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# Complex roots of the characteristic equations 2

## Video transcript

so where we left off we I had given you the question you know it's these these these type of equations are fairly straightforward when we have two real roots then this is the general solution and if you have your initial conditions you can solve for C 1 and C 2 but the question I'm asking is what happens when you have two complex roots or essentially when you're trying to solve the characteristic equation when you're trying to solve that quadratic the B squared minus 4ac that that's negative so you get the two roots end up being complex conjugates and we said okay let's say that our two roots are lambda plus or minus mu I and we just did a bunch of algebra to sit we said well if those are the roots and we substitute it back into this formula for the general solution we get all of this and we kept simplifying it all the way until we got here where we said e is equal to e you know y is equal to e to the lambda X plus c1 X etc and we said can we simplify this further and that's where we took out Euler's equation or Euler's formula or Euler's definition depending what you want which I I'm always in awe of every time I see it or use it but we've talked a lot about that in the in the calculus playlist but we could use this to maybe further simplify it so I could wrote e to the MU X I as cosine mu X plus I sine mu X and I wrote e to the minus mu X I is cosine minus mu X plus I sine minus you at MU X and now we could use a little bit about what we know about trigonometry right cosine of cosine of minus theta is equal to cosine of theta and we also know that sine of minus theta is equal to minus sine of theta so let's use these identities to simplify this a little bit more so we get Y is equal to e to the lambda x times and let's see we could actually distribute the c1 to so times c1 cosine of MU X plus I times c1 sine of MU X plus plus all of this is in this parenthesis right here plus c2 instead of cosine of negative mu X we know this identity so we can just write this as cosine of MU X as well because cosine of minus X is the same thing as cosine of X plus C I times c2 sine of minus mu X that's the same thing as minus sine of X well actually let's take this let's make this to take the minus sign out there so minus sine of MU X and let's see it seems like we're getting to a point that we can simplify it even more we could add the two cosine terms so we get the general solution and I know this is this is a hi this problem requires a lot of algebraic stamina but as long as you don't give me careless mistakes you'll find it reasonably rewarding because you'll see where things are coming from so we get why the general solution is y is equal to e to the lambda x times let's add up the two cosine mu X terms so it's c1 plus c2 times cosine of MU X let's add the two sine of MU X terms so plus I we could constantly call that c1 I see one I that's that - see - I see two I times sine of MU X and we're almost done simplifying and the last thing we can simplify is well you know C 1 and C 2 are arbitrary constants so let's just define this as another constant I don't know let's call it I'll just call it C 3 I mean just to just to not confuse you by using c1 twice I'll call this C 3 and now this might be a little bit of a stretch for you but if you think about it really makes sense this is still just a constant right especially especially if I say you know what I'm not restricting this the constants to the real see could be an imaginary number so if C is an imaginary number or some type of complex number we don't even know whether this is necessarily an imaginary number so but you know we're not going to make any assumptions about it let's just say that this is some other arbitrary constant call this C four and we can worry about it when we're actually given the initial conditions but what this gives us if we make that simplification we actually get a pretty straightforward general solution to our differential equation where the characteristic equation has complex roots and that I'll do in a new color that is y is equal to e to the lambda X e to the lambda X times some constant I'll call it C 3 it could be C 1 could be C 100 whatever some constant times cosine of mu of X plus some other constant I called it C 4 doesn't have to be C 4 I just want to confuse it with these but some other constant times the sine of mu of X so there's there's there's really two things I want you to realize one is we haven't done anything different at the end of the day we still just took the two roots and substituted it and substitute it back into these equations for R 1 and R 2 the difference is we just kept algebraically simplifying it so that we got rid of the eyes that's all we did there was really nothing new here except some except for some algebra and the use of Euler's formula but when we when we had when these when R 1 and R 2 involved complex numbers we got to this simplification so in general if you get the characteristic equation and your two roots are mu plus or minus oh sorry no your two roots are lambda plus or minus mu I then the general solution is going to be this and it's not to you know if you had to memorize that although I don't want you to you should be able to derive this on your own but it's not too hard to think and actually if you ever forget it just whenever you know solve your solve your characters equation get your complex numbers and just substitute it right back in this equation and then with the real numbers instead of the lambda and mu with the real numbers just do the simplification we did and you'll get to the exact same point but if you're taking an exam and you know you don't want to waste time and you want to be able to do something fairly quickly you can just remember that if I have a complex root or F complex roots to my characteristic equation lambda plus or minus mu I I can just I then my general solution is e to the lambda X times some constant times cosine of MU X plus some constant times sine of MU X and let's see if we can we can do a problem real fast that involves that so let's say I had the the differential equation see Y prime prime plus the first derivative plus y is equal to zero so our characteristic equation is R squared plus R plus one is equal to zero let's break out the quadratic quadratic formula so the roots are going to be negative B so it's negative 1 plus or minus the square root of B squared B squared is 1 minus 4 times AC well a and C are both 1 so it's just minus 4 all of that over 2 right 2 times a all of that over 2 so the roots are going to be negative 1 plus or minus the square root of this is negative 3 over 2 or we could rewrite this as the roots are R is equal to negative 1/2 plus or minus well we could rewrite this as I times the square root of 3 or square root of 3i over 2 or we could write this as square root of 3 over 2 times I actually that's the best way to write it write square root of you just take the I out so it takes the negative 1 out and you have left with square root of 3 over 2 so these are the roots and now if we want the general solution we just have to throw this right back into that and we'll have our general solution so right that right down here so our general solution will be y is equal to e to the real part of our complex conjugates so e to the minus 1/2 times X right this is our lambda times some constant I'll write c1 now c1 times cosine of of D you could kind of the of the kind of the imaginary part without the I so cosine of square root of 3 over 2 X plus c2 times sine of square root over square root of 3 over 2 X not too bad we had complex roots and it really didn't take us any more time than when we had two real roots you just have to realize this and then you have to just find use a quadratic equation to find the complex roots of the characteristic equation and realize that this is lambda this minus 1/2 is lambda and that the square root of 3 over 2 is equal to MU and then substitute back into this solution that we got anyway in the next video I'll do another one of these problems and we'll actually have initial conditions so we can solve for C 1 and C 2 see you in the next video