Solving this Recurrence Relation in terms of previous values. What will be the value of $X(n)$ and $Y(n)$ in terms of given $n,X(0),Y(0)$.
$$
X(n) = X(n-1) + Y(n-1) \\
Y(n) = 2X(n-1) + Y(n-1)
$$

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. 
