# Solve in series x^2y''+xy'+(x^2-k^2)y=0?

Jul 12, 2018

The solutions of the equation are the Bessel functions.

#### Explanation:

Suppose we can express $y \left(x\right)$ as the sum of a power series:

$y \left(x\right) = {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n}$

for $x \in I$, and that we can differentiate term by term, so that:

$y ' \left(x\right) = {\sum}_{n = 1}^{\infty} n {c}_{n} {x}^{n - 1}$

$y ' ' \left(x\right) = {\sum}_{n = 2}^{\infty} n \left(n - 1\right) {c}_{n} {x}^{n - 2}$

Substitute now in the original equation:

${x}^{2} {\sum}_{n = 2}^{\infty} n \left(n - 1\right) {c}_{n} {x}^{n - 2} + x {\sum}_{n = 1}^{\infty} n {c}_{n} {x}^{n - 1} + \left({x}^{2} - {k}^{2}\right) {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n} = 0$

${\sum}_{n = 1}^{\infty} n \left(n - 1\right) {c}_{n} {x}^{n} + {\sum}_{n = 2}^{\infty} n {c}_{n} {x}^{n} + \left({x}^{2} - {k}^{2}\right) {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n} = 0$

${\sum}_{n = 1}^{\infty} n \left(n - 1\right) {c}_{n} {x}^{n} + {\sum}_{n = 2}^{\infty} n {c}_{n} {x}^{n} + {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n + 2} - {k}^{2} {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n} = 0$

Extract the terms with $n < 2$:

${\sum}_{n = 2}^{\infty} n \left(n - 1\right) {c}_{n} {x}^{n} + {c}_{1} x + {\sum}_{n = 2}^{\infty} n {c}_{n} {x}^{n} + {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n + 2} - {k}^{2} {c}_{0} - {k}^{2} {c}_{1} x - {k}^{2} {\sum}_{n = 2}^{\infty} {c}_{n} {x}^{n} = 0$

and scale the index of the the third sum:

${\sum}_{n = 2}^{\infty} n \left(n - 1\right) {c}_{n} {x}^{n} + {\sum}_{n = 2}^{\infty} n {c}_{n} {x}^{n} + {\sum}_{n = 2}^{\infty} {c}_{n - 2} {x}^{n} - {k}^{2} {\sum}_{n = 2}^{\infty} {c}_{n} {x}^{n} = - {c}_{1} x + {k}^{2} \left({c}_{0} + {c}_{1} x\right)$

Group now the terms in a single sum:

${c}_{1} x - {k}^{2} \left({c}_{0} + {c}_{1} x\right) + {\sum}_{n = 2}^{\infty} \left({n}^{2} {c}_{n} - n {c}_{n} + n {c}_{n} + {c}_{n - 2} - {k}^{2} {c}_{n}\right) {x}^{n} = 0$

${c}_{1} x - {k}^{2} \left({c}_{0} + {c}_{1} x\right) + {\sum}_{n = 2}^{\infty} \left(\left({n}^{2} - {k}^{2}\right) {c}_{n} + {c}_{n - 2}\right) {x}^{n} = 0$

As the sum is null the coefficient of each degree must be null, so we get:

$\left\{\begin{matrix}{k}^{2} {c}_{0} = 0 \\ \left(1 - {k}^{2}\right) {c}_{1} = 0 \\ \left({k}^{2} - {n}^{2}\right) {c}_{n} = {c}_{n - 2}\end{matrix}\right.$

which allows to determine the coefficients recursively.

Note that:

1) All the coefficients are zero for $n < k$

2) The coefficient for $n = k$ can be assigned arbitrarily, and because all non null coefficient are proportional this is equivalent to scaling the function.

3) For $n > k$ as ${c}_{n}$ depend only on ${c}_{n - 2}$ all the coefficients with the same parity as $k$ are non null and all the coefficients with the different parity are null.