00:01
Hi everyone, so today we are going to solve this given system of differential equations in which the first equation is d x divided by dd is equal to 2y plus exponential t and the second equation is d y by d t is equal to 8x negative t fine and the given conditions are that x of 0 is equal to 1 and y of 0 is equal to 1.
00:43
Now we are required to use the lapless transform in order to solve this given a system of differential equations.
00:59
Fine.
01:00
Now how will we do that? the solution for this question is very easy.
01:07
Fine.
01:08
So first what we are going to do, we are going to take the laplace transform of these equations on both sides.
01:22
Fine.
01:23
So we're going to take the laplace transform on both sides.
01:37
Fine.
01:40
So now we have that if x of s is a equal to l x of t and and y of s is equal to y of d.
02:17
Then after transforming the both sides of the equations we get that the first equation would become d this is the first equation right d x by d t is equal to 2 i plus exponential plus exponential t fine now we can write it as x prime is equal to 2y plus exponential t next we are going to shift this term over here so i get that x prime negative 2y is equal to positive exponential t fine next i'm going to take the lap plus transform.
03:09
So i get s x of s negative 1, negative 2, y of s is equal to 1 over s minus 1.
03:26
Next i'm going to shift this term over here.
03:31
So i get that s, x of s negative 2, y of s, negative 2, y of 1.
03:39
Of s is equal to 1 divided by s minus 1 plus 1 let's say that this is equation number 1 fine next at the given differential equation is d y by d t that is equal to 8x negative t i can write it as y prime is equal to 8 x negative t next i'm going to shift this 8 x over here so i get y prime negative 8x is equal to negative t fine and the next step is to apply the lap plus transform so i get s y of s negative y of 0 negative 8 x of s is equal to negative 1 over s now we have that y of 0 is equal to 1 fine.
04:53
So we have s y of s negative 1, negative 8 x of s that is equal to negative 1 divided by s next i'm going to shift this negative 1 over here so i get s y of s negative 8 x of s is equal to 1 negative 1 divided by s square.
05:20
Let's say that this is equation number two.
05:23
Fine.
05:25
So next what we are going to do over here is that we are going to multiply equation one.
05:40
Now which one is equation one? so i'm going to multiply this whole equation with eight.
05:51
Fine.
05:52
And so i can write it as.
05:54
I'm going to multiply equation 1 with 8 and next i will multiply equation 2 with s.
06:09
Now which one is equation 2? this one is equation 2 so i'm going to multiply with s and then i'm going to add both of these equations.
06:23
Adding both equations.
06:27
Fine.
06:28
So let's multiply.
06:32
The first equation with 8, so i get 8 times x x of s negative to y of s plus s, s y of s, negative s y of s negative 8 x of s.
07:01
So i'm running out of space over here, so i'm going to write it over here is equal to 8 times 1 plus 1 divided by s minus 1.
07:23
Fine.
07:24
Plus i have s times 1 minus 1 divided by s square.
07:33
Now what i have done in this step is that i have multiplied the first equation with 8.
07:41
Over here you can see that this is the first equation that has been multiplied with 8.
07:46
And over here you can see that this is the second equation and it has been multiplied with s.
07:51
Now this is the right hand side of the first equation that is being multiplied with s.
07:55
And this is the second, sorry, this is the right hand equation.
07:59
So right hand of the second equation that is being multiplied with s.
08:05
Fine.
08:06
So now i'm going to open the parentheses and let's see what it looks like...