This is clearest when viewed from the point of view of linear algebra.
Let $V=\Bbb C^{\Bbb N}$, the vector space of infinite sequences of complex numbers with componentwise addition and scalar multiplication. Consider the $r$-th order homogeneous recurrence
$$x_n=a_1x_{n-1}+a_2x_{n-2}+\ldots+a_rx_{n-r}\tag{1}$$
with constant coefficients. A sequence $\alpha=\langle\alpha_n:n\in\Bbb N\rangle\in V$ satisfies $(1)$ if and only if
$$\alpha_n=a_1\alpha_{n-1}+a_2\alpha_{n-2}+\ldots+a_r\alpha_{n-r}$$
for all $n\ge r$. If $\alpha\in V$ satisfies $(1)$, we call $\alpha$ a solution to $(1)$. The set of solutions to $(1)$ is a subspace of $V$.
Proposition. Let $H=\{\alpha\in V:\alpha\text{ satisfies }(1)\}$; then $H$ is a subspace of $V$.
Proof. The proof is entirely straightforward. If $\alpha,\beta\in H$ and $b,c\in\Bbb C$, let $\gamma=b\alpha+c\beta$. Then for each $n\ge r$ we have
$$\begin{align*}
\gamma_n&=b\alpha_n+c\beta_n\\
&=b(a_1\alpha_{n-1}+\ldots+a_r\alpha_{n-r})+c(a_1\beta_{n-1}+\ldots+a_r\beta_{n-r})\\
&=a_1(b\alpha_{n-1}+c\beta_{n-1})+\ldots+a_r(b\alpha_{n-r}+c\beta_{n-r})\\
&=a_1\gamma_{n-1}+\ldots+a_r\gamma_{n-r}\;,
\end{align*}$$
and hence $\gamma\in H$. $\dashv$
It turns out that $\dim H=r$. (This isn’t very hard to show, but doing so would take me further afield than is really necessary.) $H$ therefore has a basis $B=\{\beta^1,\dots,\beta^r\}$ (where the superscripts are just indices, not exponents). Thus, each $\alpha\in H$ can be written uniquely in the form
$$\alpha=b_1\beta^1+b_2\beta^2+\ldots+b_r\beta^r$$
for some $b_1,b_2,\dots,b_r\in\Bbb C$. The technique that you’re learning with the characteristic equation is a way to find such a basis.
In your particular problem $r=2$, and you’ve found that the sequences $\alpha$ and $\beta$ given by $\alpha_n=2^n$ and $\beta_n=(-3)^n$ are in $H$. It’s also clear that they are linearly independent: neither is a scalar multiple of the other. Since in this case $\dim H=2$, $\{\alpha,\beta\}$ must be a basis for $H$, and therefore every solution $\gamma$ to your recurrence must have the form $\gamma=a\alpha+b\beta$, i.e.,
$$\gamma_n=a\alpha_n+b\beta_n=a2^n+b(-3)^n\quad\text{ for all }n\in\Bbb N\;,\tag{2}$$
for some constants $a$ and $b$. The initial conditions are then used to pin down the specific values of $a$ and $b$. Since every linear combination of $\alpha$ and $\beta$ is a solution of the recurrence, not just the simple multiples $a\alpha$ and $b\beta$ of the solutions $\alpha$ and $\beta$, there’s no guarantee that your initial conditions will be satisfied by one of those simple multiples.
A simple analogy: every vector in $\Bbb R^2$ is a linear combination of the basis vectors $\langle 1,0\rangle$ and $\langle 1,1\rangle$, but if you need one whose first component is $0$, you can’t get it as a scalar multiple of $\langle 1,0\rangle$ or of $\langle 1,1\rangle$: you have to use a linear combination of both basis vectors.
