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I'm having a bit of a problem with recurring relations. I have a question that goes: "Solve the recurrence relation:" $$ a_{k+2}-4a_k=k+2 ; a_0=1 ; a_1=0 $$

And so far I've done $$ a_{k+2} -4a_k=0$$ $$r^2-4=0$$

$$ r^2 = 4$$ $$r = \pm\sqrt4$$

And this is where I lose what I'm supposed to do. My teacher gave us a sample where he did $$a_{k+2} +a_k = k$$ $$a_{k+2}+a_k=0$$ $$r^2+1=0$$ $$r^2=-1$$ $$ r=+-\sqrt1 = i=+-i =>u=0;v=1 ->|r|=1;\alpha=\frac{\pi}{2} $$ (This is not the whole example of course) Where did he get the pi per two? And how I'm supposed to continue from where I am?

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    Note that it is a mistake to conclude from $r^2=4$ that $r=\sqrt 4$, because this loses one of the roots. Specifically, $\sqrt 4 = 2$, so you should have $r = \pm \sqrt 4$, i.e., $r = \pm 2$. – 2017-01-17
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    Yeah I was thinking about that but wasn't completely sure. I will edit it to the main post. – 2017-01-17
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    It's still not entirely correct: $\sqrt 4$ is not the same as $\pm 2$! The square root symbol refers to the *principal square root*, that is, the positive root, so $\sqrt 4$ just means $2$. To correct this, either remove the line $r = \sqrt 4$, or change it to $r = \pm \sqrt 4$. – 2017-01-17

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For the first part you are solving the homogeneous part of the recurrence $a_{k+2}-4a_k=0$ and you find the solutions to be $c_12^k$ and $c_22^{-k}$. You can add either of these to any solution to get a new solution. You now need to find one solution to the whole equation. I don't understand what you are doing there. I would assume a linear function $a_k=dk+e$. Now substitute that in, finding $$d(k+2)+e-4(dk+e)=k+2\\-3dk+2d-3e=k+2\\d=-\frac 13\\e=-\frac 89$$ where the last two lines come from the second by insisting that it be true for all $k$. The full solution is then $a_k=c_12^k+c_22^{-k}-\frac k3 -\frac 89$ and you evaluate $c_1,c_2$ from the initial conditions you are given.