So I have read the posts on this site involving recurrence relations, however this problem is a little different, because of the constant a involved with the recursive portion. I'm trying to solve this recurrence relation by expanding it out and here is what I have so far:
I'm not sure where to go from here and how to find the asmpytotic bounds.
The first four lines of your argument were correct; all you missed was the proper generalization. In general, we see that $$ T(n) = c+c\alpha+c\alpha^2+\dotsb+c\alpha^{j-1}+\alpha^j T(n-2j) $$ Now we want to drive $T(n-2j)$ down to a value we know, namely $T(1)=c$. To do this we'll need $n-2j=1$ and so we set $j=(n-1)/2$, giving us $$\begin{align} T(n) &= c+c\alpha+c\alpha^2+\dotsb+c\alpha^{j-1}+\alpha^j T(n-2j)\\ &= c+c\alpha+c\alpha^2+\dotsb+c\alpha^{(n-1)/2-1}+\alpha^{(n-1)/2} c\\ &= c(1+\alpha+\alpha^2+\dotsb+\alpha^{(n-1)/2}) \end{align}$$ and this is just a geometric series, so I'll leave the last steps to you.
The trick here is to solve a modified recurrence in which there is no constant. Let $S(n) = T(n) - C/(1-\alpha)$. Then $$ S(n) = T(n) - C/(1-\alpha) = C + \alpha T(n-2) - C/(1-\alpha) = \\ C + \alpha (S(n-2) + C/(1-\alpha)) - C/(1-\alpha) = \alpha S(n-2). $$ (I found the constant $C/(1-\alpha)$ by solving linear equations.)
The solution to $S(n)$ with initial conditions $S(0),S(1)$ is easily found to be $$ S(2n) = \alpha^n S(0), \qquad S(2n+1) = \alpha^n S(1). $$ It follows that $$ T(2n) = \alpha^n (T(0) - C/(1-\alpha)) + C/(1-\alpha), \\ T(2n+1) = \alpha^n (T(1) - C/(1-\alpha)) + C/(1-\alpha). $$
