= Solution
The homogeneous sequences are again $(-2)^k$ and $(-3)^k$. A particular solution $D2^k$ has left-hand side $10D2^k$, so $D=1/10$. Consequently $y_k=A(-2)^k+B(-3)^k+2^k/10$. The two <initial conditions> require $A+B=9/10$ and $-2A-3B=4/5$, giving $A=7/2$ and $B=-13/5$. Thus
$$
\boxed{y_k=\frac{35(-2)^k-26(-3)^k+2^k}{10}.}
$$
The <linear recurrence relation> can be rearranged as $y_{k+1}=2^k-5y_k-6y_{k-1}$. Starting with the two integer values, <mathematical induction> proves that every $y_k$ for $k\geq0$ is an integer. For $n\geq1$, the closed form gives
$$
2^n-26(-3)^n=10y_n-35(-2)^n.
$$
The first term on the right is divisible by ten; the second is too, because $(-2)^n$ supplies a factor two and $35$ supplies a factor five. Therefore \b[$10$ divides $2^n-26(-3)^n$ for every positive integer $n$]. The positivity condition matters: at $n=0$ the expression is $-25$.
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