00:01
In this question, basically, we look at two functions, right? x of a function of t and one of a function t, right? and this x and one are supposed to satisfy this first order linear differential equations, right? and we can use to solve this type of equations by writing xt equals, let's say, a times, i would like to call it lambda 1 times t, and y equals b times e, number two times t.
00:27
Then we can substitute this into this.
00:30
Expression we have found this to be given by.
00:34
To be, actually, we found this actually to be given by.
00:36
If you do the, if you put things in, you'll find it's very clear that lambda 1 and lambda 2 must be the same, right? because otherwise, this equation cannot be satisfied.
00:49
So we call it lambda.
00:50
Well, i just called lambda instead of called lambda 1, number 2, right? so if you put in the a, you have found that if from the first equation, you'll find this to be, you put it in, you'll find it.
01:01
And this take the derivative to get a times lambda right.
01:05
And then this will be equal to minus 0 .4 and plus y, which is you get a b, right? and then the second equation, you take the derivative y, which you get b lambda, and that gives you minus a, right, and then minus 2 .4 times b, right? and that would be it.
01:28
Oh, this is your here as well, right? so that would be what we would, i think it would define with these equations.
01:36
So, and of course you put, you rearrange this equation, you will find that, you rearrange in these equations, we have found this actually lambda plus 0 .4 and then 1, right? and then for this, you will find this 1 and then lambda plus 2 .4, right? i think that would be it.
02:04
Yes, lambda and b, it should be minus one, actually.
02:10
I got wrong, and a you put there and should be like that.
02:13
So you'll find this to be zero, right? you will get an equation like this...