Recurrence relation (linear, second-order, constant coefficients)
The general approach to solving recurrence relations is the following: given a recurrence relation $$a_n+\alpha_1a_{n-1}+...+\alpha_ra_{n-r}=\beta(n) \; .$$
I) First you solve the homogeneous part $a_n^{(h)}+\alpha_1a_{n-1}^{(h)}+...+\alpha_ra_{n-r}^{(h)}=0$:
$\quad$ a) By solving the characteristic equation: $x^r+\alpha_1x^{r-1}+...+\alpha_r=0$ you get $x_1,...,x_r$ the roots of the equation.
$\quad$ b) If all $x_1,...,x_r$ are different then the $a_n^{(h)}=A_1x_1^n+...+A_rx_r^n$, where $A_1,...,A_r$ are the coefficients.
$\quad$ c) If you have a root $x_j$ is of multiplicity $k$ then you replace the term $A_jx_j^n$ with $(B_1n^{k-1}+...+B_k)x_j^n$
II) Now you find a particular solution to the equation $a_n^{(p)}+\alpha_1a_{n-1}^{(p)}+...+\alpha_ra_{n-r}^{(p)}=\beta(n)$. Each $\beta(n)$ needs a different approch to guess $a_n^{(p)}$. For example, if $\beta(n)=k^nf(n)$, where $k$ is a number and $f(n)$ is a polynomial in $n$, then $a_n^{(p)}=k^nn^sg(n)$ where $k$ is the same $k$, $s$ is the multiplicity of $k$ as a root of the characteristic equation in the homogeneous part and $g(n)$ is a polynomial of the same degree as $f(n)$.
Finally, $a_n=a_n^{(h)}+a_n^{(p)}$
The general second order homogeneous linear recurrence/difference equation with constant coefficients
$$x_{n}+c_{1}x_{n-1}+c_{2}x_{n-2}=0\qquad (1)$$
has two fundamental solutions $(\lambda _{1}^{n})_{n\geq 0}$ and $(\lambda _{2}^{n})_{n\geq 0}$, where $\lambda _{1},\lambda _{2}$ are the two zeroes of the characteristic polynomial
$$\lambda ^{2}+c_{1}\lambda +c_{2}\qquad (2)$$
Let's confirm.
$$\begin{eqnarray*} \lambda _{1}^{n}+c_{1}\lambda _{1}^{n-1}+c_{2}\lambda _{1}^{n-2} &=&\lambda _{1}^{n-2}\left( \lambda _{1}^{2}+c_{1}\lambda _{1}+c_{2}\right) \equiv 0 \\ && \\ \lambda _{2}^{n}+c_{1}\lambda _{2}^{n-1}+c_{2}\lambda _{2}^{n-2} &=&\lambda _{2}^{n-2}\left( \lambda _{2}^{2}+c_{2}\lambda _{2}+c_{2}\right) \equiv 0 \end{eqnarray*}$$
If $\lambda _{1}\neq \lambda _{2}$ the general solution of $(1)$ is a linear combination of $\lambda _{1}^{n}$ and $\lambda _{2}^{n}$
$$x_{n}=A\lambda _{1}^{n}+B\lambda _{2}^{n}\qquad (3)$$
as you can confirm substituting $(3)$ in $(1)$. Let's apply this result to your second difference equation $a_{n}-a_{n-1}-2a_{n-2}=0$. The characteristic polynomial $\lambda ^{2}-\lambda -2$ has the zeroes $\lambda _{1}=-1,\lambda _{2}=2$.
Thus $(3)$ becomes
$$a_{n}=A(-1)^{n}+B2^{n}$$
The constants $A$ and $B$ are determined by using the initial conditions $a_{0}=2$, $a_{1}=4$.
Added: Determination of $A$ and $B$:
$$a_{0}=A(-1)^{0}+B2^{0}=A+B=2\Leftrightarrow B=2-A$$
Hence
$$a_{n}=A(-1)^{n}+\left( 2-A\right) 2^{n}$$
and
$$a_{1}=A(-1)^{1}+\left( 2-A\right) 2^{1}=-A+4-2A=-3A+4=4\Leftrightarrow A=0.$$
Since $A=0$ and $B=2-A=2$ the solution is
$$a_{n}=2\cdot 2^{n}=2^{n+1}\qquad n\geq 2$$
As for your first equation it is not homogenous because the RHS is not zero. For solving it, please see the explanation by Dennis Gulko in his answer.
Consider the recurrence relation for $n\ge 1$ given by \begin{equation} \tag{1} \label{1} A_{n+1}=\alpha A_n+\beta A_{n-1}+f_n, \end{equation} where $\alpha$ and $\beta$ are constants (not dependent on $n$), and $f_n$ is a given function of $n$. Eqn.~\eqref{1} is to be solved either for specified $A_0$, $A_1$, or two constraints of the form \begin{align}\tag{2}\label{2} \sum_{n=1}^{\infty} c_nA_n&=h_n, \\ \sum_{n=1}^{\infty} d_nA_n&=g_n. \end{align}
We can write Eqn.~\eqref{1} as \begin{equation} \tag{3} \label{3} \begin{bmatrix} A_{n+1} \\ A_n \end{bmatrix}=T\begin{bmatrix} A_n \\ A_{n-1} \end{bmatrix}+\begin{bmatrix} f_n \\ 0 \end{bmatrix}, \end{equation} where \begin{equation*} T=\begin{bmatrix} \alpha & \beta \\ 1 & 0 \end{bmatrix}. \end{equation*}
First consider the case where the eigenvalues of $T$ are distinct, and given by \begin{align} \tag{4} \label{4} \lambda_1&=\frac{\alpha-\sqrt{\alpha^2+4\beta}}{2}, \\ \lambda_2&=\frac{\alpha+\sqrt{\alpha^2+4\beta}}{2}. \end{align} We deal with the case of repeated eigenvalues where $\beta=-\alpha^2/4$, and $\lambda_1=\lambda_2=\alpha/2$ later by taking appropriate limits. Since the eigenvalues are distinct, $T$ is diagonalizable, and can be written as \begin{equation} \tag{5} \label{5} T=\lambda_1 P_1+\lambda_2 P_2, \end{equation} where $P_1$ and $P_2$ are projections given by \begin{align*} P_1&=\frac{T-\lambda_2 I}{\lambda_1-\lambda_2}, \\ P_2&=\frac{T-\lambda_1 I}{\lambda_2-\lambda_1}. \end{align*} Thus, we have \begin{equation} \tag{6} \label{6} T^n=\lambda_1^n P_1+\lambda_2^n P_2. \end{equation}
We now have \begin{align} \tag{7} \label{7} \begin{bmatrix} A_{n+1} \\ A_n \end{bmatrix}&=T^n\begin{bmatrix} A_1 \\ A_0 \end{bmatrix}+\sum_{k=1}^{n-1}T^{n-k}\begin{bmatrix} f_k \\ 0 \end{bmatrix} +\begin{bmatrix} f_n \\ 0 \end{bmatrix} \notag \\ &=(\lambda_1^n P_1+\lambda_2^n P_2)\begin{bmatrix} A_1 \\ A_0 \end{bmatrix}+\sum_{k=1}^{n-1}(\lambda_1^{n-k} P_1+\lambda_2^{n-k} P_2) \begin{bmatrix} f_k \\ 0 \end{bmatrix}+\begin{bmatrix} f_n \\ 0 \end{bmatrix}, \end{align} which leads to the explicit formula \begin{equation} \tag{8} \label{8} A_n=\frac{1}{\lambda_2-\lambda_1}\left[(\lambda_2^n-\lambda_1^n)A_1-\lambda_1\lambda_2\left(\lambda_2^{n-1}-\lambda_1^{n-1}\right)A_0 +\sum_{k=1}^{n-1} \left(\lambda_2^{n-k}-\lambda_1^{n-k}\right)f_k\right]. \end{equation} If $f_n=f_0$ is a constant, Eqn.~\eqref{8} simplifies to \begin{equation} \tag{9} \label{9} A_n=\frac{1}{\lambda_2-\lambda_1}\left[(\lambda_2^n-\lambda_1^n)A_1-\lambda_1\lambda_2\left(\lambda_2^{n-1}-\lambda_1^{n-1}\right)A_0 +\left(\frac{\lambda_2(\lambda_2^{n-1}-1)}{\lambda_2-1}-\frac{\lambda_1(\lambda_1^{n-1}-1)}{\lambda_1-1}\right)f_0\right]. \end{equation} If $\lambda_1=1$, say, then by taking the limit in the above equation, we get \begin{equation} \tag{10} \label{10} A_n=\frac{1}{\lambda_2-1}\left[(\lambda_2^n-1)A_1-\lambda_2\left(\lambda_2^{n-1}-1\right)A_0 +\left(\frac{\lambda_2(\lambda_2^{n-1}-1)}{\lambda_2-1}-(n-1)\right)f_0\right]. \end{equation} In case the constraints in Eqn.~\eqref{2} are imposed in place of prescribed values for $A_0$ and $A_1$, then we determine $A_0$ and $A_1$ by substituting the solution given by Eqn.~\eqref{8} into Eqns.~\eqref{2}.
The solution for the case where the eigenvalues of $T$ are repeated is obtained by taking the limit $\lambda_2\rightarrow \lambda_1$ in Eqn.~\eqref{8}, and setting $\lambda_1=\alpha/2$ as \begin{equation} \tag{11} \label{11} A_n=\left(\frac{\alpha}{2}\right)^{n-1}\left[nA_1-\frac{\alpha(n-1)A_0}{2}\right]+\sum_{k=1}^{n-1}(n-k)f_k\left(\frac{\alpha}{2}\right)^{n-k-1}. \end{equation} If $f_n=f_0$ is a constant, Eqn.~\eqref{11} simplifies to \begin{align*} A_n&=\left(\frac{\alpha}{2}\right)^{n-1}\left[nA_1-\frac{\alpha(n-1)A_0}{2}\right]+\frac{4\left[2^n+(n-1)\alpha^n-2n\alpha^{n-1}\right]f_0}{(\alpha-2)^22^n}, && (\alpha\ne 2), \\ &=nA_1-(n-1)A_0+\frac{n(n-1)f_0}{2}, && (\alpha=2). \end{align*}
As examples, consider the following:
- The Fibonacci sequence defined by $A_{n+1}=A_n+A_{n-1}$ with $A_0=0$, $A_1=1$, so that $\alpha=1$, $\beta=1$, $\lambda_1=(1-\sqrt{5})/2$, $\lambda_2=(1+\sqrt{5})/2$, $f_n=0$, and \begin{equation*} A_n=\frac{\lambda_2^n-\lambda_1^n}{\lambda_2-\lambda_1}. \end{equation*}
- The recurrence \begin{equation*} A_{n+1}=4A_n-4A_{n-1}+4^{n-1}, \end{equation*} with $A_0=0$, $A_1=1$, so that $\alpha=4$, $\beta=-4$, $\lambda_1=\lambda_2=2$, $f_n=4^{n-1}$. From Eqn.~\eqref{11}, we get \begin{equation*} A_n=n2^{n-1}+2^{n-2}(2^n-n-1). \end{equation*}
- The solution to your Q1 is given by Eqn.~\eqref{10} with $\lambda_2=3$ and $f_0=6$.
The above procedure can obviously be generalized to higher-order recurrences, by considering the eigenvalues of $T$ to be distinct, and then deriving various subcases for repeated eigenvalues by taking appropriate limits as in the development above.