00:01
In the question it is given that we have dx by dt is equal to minus x plus y and we have to use laplace transform to solve this.
00:10
So from here dy by dt will be equal to 2x and it is given that x of 0 is equal to 0 and y of 0 is equal to 1.
00:22
So let l of x is equal to x of s and l of y is equal to y of s.
00:34
So on applying l to both the equations we will get l of dx by dt will be equal to minus l of x plus l of y.
00:46
So from here further l of dy by dt will be equal to 2l of x.
00:54
So now we have l of dx by dt will be equal to s x of s minus x of 0 which is equal to s x of s and l dy by dt will be equal to s y of s minus y of 0 which is equal to s y of s minus 1.
01:23
Now we have the system that is s x will be equal to minus x plus y.
01:30
So on taking x common we will have s plus 1 in the bracket minus y is equal to 0 and the second one we have s y minus 1 is equal to 2x.
01:41
So from here minus 2x plus s y will be equal to 1.
01:47
Now from the first equation we will have y will be equal to s plus 1 x.
01:54
So we will put this value in minus 2x plus s y is equal to 1.
02:05
We will get minus 2x plus s s plus 1 x is equal to 1.
02:09
Further we can write this as minus 2 plus s s plus 1 x is equal to 1 on taking x common.
02:19
Now we will solve for x.
02:23
So we will have x is equal to 1 upon s square plus s minus 2.
02:30
So this is further equal to 1 upon s plus 2 multiplied by s minus 1.
02:36
We have factorized this quadratic equation.
02:39
So now we will solve for y.
02:42
Since we know that since y is equal to s plus 1 x we will have y is equal to s plus 1 upon s plus 2 multiplied by s minus 1.
02:57
Now we will have both the values for x and y...