For the given differential equation below, find the complementary function, particular integral of the function and general solution of the motion.

The $(n+1)$-th term is $u_{n+1} = (-1)^{n+1} \frac{1}{(n+1)+1}(x-3)^{n+1} = (-1)^{n+1} \frac{1}{n+2}(x-3)^{n+1}$.

$$\left| \frac{u_{n+1}}{u_n} \right| = \left| \frac{(-1)^{n+1} \frac{1}{n+2}(x-3)^{n+1}}{(-1)^n \frac{1}{n+1}(x-3)^n} \right|$$ $$= \left| (-1) \cdot \frac{n+1}{n+2} \cdot (x-3) \right|$$ $$= \frac{n+1}{n+2} |x-3|$$

$$\lim_{n \to \infty} \left( \frac{n+1}{n+2} |x-3| \right) = |x-3| \lim_{n \to \infty} \left( \frac{1 + \frac{1}{n}}{1 + \frac{2}{n}} \right)$$ $$= |x-3| \cdot \frac{1+0}{1+0} = |x-3|$$

For the series to converge, the limit must be less than 1:

$$|x-3| < 1$$

This inequality can be written as:

$$-1 < x-3 < 1$$

Add 3 to all parts of the inequality:

$$-1+3 < x < 1+3$$ $$2 < x < 4$$

At $x=2$:
Substitute $x=2$ into the original series:

$$\sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}(2-3)^n = \sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}(-1)^n$$ $$= \sum_{n=0}^{\infty} \frac{1}{n+1}(-1)^{2n} = \sum_{n=0}^{\infty} \frac{1}{n+1}$$

This is the harmonic series (starting from $n=0$, so it's $1 + \frac{1}{2} + \frac{1}{3} + \dots$), which is known to diverge.

At $x=4$:
Substitute $x=4$ into the original series:

$$\sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}(4-3)^n = \sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}(1)^n$$ $$= \sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}$$

This is the alternating harmonic series. We can apply the Alternating Series Test:
Let $b_n = \frac{1}{n+1}$.

1. $b_n > 0$ for all $n \ge 0$.
2. $b_n$ is a decreasing sequence: $\frac{1}{n+1} > \frac{1}{n+2}$.
3. $\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{n+1} = 0$. Since all conditions are met, the series converges at $x=4$.

Combining the results from the ratio test and the endpoint checks, the series converges for $x$ values such that $2 < x \le 4$.

The interval of convergence is $\boxed{(2, 4]}$.