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‫Welcome back.

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‫So I hope that you've tried to solve this differential equation yourself and perhaps some of you got

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‫stuck there, but that was the point, because now I'm going to teach you something new, something

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‫that you might not have known before.

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‫Again, we assume that our general solution has the form error as a function of time equals and then

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‫a constant the oil and no to the power of London Times t and this form is a more general form compared

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‫to the other form that we had.

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‫Which was this one?

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‫This one is more specific because here you already know that you have to Londis number one.

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‫And number two, because we have a second order differential equation.

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‫However, this solution here, it's more general, it's more general, because here you don't know how

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‫many lambdas you have.

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‫You will find that out later once you apply it to your differential equation.

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‫So this general form, it is suitable to all LTI homogeneous differential equations.

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‫So if you have a fifth or differential equation that is LTI and then homogeneous, then you would also

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‫apply this solution to that differential equation.

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‫And then since it has five orders there, that means that you will have five lambdas in the end.

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‫OK, let's take the derivative of the solution.

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‫So error dot as a function of time equals and then see times lambda the oil and number to the power

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‫of Lambda Times t.

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‫And this would be the second derivative of the solution.

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‫And now we put these things into our differential equation here, and so if you do that, then it will

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‫look like this and now you can factor out see times, the oil, the number to the power of London Times

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‫T and then in the brackets you will have this expression, which equals to zero.

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‫Of course.

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‫So just like before, we assumed that C does not equal to zero.

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‫And this part here, it cannot be zero either.

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‫That means that this one here in the brackets, this one equals to zero.

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‫So this expression here, this quadratic function, it goes to zero.

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‫And now since it's a quadratic function, now we see that it has two lambdas because once we solve this

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‫quadratic equation, you will have to love this.

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‫You will have two roots.

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‫And that's why differential equations that are second order, they have two root.

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‫So to lambdas and you already know the formula for quadratic function.

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‫So it's going to be Lunda one and two.

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‫So two landers equals and then you have minus four here.

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‫So you have the opposite side.

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‫So four divided by two.

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‫And since you have two lambdas then you will have plus here and minus here.

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‫And so here you will have square root and four divided by two equals two.

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‫So you can just put here two squared and minus 13 because you have plus 13 here.

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‫And the sine changes here is going to be minus 13.

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‫So lambda one equals four, divided by two equals two and then plus and then here in the square root

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‫you will have four minus thirteen which is minus nine and then Lunda two equals again two.

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‫But now you have minus and square root minus nine.

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‫So now we have a negative number in the square root, but that's not a problem.

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‫Luckily, we have imaginary numbers.

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‫We have I equals square root minus one.

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‫In other words, I squared equals minus one.

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‫So this is an imaginary number.

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‫And so your roots are then the following you have number one equals two plus.

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‫And then instead of square root minus nine, you can write it down like this, minus one times nine

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‫or in other words, it's going to be two plus square root minus one times square root nine in the same

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‫thing with lambda to it's going to be now two minus and then minus one times nine in the square root.

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‫And that means that you have to minus then square root minus one.

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‫Times square root nine.

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‫So number one equals and then you have two plus and now squared minus one, that's your eye and square

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‫root of nine.

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‫That's your three.

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‫And then you alarm the two will be two minus eye times three.

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‫So these here they are complex roots.

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‫They're complex numbers and complex numbers have two parts.

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‫They have a real part, which is this one and this one.

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‫So real part, it's here and also here.

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‫And then this part here and this part here.

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‫These are your imaginary parts.

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‫All right.

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‫So this one and also this one.

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‫So your general solution is error is a function of time equals the constant times, the oil, the number

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‫to the power of two plus eight times, three times T and then plus the B constant and then the oil or

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‫number to the power of two minus I times three times t we can also rewrite it like this error as a function

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‫of time equals then the a constant and then the oilier number to the power of two times t times the

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‫oilor number to the power of I times three times T and the same thing here.

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‫B times the oil, the number to the power of two times T times another oil, the number to the power

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‫of minus eight times.

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‫Three times T.

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‫So you can separate them like this.

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‫But now there is something called an Euler's formula.

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‫It says this, the oilor number to the power of I times X equals cosine X plus I times sine X and also

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‫the Oilor number to the power of minus I times X can be written like this, the oil number to the power

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‫of AI and then times minus X like this.

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‫And so you can write it down like this, cosine minus X plus I times sine minus X.

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‫Now cosine is an even function and that's why this cosine minus X it will be simply cosine x.

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‫However, sine is an odd function and therefore sine minus X will be minus sine X.

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‫And that's why here you will have minus I times sine X like this, and this relationship can be proven

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‫using Taylor expansion.