‫Welcome back.
‫So now we are going to derive a relationship between the polls and the key values in our differential
‫equation that looked like this error, double that minus K, two times error, Dott minus K one times
‫the error equals zero, essentially by choosing the polls that make the error behave like we want to,
‫and then by calculating the key values, we construct a differential equation suitable for us.
‫So our general solution is error as a function of time equals and then a random constancy and then the
‫oil number to the power of Lambda Times t this is our general solution.
‫Then we take the derivative of it and we will have error dot as a function of time equals and then C
‫times lambda times the oil or no to the power of Lambda Times T and then after we have taken the first
‫derivative of it, we're going to take the second derivative of it.
‫So e double that as a function of time equals C times lambda squared times the oilor number to the power
‫of Lambda Times T and now we're going to put all those three things into our differential equation.
‫First of all, the error double dot is this one here.
‫So C times lambda squared times the oil and number to the power of Lambda Times T then minus K two that
‫you have here times the error dot which is this one here.
‫So C times lambda times the oil number to the power of Lambda Time C and then finally minus K one that
‫you have here and then you multiplied by the error itself.
‫So times C times the oil, the number to the power of Lambda Times T and all that of course equals to
‫zero.
‫You can factor out C and also the oil, a number to the power of the time D and then you will have in
‫the brackets this expression lambda squared minus K two times lambda minus K one.
‫And all that equals to zero, assuming that the constancy cannot equal to zero and also the oil and
‫number to the power of the time.
‫C cannot be zero either.
‫That means that this equation can only be zero if what you have in the brackets is zero.
‫So lambda squared minus K two times lambda minus K one that must equal to zero.
‫And now we get the lambda or the roots or the polls.
‫So you see you have three names for the same thing.
‫Route's polls and lambdas and power constants.
‫Eventually it's the same thing.
‫So lambda one and two equals.
‫And then according to the formula, in order to find the roots for this quadratic equation, you have
‫K two divided by two, then plus minus and then square root and then K two over to all that is taken
‫to the power of two.
‫So you square it and then you're going to have plus K one here.
‫So that will be your lambda one and that will be your lambda too.
‫So you have expressed your lambdas in terms of case you have lambda one as a function of K one and K
‫to and then you have lambda two as a function of K one and K two and now you will have an exercise.
‫Now you simply have to express the case in terms of Londis.
‫So you need to have an expression for K one as a function of.
‫The one and Lunda two, and then you need to have an expression for K two as a function of Lunda one
‫and Lunda to.
‫And so your exercise now is to rearrange these equations in order to get this and this right now you
‫have this and this, but you need to rearrange these equations to get this and this.
‫That will be your exercise.
‫So give it a try and we'll talk about it in the next video.
‫Thank you very much.