#Linear Harmonic Oscillator
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Consider the linear oscillator equation
$$ \tag{LHO} y'' + \omega^2 y = 0. $$
We want to solve this equation using a series expansion around the origin.
##Frobenius Solution Method
Start with the Ansatz
$$ y(x) = x^\rho \sum_{n=0}^\infty c_n x^n, $$
where $c_0 \neq 0$ by definition (otherwise we would just rename the coefficients and absorb into $\rho$ the lowest power of $x$). Using this into the original differential equation gives the condition
$$ \sum_{n=0}^\infty \left[ (\rho + n)(\rho+n-1) c_n x^{\rho+n-2} + c_n x^{\rho + n} \right] = 0. $$
Imposing that the coefficients of all the powers of $x$ must vanish we have
| power of $x$ $\qquad$
|
coefficients $\qquad \qquad \qquad \qquad \qquad$
|
| $x^{\rho-2}$ |
$\rho(\rho-1) c_0$ $=0$
|
| $x^{\rho-1}$ |
$\rho(\rho+1) c_1$ $=0$
|
| $x^{\rho}$ |
$(\rho+2)(\rho+1) c_2 + \omega^2 c_0 = 0$ |
| $x^{\rho+1}$ |
$(\rho+3)(\rho+2)c_3 + \omega^2 c_1 = 0$ |
and so on.
In particular the coefficient of $x^{\rho-2}$, together with the fact that $c_0 \neq 0$ gives the condition (indicial equation)
$$ \rho(\rho-1) = 0 \Longrightarrow
(\rho = 0 \text{ or } \rho = 1) $$
###$\rho=0 $
If $\rho = 0$ then
$$ y(x) = \sum_{n=0}^\infty c_n x^n, $$
and the recurrence relation for the coefficients of the even powers of $x$ gives
$$ c_2 = - \frac{\omega^2}{1 \cdot 2} c_0, $$
$$ c_4 = - \frac{\omega^2}{2 \cdot 3} c_2 = (-1)^2 \frac{\omega^4}{4!} c_0, $$
$$ c_{2n} = (-1)^n \frac{\omega^{2n}}{(2n)!} c_0. $$
At the same time, the odd coefficients $c_{2n+1}$ are dependent on $c_1$ as follows:
$$ c_{2n+1} = (-1)^n \frac{\omega^{2n}}{(2n+1)!} c_1 $$
Setting $c_1=0$ thus gives the solution
$$ y(x) = \sum_{n=0}^\infty (-1)^n \frac{(\omega x)^{2n}}{(2n)!} c_0 = c_0 \cos(\omega x). $$
while if $c_1 \neq 0$ we have
$$ y(x) = c_0 \cos(\omega x) + \frac{c_1}{\omega} \sin(\omega x). $$
###$\rho=1 $
If $\rho=1$ then
$$ y(x) = \sum_{n=0}^\infty c_n x^{n+1}, $$
and we have the following relations for the coefficients
$$ c_1 = 0,$$
$$ c_{2n} = (-1)^n \frac{\omega^{2n}}{(2n+1)!}, $$
corresponding to the solution
$$ y(x) = \frac{c_0}{\omega} \sin(\omega x).$$
