00:01
We are given with the difference equation that is y x plus 2 plus 4 y x plus 1 plus 3 y x is equals to 3 power x.
00:22
Conditions are given as y 0 is 0 y 1 is 1.
00:29
We have to use the z transformation for solving.
00:32
For that multiply both side by z.
00:36
So y x plus 2 plus 4 y x plus 1 plus 3 y x is equals to z of 3 power x.
00:50
It is the transformation z of 3 power x will be z over z minus 3.
01:00
If there is a then that will be a.
01:02
Now opening this it will come z square into y at z minus y at 0 minus y at 1 into z inverse plus it is a z is a linear function so 4 can be taken out common and z into y of z minus y of 0 plus 3 is common and z of y is z y.
01:40
It is a y of z is equals to z of 3 power x.
01:45
We have already write z upon z minus 3.
01:49
Now put the value of y 0 and y 1.
01:51
Y 0 is 0 y 1 is 1.
01:53
If we solve and multiply this we get z square plus 4 z plus 3 into y at z is equals to z over z minus 3.
02:14
This term will be left and shifting to right.
02:18
This term this term will be left and shifting to right.
02:23
We get only z here plus z.
02:27
Now solving this y of z is equals to it will be z square minus 2 divide with z minus 3 and if we factorize this we get z plus 1 and z plus 3.
02:47
Now losing partial friction from here it is z square plus 2 z...
