00:01
Let's solve this initial value problem by using laplace transform.
00:05
So here we have this differential equation.
00:08
Y double prime minus 6 y prime plus 8y equals 0.
00:12
And we also given this initial conditions.
00:14
That is y of 0 equals 2 and y prime of 0 equals 4.
00:18
So first i'm going to apply laplace transform on both sides.
00:23
So when i do that, i will get laplace transform of y double prime minus 6y prime prime plus 8 we apply laparce transform to both sides so the plus transform of zero now we can apply the linearity property on the left side when we do that we have to apply laparce transform to each terms inside the brackets so this is laparce transform of y double prime minus la plus la plus transform of 8y and this equals 0 because la plus transform of 0 equal 0.
01:00
So i just put 0 over here.
01:02
Now i'm going to use this laplace transform formula.
01:06
That is, we have formula for laplace transform of y prime as plus laplace transform of y double prime.
01:13
So i'm going to replace laplace transform of y double prime first.
01:17
This is replaced as this equation.
01:21
So it is s squared laplace transform of y minus s times of y of 0.
01:31
Minus y prime of 0.
01:34
So we have replaced for laplace transform of y double prime.
01:39
Now let's replace for laplace transform of y.
01:42
So this can be factored out that is we have six here.
01:46
So i can add down this as basically this laplace transform of 6 y prime.
01:51
So this equals so we can factor out the six.
01:54
So it is six times of laplace transform of y prime.
01:57
So i'm going to replace this laplace transform y prime as this equation that is s times of laplace transform of y so i put a bracket s times s times of laplace transform of y minus y of zero i close this bracket and finally we have laplace transform of eight y once again we can factor out the eight out of the la plus transform and then take only the laplace transom of y so this equals zero now let's simplify this using the initial value conditions that is given.
02:37
So we are given y of 0 equals 2.
02:40
So wherever we have y of 0, we can replace this as 2.
02:44
So therefore this will be 2 and this will also we can replace it as 2.
02:49
And we have y prime of 0 that we have to replace it as 4.
02:53
So therefore this can be replaced as 4.
02:56
We have only 1 y prime of 0.
02:59
So let's clean this up.
03:01
So therefore this will be s squared laplace transform of y minus this is s times of 2 is 2 s negative 2 s and then this is negative 4 minus we have 6 times of the plus transform of y minus 2.
03:23
In fact we can do the distribution here.
03:26
So let me do the distribution.
03:27
So we have 6 times of s times of laplace transform of y, that is negative 6 s, the plus transform of y.
03:40
And then negative 6 times of negative 2 is pos 212.
03:45
And finally we have 8 times of laplace transform of y.
03:49
And so this equals 0.
03:51
Now let's factor out all the laplace transform of y's.
03:55
So we have three laplace transform of y terms.
03:58
That is this term.
03:59
This term as plus this term.
04:01
I'm going to factor only the laplace transform of y.
04:05
And when i do that, i'm left with the s squared from here.
04:09
And i'm left with the negative 6s.
04:12
And i'm left with the positive 8 over here.
04:16
And finally, we write down the remaining terms.
04:18
That is, we have negative 2s...