‫And now you might think that, wait a minute, Setu had an imaginary number in it, it had this eye
‫in it, right.
‫So this was your CI one and this is your C two and C one was one, like you can see here.
‫And then C two ended up being minus two.
‫Over three.
‫So how can see to be a real number, minus two, over three, if it's also I times A minus B.
‫You might think that it's a complex number because it has an eye here, but it turns out to be a real
‫number.
‫So how can it be?
‫Well, that just means that the A or B, Constance or both have also an imaginary part in them.
‫And if you put them all together, then the imaginary parts cancel out.
‫Let's see that.
‫So you know that A plus B equals one because that's your C one.
‫Right.
‫And that means that you can rewrite it like this.
‫A equals one minus B, then C two was eight times.
‫And in the parentheses you will have A minus B and that equals minus two.
‫Over three.
‫And now what I'm going to do instead of this A, I'm going to put here one minus B like this and then
‫another B here.
‫So what you have here is I times one minus to be in that equals minus two over three.
‫So I can write it down like this.
‫I'm going to multiply by one and by minus two times B, so I'm going to have I minus two times I times
‫B equals minus two, over three.
‫I can rewrite this entire thing like this I plus two over three because I'm going to put this minus
‫two over three on the other side of the equation sine and so all that equals two times I times B and
‫that means that B equals I plus two over A three and then all that divided by two times I.
‫I can also rewrite B like this.
‫So you're going to have I divided by two times I and then plus two over and then two times I times three
‫like this.
‫So I can just rewrite this form in this form.
‫I can cancel out the eyes here and then I can cancel out the two here.
‫And so B equals one half plus one over three times I and since A equals one minus B that means that
‫it equals one minus.
‫And then you have here one half plus one over three times I and so we can rewrite a like this one minus
‫one half is going to be one half and then minus one over three times I.
‫And so that is your a.
‫And now let's do the reverse C one equals A plus B, you can see it from here.
‫And so you're A is one half minus one over three times I.
‫And then plus B, your B is plus one 1/2 plus one over three times I, so you can see that you can get
‫rid of this term and this term because they amount to zero.
‫And then here you have one half plus one half, which equals one and then C two equals I times A minus
‫B. In other words I times A was one half minus one over three times I and now you have a minus sign
‫here and your B was one 1/2 plus one over three times I.
‫So because of this minus sign you will have here minus one half, minus one over three times.
‫I like this.
‫And now you can get rid of one half and minus one half because they amount to zero and so you will have
‫eight times and then if you add these two terms together you will have minus two over three times I
‫you can cancel out the eyes and then you will have minus two.
‫Over three.
‫That was your C two.
‫So you see the imaginary numbers are hidden inside the A and B constants in such a way that in total
‫the imaginary parts cancel out and you're left with a real number constant.
‫So you'll C one and see two.
‫They're real numbers in the end.
‫And by the way, when you have the second order differential equation and you have complex routes,
‫then those complex routes are always like this X plus minus and then you will have eight times Y, so
‫you have to lend us.
‫But both of those lenders have the same real number and then one of them has a plus eight times y imaginary
‫part and then the other lambda has minus eight times Y imaginary part.
‫So they are always conjugates.
‫So in our case it was two plus minus and then I times three you see you have the same real part, but
‫then you have plus I times three imaginary part for one lambda and then minus I times three for another
‫lambda lambda one equals this and lambda two equals this right here and now as an exercise, take this
‫solution that is particular to our initial conditions and check if it really is the right solution for
‫this differential equation here.
‫For this one again, take the first derivative of it and then take the second derivative of it and then
‫put them inside your differential equation, including the solution itself, and see if zero actually
‫equals zero.
‫Try doing it exactly like before and see if you can check if the solution is truly correct.
‫So thank you very much and I'll see you in the next video.