Solving this Recurrence Relation in terms of previous values. What will be the value of $X(n)$ and $Y(n)$ in terms of given $n,X(0),Y(0)$.
$$
X(n) = X(n-1) + Y(n-1) \\
Y(n) = 2X(n-1) + Y(n-1)
$$
 A: Observe that:
$$
\begin{matrix}
x_n - x_{n-1} = y_{n-1} \\
y_n - y_{n-1} = 2x_{n-1}
\end{matrix}
$$
by subtracting the first relation for $y_n$ and $y_{n-1}$:
$$
x_{n+1} - x_n - x_n + x_{n-1} = y_n - y_{n-1} = 2x_{n-1}
$$
Thus:
$$
x_{n+1} - 2x_n - x_{n-1} = 0
$$
which is a linear one variable recurrence.
A: This relation can be written as
$$\begin{pmatrix}X(n) \\ Y(n)\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix} \begin{pmatrix}X(n - 1) \\ Y(n - 1)\end{pmatrix}. $$
If you define $A = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}$, this yields 
$$\begin{pmatrix}X(n) \\ Y(n)\end{pmatrix} = A^n \begin{pmatrix}X(0) \\ Y(0)\end{pmatrix}. $$
Now you only need to calculate the powers of $A$, this works straightforward via diagonalzation.
A: This linear recurrence may be represented matricially:
$$\begin{bmatrix}X_n\\Y_n\end{bmatrix} = \begin{bmatrix}1&1\\2&1\end{bmatrix} \begin{bmatrix}X_{n-1}\\Y_{n-1}\end{bmatrix},$$
hence
$$\begin{bmatrix}X_n\\Y_n\end{bmatrix} = \begin{bmatrix}1&1\\2&1\end{bmatrix}^n\begin{bmatrix}X_0\\Y_0\end{bmatrix}.$$
Set $\;A= \begin{bmatrix}1&1\\2&1\end{bmatrix}$. To compute $A^n$, you have to find the Jordan normal form of $A$: $J=D+N$, where $D$  is a diagonal matrix, with the eigenvalues of $A$ on the diagonal, and $N$ is a nilpotent matrix which commutes with $D$. It is then easy to ompute powers of $J$, as we can use the binomial formula, and $N$ has a finite number of non-zero powers. The last step consists in going back from $J^n$ to $A^n$ by the change of basis formula: if $J=P^{-1}AP$,we also have $J^n=P^{-1}A^nP$, whence $\;A^n=PJ^nP^{-1}$.
In the present case it is simple: $A$ is diagonalisable:  its eigenvalues are $1\pm\sqrt2$, the corresponding eigenvectors are $\;\begin{bmatrix}1\\\mp\sqrt2\end{bmatrix}$, so that the change of basis matrix is
$$P=\begin{bmatrix}1&1\\-\sqrt2&\sqrt2\end{bmatrix},\quad D=\begin{bmatrix}1+\sqrt2 & 0\\-0&1-\sqrt2\end{bmatrix},\quad P^{-1}=\frac14\begin{bmatrix}2&-\sqrt2\\2&\sqrt2\end{bmatrix}, $$
whence, if I'm not mistaken:
$$A^n=\frac14\begin{bmatrix}2\bigl((1+\sqrt2)^n+((1-\sqrt2)^n\bigr) &\sqrt2\bigl((1-\sqrt2)^n-((1+\sqrt2)^n\bigr)\\2\sqrt2\bigl((1-\sqrt2)^n-((1+\sqrt2)^n\bigr)&2\bigl((1+\sqrt2)^n+((1-\sqrt2)^n\bigr)\end{bmatrix}.$$
A: One instructive solution is to use generating functions. Define:
$\begin{align}
f(z)
  &= \sum_{n \ge 0} X(n) z^n  \\
g(z)
  &= \sum_{n \ge 0} Y(n) z^n
\end{align}$
Shift indices to get:
$\begin{align}
X(n + 1)
  &= X(n) + Y(n) \\
Y(n + 1)
  &= 2 X(n) + Y(n)
\end{align}$
Multiply the recurrences by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$\begin{align}
\frac{f(z) - X(0)}{z}
  &= f(z) + g(z) \\
\frac{g(z) - Y(0)}{z}
  &= 2 f(z) + g(z)
\end{align}$
This gives:
$\begin{align}
f(z)
  &= \frac{X(0) + (Y(0) - X(0)) z}{1 - 2 z - z^2} \\
g(z)
  &= \frac{Y(0) - 2 (Y(0) - 2 X(0)) z}{1 - 2 z - z^2}
\end{align}$
This can be written as partial fractions, the denominator factors as $(1 - (1 - \sqrt{2}) z) (1 - (1 + \sqrt{2}) z)$. The result is quite ugly, but you'll get terms of the form:
$$
\frac{1}{1 - \alpha z}
  = \sum_{n \ge 0} \alpha^n z^n
$$
so that the coefficients of $f$ and $g$ are immediate. 
