Find the closed form of the following recurrence relation:
$$T(n)=\begin{cases}c&\text{ if }n=1\\2T\left(\frac n2\right)+\log n&\text{ otherwise.}\end{cases}$$
Can someone give me a step by step method on how to find the closed form of this specific relation? I have tried following other posts but can't see how they relate to this equation.
$$ T(2^{\log_2 n})=2T(2^{\log_2\frac n2})+\ln n $$
now calling $T'(u) = T(2^u)$ with $u = \log_2 n$ we have the linear recurrence
$$ T'(u)=2T'(u-1) + u\ln 2 $$
which can be solved as
$$ T'(u) = T'_h(u)+T'_p(u) $$
with
$$ T'_h(u)-2T'(u-1)=0\\ T'_p(u)-2T'_p(u-1) = u\ln 2 $$
The homogeneous solution is direct
$$ T'_h(u) = C 2^{u-1} $$
now making $T'_p(u) = C(u)2^{u-1}$ and substituting into the particular we have a recurrence for $C(n)$
$$ C(n)-C(n-1) = 2^{-u}u\ln 4 $$
and solving for $C(n)$ we get
$$ C(n) = \left(2-2^{-u}(u+2)\right)\ln 4 $$
and finally
$$ T'(u) = C 2^{u-1} + \left(2-2^{-u}(u+2)\right)\ln 4 2^{u-1} = C 2^{u-1}+\left(2^u-1\right) \ln 4-u \ln 2 $$
hence
$$ T(n) = \frac{1}{2} n (C+\ln 16)-\ln (4 n) $$
imposing now the initial conditions we have $C = \ln 16-2\ln 4 = 0$ and finally
$$ T(n) = n\ln 4-\ln (4n) $$