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‫Welcome back.

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‫So in the previous videos, we found our roots here, London, one and two.

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‫And remember, the roots here are actually the power constants here.

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‫So the oilor number to the power of lamda times t and since your differential equation is a second order

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‫differential equation because of that fact, you will have to Londis London one and Lunda two.

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‫And that means that for this differential equation, the general solution is error as a function of

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‫time equals and then the a constant times the oil or no to the power of the first London that you found

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‫here, which is two, we're going to put it here times T and then plus another constant let's call it

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‫B.

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‫And then again, you have this Oilor number here to the power of minus three times T.

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‫So this to here deaths your loved one and then this minus three, that's your lump, the two like this

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‫and now we have to find the constants A and B to do that.

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‫We need additional information.

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‫We need to know the initial conditions and what are the initial conditions.

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‫They essentially tell you what the error and error dot are at.

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‫Time equals zero seconds.

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‫So at the very beginning, when time equals zero seconds, you want to know your error and then your

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‫error time derivative.

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‫So let's say that at time equals zero seconds.

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‫Your error at time equals zero seconds will be one meter.

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‫And let's say that your error dot at time equals zero seconds will be zero meters per second.

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‫And now we put this information into our general solution.

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‫So error at zero seconds equals a times.

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‫The oil, the number to the power of two times zero because your time equals zero seconds and then plus

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‫B and again, the oil number to the power of minus three times zero seconds.

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‫And then all that equals one.

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‫So this thing here will become one and then this thing here will become one because you will have zero

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‫in the exponent here.

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‫So that means that this thing and this thing will be once.

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‫So you will have A plus B equals one.

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‫And now if you take this general solution and you take the derivative of it, then it will be like this

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‫error dot as a function of time equals.

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‫And now you will have a times, two times the oil and number to the power of two T and then plus B times

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‫minus three, the oilor number to the power of minus three times T.

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‫And that would be your time derivative of your general solution and then your IDOT at zero seconds will

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‫be a times two times the oil no to the power of two times zero and then minus B times three times the

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‫oil a number and then to the power of minus three times zero.

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‫And then all that will equal zero.

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‫Again, this part here will become one and this part here will become one.

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‫So you will be left with two times A minus three times B equals zero.

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‫And then you also have this equation A plus B equals one.

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‫So you have two equations and two unknowns and your unknowns are A and B, so I can say that my A equals

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‫one minus B, that's from this equation here.

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‫And then I'm going to substitute A with this one minus B in this equation.

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‫So two times one minus B and then minus three B equals zero.

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‫Or in other words, I have two minus two times B minus three times B equals zero.

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‫So I have two minus five times B equals zero.

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‫So five times B equals two.

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‫And that means that B equals two over five and six A equals one minus B, that means that A equals one

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‫minus two over five.

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‫In other words, A equals five over five, which is one minus two over five.

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‫And the answer is three over five.

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‫So you're a equals three over five and your B equals two over five.

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‫And in conclusion, for these initial conditions here, this one and this one, your general solution

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‫will be error as a function of time, just like here.

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‫But now you know, you're A and B constants and you're A was three over five times the Euler number

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‫to the power of two times T plus.

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‫And now your B was two over five times the oil or no to the power of minus three times T..

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‫So that's the solution for our differential equation here when these are your initial conditions.

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‫So these constants, they depend on your initial conditions, the lambdas here too.

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‫And minus three, they do not depend on your initial conditions, but the constants, A and B, they

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‫do depend on your initial conditions, your lambdas two and minus three.

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‫They depend what kind of constants you have here.

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‫Here we have one times IDOT and then minus six times E, and in general you see your error as a function

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‫of time.

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‫Now only depends on time.

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‫Right?

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‫So we have solved the differential equation because we have a function in which our dependent variable

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‫error only depends on time.

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‫And you don't have any derivatives here, right.

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‫You don't have any error time derivatives.

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‫And by the way, if you had a homogeneous third order LTI differential equation, which would look like

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‫this, now you see you have a third derivative of error with respect to time and also the second derivative,

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‫the first derivative and the dependent variable itself.

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‫Then the general solution for this differential equation would be the following error as a function

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‫of time equals that would be your first term plus that would be your second term.

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‫And then plus and that would be your third term.

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‫You see now you have three power constants, you have Lunda one, and then you have LAMDA two, and

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‫then you have LAMDA three.

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‫Or in other words, you have three routes.

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‫But OK, let's go back to our differential equation and for that differential equation, the solution

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‫was this one here and now as an exercise, I'd like you to take this solution and check if it's correct.

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‫Take the first and second time derivative of it and then put it in the original differential equation.

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‫So now that you know these constants, take your Avodart and then erodable dot and then put everything

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‫into these variables here, erodable dot dot and then the error itself put all that into this differential

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‫equation and then see if the left side of the equation sine equals the right side of the equation sine.

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‫So try it out and I'll see you in the next video.

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‫Thank you very much.

