<s>[INST] what to do next recurrence relation when solving exponential function? find gernal solution of :$a_n = 5a_{n– 1} – 6a_{n –2} + 7^n$
Homogeneous solution:
$$a_n -5a_{n– 1} + 6a_{n –2} = 7^n$$
put $a_n=b^n$:
$$b^n -5b^{n– 1} + 6b^{n –2} =0
\\b^{n-2} (b^2-5b^{} + 6b) =0
\\b^2-5b^{} + 6b =0
\\(b-2)(b-3)=0\\
b=2,3$$
$$a^h_{(n)} = C_1 3^n+ C_2 2^n$$
Particular solution:
Since RHS is exponent so $a^p_{(n)} = da^n$
put $a^p_{(n)}$ in $a_n -5a_{n– 1} + 6a_{n –2} = 7^n$
$$da^n -5da^{n– 1} + 6da^{n –2} =7^n$$
 [/INST] Your computation of the general solution of the homogeneous equation $a_{n}=5a_{n-1}-6a_{n-2}$ is correct. Note that the homogeneous equation does not have $7^n$ on the right. 
We want to find a particular solution of the full equation 
$$a_{n}=5a_{n-1}-6a_{n-2}+7^n.\tag{1}$$ 
We look for a solution of the shape $a_n=(k)7^n$, where $k$ is a constant. Substituting in (1), we get
$$(k)7^n =(5k)7^{n-1}-(6k)7^{n-2}+7^n.$$
Dividing both sides by $7^{n-2}$, we get
$$49k=35k-6k+49.$$
Solve for $k$. We get $k=\frac{49}{20}$.
Thus the general solution of (1) is
$$C_1\cdot 3^n+C_2\cdot 2^n +\frac{1}{20}7^{n+2}.$$
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