series solutions of differential
equations
a(x)y''+b(x)y'+c(x)y=f(x)
Solution will be a power series: y =
\sum_{i=0}^\infty a_i(x - x_0)^i
The sums coming from y'' and
y' should have a common x^i factor to be able to join them. Then a
recursive formula for a_i can be
derived.
Example:
y'' + 2xy' - y = 0,
y(0) = 0, y'(0) = 1
- Let y = \sum_{i=0}^\infty a_ix^i
- Then y' = \sum_{i=0}^\infty
ia_ix^{i-1}
- Then y'' = \sum_{i=0}^\infty
i(i-1)a_ix^{i-2}
- Create a common x^i: y'' = \sum_{j=0}^\infty
(j+2)(j+1)a_{j+2}x^j (we leave y' with a x^{i-1} because b(x) = 2x)
- DE takes form of: \sum_{j=0}^\infty
(j+2)(j+1)a_{j+2}x^j + \sum_{i=0}^\infty 2ia_ix^i - \sum_{i=0}^\infty
a_ix^i = 0
- Simplify: \sum_{i=0}^\infty
\big[(i+2)(i+1)a_{i+2} + (2i - 1)a_i \big] x^i = 0
- Which means: (i+2)(i+1)a_{i+2} + (2i -
1)a_i = 0 \implies a_{i+2} = -\frac{2i - 1}{(i+2)(i+1)}a_i
- After calculating some a terms we
get: y = a_0(1 + \frac{x^2}{2} -
\frac{3x^4}{4!} + \frac{21x^6}{6!} - \cdots) + a_1(x - \frac{x^3}{3!} +
\frac{5x^5}{5!} - \frac{45x^7}{7!} + \cdots)
- Using the initial conditions we get a_0 =
0 and a_1 = 1
the Frobenius method
When we cannot get a recursive formula for a_i
Then we consider a solution in a form of y
= x^r \sum_{i=0}^\infty a_i