Solving recursive equation

Question

$$T(n) = 13T(n-1)-42T(n-2)+9+15n\cdot8^n + 5n^2 \cdot 19^n + 16 n^3 + 8^{n+1} + 18n^4$$

I am really stuck here guys. I don't even have a value for $T(0)$. I have tried simplyfing it, but there was no point in doing that. Any help ?

Answer 1

The general solution to $T(n)-13T(n-1)+42T(n-2)=0$ is $T(n)=A 6^n+B 7^n$. In order to obtain the particular solution to \begin{eqnarray*} T(n)-13T(n-1)+42T(n-2)= \color{green}{9+16n^3+18n^4}+\color{blue}{(15n+8)8^n}+\color{red}{5n^2 19^n} \end{eqnarray*} Use \begin{eqnarray*} T(n)= \color{green}{\alpha+\beta n +\gamma n^2 +\delta n^3 + \epsilon n^4}+\color{blue}{(\zeta + \eta n)8^n}+\color{red}{(\lambda +\mu n +\nu n^2) 19^n} \end{eqnarray*} The values of $\alpha,\beta , \gamma ,\delta , \epsilon ,\zeta , \eta ,\lambda ,\mu$ & $\nu$ can be evaluated using linear algebra ... but it will be quite a grind.

EDIT : look at this ... http://people.uncw.edu/tompkinsj/133/recursion/homogeneous.htm It is standard to assume an exponential solution & then the solution is a linear combination of these.