00:01
In this question, we're asked to solve a certain initial value problem while using laplace transforms and the second shifting theorem.
00:09
I've written down the second shifting theorem already, as well as the representation for f of t.
00:16
The reason why it's as such, the first term explains this part here.
00:23
It's just a simple shift upward from up to four units for the first second.
00:31
And i shift up two units more.
00:33
After one second.
00:36
Hence, this is the second part.
00:42
Now that we have this information, we're now allowed to use, we're now allowed to use it and figure out the laplace transforms we want.
00:50
So let's copy down our initial value problem once more, and take the laplace transform on both sides.
01:08
And for convenience, we're going to let the laplace of y be equal to f of s.
01:17
So what that does is that will create us, s squared f of s minus s y not minus y dot of zero minus two times s times the laplace transform of y minus y of zero and that will equal utilizing the laplace transform of the unit step function we have this is equal to four e to negative s or because t equals zero here it will be four over s plus two e to the negative s over s.
02:01
At this point, we're going to substitute in our initial conditions.
02:05
So what we have is we have s squared f of s minus y of zero is six.
02:12
So we're going to have a six s minus one, minus two, s times f of s, minus times minus is plus, minus two, y not.
02:26
Or that will be a plus two, why not.
02:27
But y not is equal to negative 6 so that will equal a negative that will be equal to 4 over s plus 2 e to negative s over s lastly what we're going to do is we're going to put the f of s on both sides so we have s squared plus s squared minus 2 s f of s will be equal to 6s or not 6 s it will be equal to 13 13 minus 6 s plus 4 over s plus 2e to the negative s over s.
03:16
And finally, performing out our division gives us f of s will be equal to 13 over s squared minus 2s, minus 6 s over s squared minus 2s, plus 4 over s times s squared minus 2s, plus 2 e to the negative s over s times s squared minus 2s.
03:45
That's our laplace transform.
03:49
Now what we need to do is we need to figure out the how to transform this into the time domain again.
03:58
In order to do that, we're going to employ partial fractions.
04:02
We're first going to simplify this.
04:05
First off, in the second term, the s's can cancel.
04:11
In the first term, we need to perform partial fractions.
04:16
On the other two terms, a little bit more difficult...