difference equations: y(n+2) - yn = 1, if y1 = y2 = 1

[dp212121]

Could anyone please help me solve the following difference equations?

a) y(n+2)-yn=1 , if y1=y2=1

b) y(n+2)+y(n)=n, if y1=y2=0

 

[soroban]

Hello, dp212121!

Solve the following difference equation
. . \(\displaystyle a)\;\;y(n+2) \:=\:y(n) +1,\quad y(1)\,=\,y(2)\,=\,1\)

Crank out a few terms and see if we can see a pattern . . .

\(\displaystyle \begin{array}{ccccccc}y(1) &=&1 \\ y(2) &=& 1 \\ y(3) &=& y(1) + 1 &=& 1+1 & =&2 \\ y(4) &=& y(2)+1 & =&1+1 &=&2 \\ y(5) &=&y(3)+1 &=&2+1 &=& 3 \\ y(6) &=& y(4)+1 &=&2+1 &=&3 \\ y(7) &=& y(5)+1 &=&3+1 &=&4 \\ y(8) &=& y(6)+1 &=& 3+1 &=&4 \end{array}\)


\(\displaystyle \text{The sequence is: }\:1,1,2,2,3,3,4,4,5,5,\hdots\)



One description of this function is:

. . \(\displaystyle y(n) \;=\;\left\{\begin {array}{ccc} \frac{n+1}{2} & \text{if }n\text{ is odd} \\ \\ \frac{n}{2} & \text{if }n\text{ is even} \end{array}\)


\(\displaystyle \text{Another is: }\;y(n) \;=\;\frac{1}{4}\bigg[(2n+1) - (-1)^n\bigg]\)