Answer

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diiimah:

1)) T(n)=7T(n-4)+n

2)) T(n)=2T(n/3)+n log n

There is no question here.
Just two equations

Complete Solution ???

diiimah:

1)) T(n)=7T(n-4)+n

2)) T(n)=2T(n/3)+n log n

I asked some of my friends to take a look at this question because i really wasn't sure.
This was Bob's reply.
I haven't looked at it yet and I am too tired to do so now, but i will post it so that you can take a look.

140226 difference equation.JPG

I did not fully understand Bob's answer so i asked for some clarification. Here is his answer.

The method I used is known as a 'trial solution method', and it is just that. You try something, if it works fine, if not try something else. With practice it's usually possible to guess what the solution looks like.
Have you done any work on second order ODE's with constant coefficients ? That's where your most likely to come across it. It's a commonly taught method for finding the Particular Integral. The general solution of such an equation will consist of the Particular Integral plus a Complementary Function (which is the solution of the corresponding homogeneous equation and which will contain two arbitrary constants). It's pretty much the same for the difference equation we are discussing, its general solution consists of two parts, a Particular Solution (which is what I've found), plus the solution of the corresponding homogeneous equation, (which I haven't).
140226 Bob's answer to 'answer'.JPG

I had to really think about the last step but this is what Bob did
-6An + 28A - 6B = n
therefore
-6A = 1 and 28A-6B = 0
A = -1/6
and with substitution
28*(-1/6) = 6B
B = -28/36 = -7/9

therefore
T _n = (-1/6)n - 7/9

I have not looked at the 2nd equation yet. If you are still around you can try it by yourself and see how you go.