‫Welcome back.
‫So I hope that you've tried to solve this differential equation yourself and perhaps some of you got
‫stuck there, but that was the point, because now I'm going to teach you something new, something
‫that you might not have known before.
‫Again, we assume that our general solution has the form error as a function of time equals and then
‫a constant the oil and no to the power of London Times t and this form is a more general form compared
‫to the other form that we had.
‫Which was this one?
‫This one is more specific because here you already know that you have to Londis number one.
‫And number two, because we have a second order differential equation.
‫However, this solution here, it's more general, it's more general, because here you don't know how
‫many lambdas you have.
‫You will find that out later once you apply it to your differential equation.
‫So this general form, it is suitable to all LTI homogeneous differential equations.
‫So if you have a fifth or differential equation that is LTI and then homogeneous, then you would also
‫apply this solution to that differential equation.
‫And then since it has five orders there, that means that you will have five lambdas in the end.
‫OK, let's take the derivative of the solution.
‫So error dot as a function of time equals and then see times lambda the oil and number to the power
‫of Lambda Times t.
‫And this would be the second derivative of the solution.
‫And now we put these things into our differential equation here, and so if you do that, then it will
‫look like this and now you can factor out see times, the oil, the number to the power of London Times
‫T and then in the brackets you will have this expression, which equals to zero.
‫Of course.
‫So just like before, we assumed that C does not equal to zero.
‫And this part here, it cannot be zero either.
‫That means that this one here in the brackets, this one equals to zero.
‫So this expression here, this quadratic function, it goes to zero.
‫And now since it's a quadratic function, now we see that it has two lambdas because once we solve this
‫quadratic equation, you will have to love this.
‫You will have two roots.
‫And that's why differential equations that are second order, they have two root.
‫So to lambdas and you already know the formula for quadratic function.
‫So it's going to be Lunda one and two.
‫So two landers equals and then you have minus four here.
‫So you have the opposite side.
‫So four divided by two.
‫And since you have two lambdas then you will have plus here and minus here.
‫And so here you will have square root and four divided by two equals two.
‫So you can just put here two squared and minus 13 because you have plus 13 here.
‫And the sine changes here is going to be minus 13.
‫So lambda one equals four, divided by two equals two and then plus and then here in the square root
‫you will have four minus thirteen which is minus nine and then Lunda two equals again two.
‫But now you have minus and square root minus nine.
‫So now we have a negative number in the square root, but that's not a problem.
‫Luckily, we have imaginary numbers.
‫We have I equals square root minus one.
‫In other words, I squared equals minus one.
‫So this is an imaginary number.
‫And so your roots are then the following you have number one equals two plus.
‫And then instead of square root minus nine, you can write it down like this, minus one times nine
‫or in other words, it's going to be two plus square root minus one times square root nine in the same
‫thing with lambda to it's going to be now two minus and then minus one times nine in the square root.
‫And that means that you have to minus then square root minus one.
‫Times square root nine.
‫So number one equals and then you have two plus and now squared minus one, that's your eye and square
‫root of nine.
‫That's your three.
‫And then you alarm the two will be two minus eye times three.
‫So these here they are complex roots.
‫They're complex numbers and complex numbers have two parts.
‫They have a real part, which is this one and this one.
‫So real part, it's here and also here.
‫And then this part here and this part here.
‫These are your imaginary parts.
‫All right.
‫So this one and also this one.
‫So your general solution is error is a function of time equals the constant times, the oil, the number
‫to the power of two plus eight times, three times T and then plus the B constant and then the oil or
‫number to the power of two minus I times three times t we can also rewrite it like this error as a function
‫of time equals then the a constant and then the oilier number to the power of two times t times the
‫oilor number to the power of I times three times T and the same thing here.
‫B times the oil, the number to the power of two times T times another oil, the number to the power
‫of minus eight times.
‫Three times T.
‫So you can separate them like this.
‫But now there is something called an Euler's formula.
‫It says this, the oilor number to the power of I times X equals cosine X plus I times sine X and also
‫the Oilor number to the power of minus I times X can be written like this, the oil number to the power
‫of AI and then times minus X like this.
‫And so you can write it down like this, cosine minus X plus I times sine minus X.
‫Now cosine is an even function and that's why this cosine minus X it will be simply cosine x.
‫However, sine is an odd function and therefore sine minus X will be minus sine X.
‫And that's why here you will have minus I times sine X like this, and this relationship can be proven
‫using Taylor expansion.