‫Welcome back.
‫So if this is your differential equation, then how would we solve it?
‫And if you remember, solving a differential equation meant that you would get rid of these derivative
‫terms here, and then you need to have error as a function of time and then you have some kind of equation.
‫And that equation only needs to contain time.
‫It cannot contain the derivative of the error.
‫So how would we find this function that would be the solution for this differential equation?
‫Well, the general solution for this kind of differential equation is the following error as a function
‫of time equals then you have some kind of random constant and then you have E field there to the power
‫of Lambda Times.
‫T note that E and E two.
‫There are two different things.
‫E is the error.
‫OK, this is the error of our system.
‫It's the reference value minus the true state.
‫So this one would be your error.
‫However, this is still there.
‫It's an oiler.
‫No, it's approximately two point seven one.
‫And actually you have an infinite amount of numbers here.
‫After the decimal point, I put a tilde here in order not to confuse those two so that you could differentiate
‫between the errors and the Euler number.
‫And again, C is just a random constant then Lambda is your power constant on top of the oilor number.
‫And this power constant is multiplied by time.
‫Let's, for example, assume that in the differential equation K one equals six and let's assume that
‫K two equals minus one.
‫So that would be your K to here then.
‫This would be your differential equation.
‫Right.
‫Minus and then minus one.
‫Which is your key to constant and then times up minus six and then six was your key one, constant times
‫the error equals zero or you can just write it down like this.
‫And so in our general solution, which is.
‫This one here.
‫We have to find the exponent power, constant lamda and then also the constancy, and for that we take
‫the first and second time derivatives of the general solution.
‫So let's do it here.
‫So this is your general solution.
‫And now if we take the first derivative of it, then you will have C, then this lambda will come down
‫here and then you will have the oil the no to the power of Lambda Times T and then your second time
‫derivative of the error will be C times lambda squared because you will have another lambda that will
‫come down here times the oilor number to the power of Lambda T, and this is just the chain rule from
‫calculus.
‫And now we put all of them in our differential equation.
‫So e double dot is this one.
‫So we're going to say that it's C lambda squared or the number Lambda Times T.
‫Then plus see Lunda, the oil no to the power of London Times T and then minus six C oil or no and then
‫again that would be your power here and all that equals zero.
‫So this is your error, double dot, then this is your error dot.
‫And then this here, that's for your error.
‫And now we can factor out the CS and then the other numbers and their powers like this C times the other
‫number that would be your power here.
‫And then here in the brackets I will have lambda squared plus lambda minus six and all that equals zero,
‫assuming that C does not equal to zero and well this oilor number to the power of Lunda time C cannot
‫equal to zero anyway, because even if Lambda is zero then you will have the oilor number to the power
‫of zero times t.
‫That means that you would have oilor number to the power of zero which would be one.
‫And so knowing these facts, that means that for this equation to be zero.
‫This thing in the brackets has to be zero.
‫So what you have here, you have here this quadratic equation, lambda squared, plus lambda minus six
‫equals zero.
‫And the lambdas here they are, your roots do this quadratic equation, since you have a quadratic equation,
‫that means that you have two roots, so you have to lambdas.
‫And so the formula to compute those lambdas is minus one 1/2, plus minus, and then you will have one
‫half here squared and then plus six.
‫That's a typical formula to find the roots for quadratic equation.
‫In a more general case, your quadratic equation looks like this.
‫Where you have some kind of constant in front of this square term as well, then in order to find the
‫roots, your formula would be minus B, so you take this constant B and you put an opposite sign here.
‫So here you have plus.
‫So that means that here you put minus if you had minus B here, then here you would have minus minus
‫B which would be plus B and so minus B plus minus and then square root B squared minus four times eight
‫times C.
‫So these are these constants here and then all that you divide by two times a.
‫Or you can also write it down like this, I divide this term by two times A and then I divide this term
‫by two times A, and so if you're a constant here, if it equals one, like in our case.
‫Because here you have one.
‫Well, that means that your a year will be one, so you'll just have to hear and you will also have
‫one here, so you will just have to hear two times one equals two.
‫So like this.
‫And that means that this a here would also be one, so that's how it will look like.
‫Then you can factor out this for under this square root sign, and so if you do that, then it looks
‫like this four times and then in the parentheses you will have B squared divided by four and then minus
‫C here.
‫And then you divide this entire thing by two.
‫And well, you know that if you have a square root and then you have some kind of constant here, A
‫and then B minus C, then you can rewrite it like this, a square root, times square root B minus C,
‫and in this case, this square root A would be square root four and then square root four would be two.
‫That means that you can cancel out this two with this four here, because once you take it out of the
‫square root, it becomes two and you can cancel them out.
‫And so in the end, you can rewrite this formula like this might just be over two plus and minus, and
‫that really means that you have two separate equations, one with the plus and the other one with a
‫minus.
‫And then you have your square root here and then be squared away before you can rewrite it like this,
‫be over two and then you take this entire thing squared and then here you have minus C, and that's
‫the formula that I have used here.
‫In this equation, A equals one, B equals one, and then C equals minus six.
‫And so if we go back to our original quadratic function, then our roots will be the following lambda
‫one will be minus one half plus.
‫And then here in the square root, you will have six point twenty five because one half squared will
‫be zero point twenty five and then plus six it will be six point twenty five and then lambda two equals
‫minus one half minus square root six point twenty five.
‫So your lambda won, your first root will be two and then your second root will be minus three.
‫And so these are your roots, two and minus three, and you can check it, for example, if you take
‫the one and you put it inside here, then, well, you will have two squared, plus two, minus six,
‫and then that equals to zero.
‫And then we can do the same thing with minus three.
‫So minus three squared.
‫Plus minus three and minus six, well, that equals nine, minus three and minus six, and that equals
‫zero.
‫So you see, you have to lambdas Lumb one and two, and it makes sense that you have two routes or two
‫power constants here.
‫It makes sense because your differential equation is a second order differential equation.
‫And in the next video, let's see how we formulate a general solution for this particular differential
‫equation.
‫Thank you very much.