00:01
Hello students, here given xi of s is equal to 1 divided by 6s square plus 43s plus 7 and f of t is equal to e power minus 5t.
00:18
Now given that xi of s is equal to y of s divided by f of s.
00:25
Here we can write y of s is equal to f of s into xi of s.
00:40
Now we need to find f of s.
00:44
Given f of t is e power minus 5t then applying lamplet's transform f of s will be 1 divided by s plus 5.
00:56
Now substitute this f of s and 5 of s given above in the y of s.
01:03
Then y of s will be equal to 1 divided by s plus 5 into 6s square plus 43s plus 7.
01:18
Split this 6s square like 6s square plus 42s plus 1s plus 7.
01:30
Here take common 6s we get s plus 7 inside plus here 1 inside s plus 7.
01:41
We had solved this fraction by separating method.
01:51
Substitute these two terms in place of this fraction.
01:58
We write s plus 5, s plus 7 and 6s plus 1.
02:07
Now we are having y of s in this form.
02:10
Write this as a divided by s plus 5 plus b divided by s plus 7 plus c divided by 6s plus 1.
02:27
Now we need to find a, b and c values.
02:31
For that let us rearrange.
02:37
Multiply s plus 7 and 6s plus 1 with a plus.
02:48
Similarly with b exclude b denominator and multiply another two denominator.
02:54
It is nothing but taking lcm and solving the equation.
03:01
Plus c into whole divided by these three terms.
03:15
Now here also we are having these three terms.
03:18
Here also both denominator will get cancelled.
03:22
Then we have 1 is equal to a into s plus 7, 6s plus 1 plus b into s plus 5, 6s plus 1 plus c into s plus 5, s plus 7...
