Finding a general solution of $A_n$ Find the general solution to
$ A_{n+1} + 4A_n = n $
I am unsure how to even start the question :S
 A: Split the problem into homogeneous and inhomogeoneous solutions $A_n^{(H)}$ and $A_n^{(I)}$, respectively.  The homogeneous solution satisfies
$$A_{n+1}^{(H)}+4 A_n^{(H)}=0$$
so that
$$A_n^{(H)} = K \cdot (-4)^n$$
where $K$ is some constant.  For the inhomogeneous solution, you guess a form that will make the left side equal to the right side; in this case, a linear form will do:
$$A_n^{(I)} = a n + b$$
where you substitute into the equation and find $a$ and $b$ by equating constant and linear terms.  The solution is then
$$A_n = A_n^{(H)} + A_n^{(I)}$$
You then find the constant $K$ by using an initial condition (which you did not specify).
A: Let $B_n = A_n + an+b$. We then have
$$B_{n+1} + 4B_n = A_{n+1} + an + a + b + 4A_n + 4an + 4b = (5a+1)n + a+5b$$
Choosing $a = -\dfrac15$ and $b = \dfrac1{25}$, i.e., setting $B_n = A_n - \dfrac{n}5 + \dfrac1{25}$, we get
$$B_{n+1} + 4 B_n = 0$$
This gives us $B_n = (-4)^n B_0$. Hence, $$A_n - \dfrac{n}5 + \dfrac1{25} = (-4)^n \left(A_0 + \dfrac1{25}\right) \implies A_n = (-4)^n \left(A_0 + \dfrac1{25}\right) + \dfrac{n}{5} - \dfrac1{25}$$
A: The answer to this kind of question comes in two parts (there is a strong analogy with linear differential equations).
Suppose $B_{n+1}+4B_{n}=0$ then you can check by linearity that $C_n=A_n+B_n$ satisfies the original equation. And you can compute $B_n$ in terms of $B_0$ without difficulty.
You are left with needing a particular solution for $A_n$ - and first off the guess for a linear expression in $n$ is a linear expression in $n$: - $A_n=pn+q$ will normally work. Sometimes, in similar problems, you will need to increase the degree so that you have terms in $n^2$. Find $p$ and $q$ and then add the sequence for $B_n$ which gives you the right initial value.
