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