Determine the interval of convergence of a given power series and show that it satisfies a differential equation

Consider the function $f(x)$ defined by the power series

$f(x) = x + \sum_{n=0}^{\infty} \frac{(3x)^{2n+1}}{(2n+1)!}.$

Determine the interval of convergence for the power series and show that $f(x)$ satisfies the differential equation

$y'' = 9 (y-x).$

First, we use the ratio test to determine the interval of convergence,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(3x)^{2n+3}}{(2n+3)!} \right) \left( \frac{(2n+1)!}{(3x)^{2n+1}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{9x^2}{(2n+2)(2n+3)} \\[9pt] &= 0. \end{align*}$

Hence, the series converges for all $x$ . Then, to show that it satisfies the given differential equation we take the first two derivatives,

$\begin{align*} && f(x) &= x + \sum_{n=0}^{\infty} \frac{(3x)^{2n+1}}{(2n+1)!} \\[9pt] \implies && f'(x) &= 1 + \sum_{n=0}^{\infty} \frac{3(2n+1)(3x)^{2n}}{(2n+1)!} \\[9pt] &&&= 1 + 3\sum_{n=0}^{\infty} \frac{(3x)^{2n}}{(2n)!} \\[9pt] \implies && f''(x) &= 3\sum_{n=1}^{\infty} \frac{3(2n)(3x)^{2n-1}}{(2n)!} \\[9pt] &&&= 9\sum_{n=1}^{\infty} \frac{(3x)^{2n-1}}{(2n-1)!}. \end{align*}$

Then we have

$\begin{align*} 9(y-x) &= 9 \left( -x + x + \sum_{n=0}^{\infty} \frac{(3x)^{2n+1}}{(2n+1)!} \right) \\[9pt] &= 9 \sum_{n=0}^{\infty} \frac{(3x)^{2n+1}}{(2n+1)!} \\[9pt] &= 9 \sum_{n=1}^{\infty} \frac{(3x)^{2(n-1)+1}}{(2(n-1)+1)!} \\[9pt] &= 9 \sum_{n=1}^{\infty} \frac{(3x)^{2n-1}}{(2n-1)!} \\[9pt] &= y''. \end{align*}$

Therefore, $f(x)$ indeed is a solution of the given differential equation.

Now, to find the sum we first need to get the general form of the solutions for the differential equation

$y'' = 9(y-x) \quad \implies \quad y'' - 9y = -9x.$

First, we find the general form of the solutions of the homogeneous equation

$y'' - 9y = 0.$

In this case we have an equation of the form $y'' + ay' + by = 0$ where $a = 0$ and $b= -9$ . From this we can compute $d = a^2 - 4b = 36$ and $k = \frac{1}{2} \sqrt{d} = 3$ . Therefore, the general form of the solutions is