00:01
Here we are going to use laplace transform to solve this initial value problem.
00:07
So we apply laplace transform to the differential equation.
00:11
We will get s -square -y plus 3sy plus 2y equal capital f.
00:34
Here since the initial condition is y of 0 equals 0, y prime of 0 equal 0, so we don't need to subtract.
00:41
The terms arising from y of 0 and y prime of 0.
00:53
So we need to know what is f of s.
00:58
So this is the laplace transform of f of t.
01:01
And f of t takes a now zero value only on the interval 0, 10.
01:07
So we integrate from 0 to 10, 1, e to the negative s t d t.
01:14
The anti -derivative of e to the negative s t is negative 1 over s, e to the negative as t.
01:27
T equals 0, t equal 10.
01:30
Plug in this value as subtract.
01:32
Then we will get negative 1 over s.
01:37
We plug in 10, so e to the negative 10s minus 1.
01:45
So it is 1 over s, 1 minus e to the negative 10 s.
01:57
So now we have capital y equal 1 over s squared plus 3s plus 2.
02:09
We divide the whole equality by s squared plus 3s plus 2.
02:16
1 minus e to the negative 10 s.
02:21
We write this part as a partial fraction.
02:25
First we need to factor it and then we write it as a partial fraction.
02:43
The partial fraction of this is a over s plus b over s plus 1 plus c over s plus 2.
02:52
You will need to solve for a, b, c and i'll just write down the result.
02:57
We will get 1 half, 1 over s minus 1 over s plus 1 half plus 1 half 1 over s plus 1 half 1 over as plus 2, 1 minus e to the negative 10 s.
03:24
Here the green underline part, this is the laplace transform of 1 1⁄2 minus e to the negative t, plus 1 half e to the negative 2 t.
03:47
So we apply the inverse la plus transform we will get y equal.
03:51
The first part is the 1 times the green underline.
03:54
This is the inverse.
03:55
Part we will get 1 half minus e to the negative t plus 1 half e to the negative to t and then minus here we have similar things to this but we need to slightly change it every t is replaced by t minus 10 and then multiply by u of t minus 10 here this u this u is the heavy side function.
05:03
So the whole thing, this y, will be equal to 1 1⁄2t, plus 1 half, e to negative 2t, if 0 0 is less than or equal to t less than 10.
05:21
If t is greater than or equal to 10, this becomes, this u of t minus 10 become 1, and we will get constant terms cancel and the e to the negative t it will become negative e to the negative t and then we'll in plus e to the negative t minus 10 plus one half e to the negative two t minus one half e to the negative two t minus 10 if t is greater than equal to 10 this is e to the 10 minus 1, e to the t...