<s>[INST] Solve the recurrence relation: $2a_n = 7a_{n-1} - 3a_{n-2}; a_0 = a_1 = 1$ $2a_n = 7a_{n-1} - 3a_{n-2}\\
a_0 = a_1 = 1$
My attempt:
$2t^2 - 7t + 3 = 0\\
t = -\frac{1}{2}, -3\\
\\
U_n = b(-\frac{1}{2})^n + d(-3)^n\\
b+d = 1 = -\frac{1}{2}b-3d\\
b = \frac{8}{5}, d = -\frac{3}{5}\\
a_n = \frac{8}{5}(-\frac{1}{2})^n - \frac{3}{5}(-3)^n\\
a_2 = -5 \neq 2 =\frac{1}{2}(7-3)$
Where did I go wrong?
Update: changing the t values to positive, new solution:
$a_n = \frac{4}{7}(\frac{1}{2})^n+\frac{3}{7}(3)^n\\
a_2 = 4$
But following the initial equation given, $2a_2 = 4$, so shouldn't $a_2$ be 2?
$a_2$ according to Wolfram Alpha is indeed 2.
 [/INST] The answer given by Wolfram Alpha does match your textbook. (Try multiplying the $2^{-n}$ through.)
Your mistake was when you changed the roots of the characteristic to positive; if 
$$a_n = A\left(\frac{1}{2}\right)^n + B\left(3\right)^n$$
then the correct method of solving for $A$ and $B$ would give $A=\frac{4}{5}$ and $B=\frac 1 5$.
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