Consider the following recurrence relation:
$$T(n)=c\cdot + 2\cdot T(n/2)$$
This is the recurrence relation for the Merge-Sort algorithm.
How can one deduce from this equation the time complexity of the algorithm which I know is: $$T(n) = O(n\cdot \log n)$$.
The "change of variable" $U(k)=2^{-k}T(2^k)$ yields the recursion $$ U(k)=U(k-1)+\frac{c}{2^k}, $$ hence, for every $k$, $$ U(k)=U(0)+c\sum_{i=1}^{k}\frac1{2^i}, $$ in particular, $$ U(0)\leqslant U(k)\leqslant U(0)+c\sum_{i=1}^{\infty}\frac1{2^i}=U(0)+c, $$ that is, $$ 2^k\cdot T(1)\leqslant T(2^k)\leqslant2^k\cdot(T(1)+c), $$ which is usually translated as $$ T(n)=\Theta(n). $$ This analysis addresses the recursion as written in the question, namely, $$ T(n)=c+2T(n/2). $$ However, this is not the recursion describing the merge-sort algorithm, which reads $$ T(n)=c\cdot n+2T(n/2). $$ The same change of variable as above reads $$ U(k)=U(k-1)+c, $$ hence, for every $k$, $$ U(k)=U(0)+c\cdot k, $$ that is, $$ T(2^k)=2^k\cdot T(1)+c\cdot k\cdot2^k=\Theta(k\cdot2^k), $$ which is usually translated as $$ T(n)=\Theta(n\cdot\log n). $$