Solve the Relation $T(n)=T(n/4)+T(3n/4)+n$ 
Solve the recurrence relation: $T(n)=T(n/4)+T(3n/4)+n$. Also, specify an asymptotic bound.

Clearly $T(n)\in \Omega(n)$ because of the constant factor.  The recursive nature hints at a possibly logarithmic runtime (because $T(n) = T(n/2) + 1$ is logarithmic, something similar may occur for the problem here).  However, I'm not sure how to proceed from here.
Even though the recurrence does not specify an initial value (i.e. $T(0)$), if I set $T(0) = 1$ some resulting values are:
  0     1
100   831
200  1939
300  3060
400  4291
500  5577
600  6926
700  8257
800  9665
900 10933

The question:  Is there a technique that I can use to solve the recurrence in terms of $n$ and $T(0)$?  If that proves infeasible, is there a way to determine the asymptotic behavior of the recurrence?

$T(n)=T\left(\dfrac{n}{4}\right)+T\left(\dfrac{3n}{4}\right)+n$
$T(n)-T\left(\dfrac{n}{4}\right)-T\left(\dfrac{3n}{4}\right)=n$
For the particular solution part, getting the close-form solution is not a great problem.
Let $T_p(n)=An\ln n$ ,
Then $An\ln n-\dfrac{An}{4}\ln\dfrac{n}{4}-\dfrac{3An}{4}\ln\dfrac{3n}{4}\equiv n$
$An\ln n-\dfrac{An}{4}(\ln n-\ln4)-\dfrac{3An}{4}(\ln n+\ln3-\ln4)\equiv n$
$\dfrac{(4\ln4-3\ln3)An}{4}\equiv n$
$\therefore\dfrac{(4\ln4-3\ln3)A}{4}=1$
$A=\dfrac{4}{4\ln4-3\ln3}$
$\therefore T_p(n)=\dfrac{4n\ln n}{4\ln4-3\ln3}$
But getting the complementary solution part is not optimistic.
Since we should handle the equation $T_c(n)-T_c\left(\dfrac{n}{4}\right)-T_c\left(\dfrac{3n}{4}\right)=0$ .
