Clarification on Solving Recurrence Relations (Linear)
The first step is to solve the easier recurrence
$$
T(n) = T(n-4) + T(n-10).
$$
This is explained in our reference question. The solution to your recurrence will likely be larger by a factor of $\log n$, roughly speaking. 

Indeed, you can consider a recursion tree in which the root is $n$, and a node $m$ has children $m-4$ and $m-10$ (you stop the recursion at some constant $m$). The solution to the recurrence $T(n) = T(n-4) + T(n-10)$ is proportional to the number of leaves, which we expect to be proportional to the number of nodes. For all $m \geq \sqrt{n}$ (say), $\log n = \Theta(\log m)$, which leads us to expect that the solution of your non-homogeneous recurrence is larger by a factor of $\log n$ than the solution of the corresponding homogeneous recurrence $T(n) = T(n-4) + T(n-10)$.

The preceding paragraph contains some intuition which will need to be proven. But whether or not the $\log n$ factor appears in the final formula makes little difference, since the growth rate of the recurrence is exponential.