How to solve this simple difference equation?

Question

$ y(k+1)-2y(k)=k2^k$

I know that theres a formula for situations where the right hand side is a geometric series, but that doesn't seem to be the case.

Answer 1

$y(k+1)-2y(k)=k2^k\\2y(k+1)-4y(k)=k2^{k+1} \\y(k+2)-2y(k+1)=(k+1)2^{(k+1)}$

We subtract both sides of the last two equations:

$y(k+2)-4y(k+1)+4y(k)=2^{k+1}$

Now you can continue based on what you know.

Answer 2

also iterating the equation should work

$$y_{k+1}=2\,y_k+k\,2^{k}$$

$$y_{k+1}=2\,(2\,y_{k-1}+(k-1)\,2^{k-1})+k\,2^{k} = 4\,y_{k-1}+2\,(k-1)\,2^{k-1}+k\,2^k $$

$$y_{k+1}=4\,(2\,y_{k-2}+(k-2)\,2^{k-2})+2\,(k-1)\,2^{k-1}+k\,2^k=...$$ $$...=8\,y_{k-2}+4\,(k-2)\,2^{k-2}+2\,(k-1)\,2^{k-1}+k\,2^{k} $$

the guess (prove by induction) is $$ y_{k+1} = 2^{k+1}y_0+\sum_{p=0}^{k-1}2^{p}\,(k-p)\,2^{k-p} $$

Answer 3

What I obtained, with an approach very similar to DP1981's one, is
y[k] = 2^(k-1) y0 + k (k-1) 2^(k-2) which can be further simplified factoring 2^(k-2).

Answer 4

It often helps to get several successive terms, to get an idea what is going on.

Say that $y(0)=c.$ Then $y(1)=0+2y(0)=2c,$ so $y(2)=2+2y(1)=2+4c,$ so $y(3)=8+2y(2)=12+8c,$ so $y(4)=24+2y(3)=48+16c,$ so $y(5)=64+2y(4)=160+32c.$

Let's look at factors and see if anything becomes clear. Note that there is no term without a $c$ in $y(0)$ or $y(1),$ which suggests that the term without $c$ will be zero when $k=0$ or $k=1.$ This suggests that said term will have $k$ and $k-1$ as factors. Let's see if we can use that idea. Indeed, we see that $$y(0)=c+0(0-1)\\y(1)=2c+1(1-1)\\y(2)=4c+2(2-1)\\y(3)=8c+2\cdot 3(3-1)\\y(4)=16c+4\cdot 4(4-1)\\y(5)=32c+8\cdot5(5-1),$$ so we seem to be on the right track. It also appears that both terms will have increasing powers of $2$ as factors. Indeed, $$y(0)=2^0c+2^{0-2}\cdot0(0-1)\\y(1)=2^1c+2^{1-2}\cdot1(1-1)\\y(2)=2^2c+2^{2-2}\cdot2(2-1)\\y(3)=2^3c+2^{3-2}\cdot 3(3-1)\\y(4)=2^4c+2^{4-2}\cdot 4(4-1)\\y(5)=2^5c+2^{5-2}\cdot5(5-1).$$

A conjecture, then: for all $k,$ we have $$y(k)=2^k\left(c+\frac{k(k-1)}4\right),\tag{$\star$}$$ where $c=y(0).$ The base case is easy. I leave the induction steps to you.

Note that you may wish to prove that it holds for all integers $k,$ not just the non-negative integers, in which case you need to prove by induction going up and down. That is, you must show the base case, and that if $$y(k)=2^k\left(c+\frac{k(k-1)}4\right),$$ then $$y(k+1)=2^{k+1}\left(c+\frac{(k+1)k}4\right)$$ and $$y(k-1)=2^{k-1}\left(c+\frac{(k-1)(k-2)}4\right).$$