cdot 4 \cdot 3} \)

\( \vdots \)

Solution:

\[ y = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots \] \[ = a_0 + a_1 x - \frac{a_0}{3 \cdot 2} x^3 - \frac{a_1}{4 \cdot 3} x^4 + \frac{a_0}{6 \cdot 5 \cdot 3 \cdot 2} x^6 + \frac{a_1}{7 \cdot 6 \cdot 4 \cdot 3} x^7 + \dots \] \[ = a_0 \left( 1 - \frac{x^3}{3 \cdot 2} + \frac{x^6}{6 \cdot 5 \cdot 3 \cdot 2} - \frac{x^9}{9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} \dots \right) + a_1 \left( x - \frac{x^4}{4 \cdot 3} + \frac{x^7}{7 \cdot 6 \cdot 4 \cdot 3} \dots \right) \]
PAGE 7

Solving Differential Equations with Power Series

We can even handle this kind of equations:

\[ y'' + \sin(x)y = 0 \quad \text{near } x = 0 \]

Near \( x = 0 \), the Taylor series for \( \sin(x) \) is:

\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \]

Substituting the series into the differential equation:

\[ y'' + \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) y = 0 \]

Replacing \( y \) and \( y'' \) with their power series representations:

\[ \sum_{n=2}^{\infty} a_n (n)(n-1) x^{n-2} + \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} \right) \sum_{n=0}^{\infty} a_n x^n = 0 \]

Distributing the polynomial terms into the second summation:

\[ x \sum_{n=0}^{\infty} a_n x^n - \frac{x^3}{3!} \sum_{n=0}^{\infty} a_n x^n + \frac{x^5}{5!} \sum_{n=0}^{\infty} a_n x^n \]