00:01
In the question we have the given system is d square plus 5 times d plus 6 times y of t is equal to d plus 4 times x of t where the input is given as x of t is equal to t square plus 1 with initial condition with the initial condition y of 0 is equal to 2 and y dash of 0 is equal to 1 now we can derive this given system as d square plus 5 times d plus 6 times y of t is equal to to d plus 4 times t squared plus 1.
01:00
So this can be written as d square.
01:04
We will simplify this right hand side as d times t square plus 1 plus 4 times t square plus 1 which simplifying gives us 2 t plus 0 plus 4 times t square plus 1.
01:21
So this is equals to we have here.
01:25
4 t squared plus 2 t plus 4 so the auxiliary equation is the auxiliary equation is given as m 2m plus 6 is equal to 0 simplifying it we have m 2m plus 3m plus 2m plus 6 is equal to 0 this this gives us m times m plus 3 plus 2 times m plus 3 is equal to 0.
02:06
Then we have m plus 2 times m plus 3 is equal to 0.
02:12
That is we have m is equal to negative 2 and m is equal to negative 3.
02:19
So let's find a particular integral which is equal to 1 over d square plus 5d plus 6.
02:28
Times 4 t squared plus 2 times t plus 4 which can be written as 1 over 6 times 1 plus d 2d over 6 times 4 t square plus 2 t plus 4 which gives us 1 over 6 in bracket 1 plus d square plus 5d over 6 inverse times 4 t squared plus 2 t plus 4.
03:06
So simplifying it we get 1 over 6 times 1 minus d square over 6 plus 5 times t over 6 plus d square over 6 plus d square times t over 6 whole square plus up to so on times 40 square plus 2 t plus 4 if we simplify this particular integral we get 1 over 6 times 40 square plus 2 t plus 4 minus 8 over 6 minus 5 over 6 8 times t plus 2 plus 25 5 over 36 times 8.
03:59
So we have 1 over 6 times 40 square plus 2 t plus 4 minus 4 over 3 minus 20 over 3 times t minus 5 over 3 plus 50 over 9.
04:18
Here we have 8...