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Current time:0:00Total duration:10:27

we learned in the last several videos that if I had a a linear differential equation with constant coefficients in a homogenous one that had the form a times the second derivative plus B times the first derivative plus C times you could say the function or the zeroth derivative equal to zero if that's our differential equation that the characteristic equation of that is a R squared plus B R plus C is equal to zero and if the roots of this characteristic equation are real let's say we have two real roots let me write that down so if the real scenario so the real scenario where the two solutions are going to be R 1 and R 2 where these are real numbers then the general solution of this differential equation and watch the previous videos if you don't remember this or if it if you haven't feel if you don't feel like it suitably proven to you the general solution is y is equal to some constant times e to the first root X plus some other constant times e to the second root times X and we did that the last several videos that we did even you did some examples now my question to you is what if the characteristic equation does not have real roots what if they are complex and just a little bit of review what do I mean by that well if I wanted to figure out the roots of this and I didn't you know if I was lazy and I just wanted to do it without having to think can i factor it I would just immediately use the quadratic equation because that always works and I would say well the roots of my of my characteristic equation are negative B negative B plus or minus the square root of B squared minus 4a see all of that all of that over 2a right so what do I mean by non real roots well if this expression right here if this b squared minus 4ac if that's a negative number but I'm going to have to take the square root of a negative number so it will actually be a imaginary number and so this whole term will actually become complex will have a real part and an imaginary part and actually the two roots are going to be conjugates of each other right we can rewrite this in the real and imaginary parts we could rewrite this as the roots are going to be equal to minus B over 2a plus or minus the square root of b squared minus 4ac over 2a and if b squared minus 4ac is less than 0 this is going to be an imaginary number so in that case what what let's just think about what the the roots look like generally and then we'll actually do some problems so let me go back up here so then the roots aren't going to be two real numbers like that the roots we can write them as two complex numbers that are conjugates of each other and I think light blue is a suitable color for that so in that situation let me write this two complex complex roots this is a complex roots scenario then the roots of the characteristic equation are going to be I know some some number let's call it lambda let's call it mu I think that's the convention that that people use actually let me see what they what they tend to use it really doesn't matter really let's say it's lambda so this number some constant called lambda and then plus or minus right plus or minus some imaginary number and so it's going to be some constant mu that's just some constant I'm not trying to be fancy but this is I think the convention used in most differential equations books so it's mu times I so this is these are the two roots and these are two roots right because we have lambda plus mu I and lambda minus mu I so these would be the two roots if if b squared minus 4ac is less than zero so let's see how that translate let's see what happens when we take these two roots and we put them into our general solution so just like we've learned before the general solution is going to be I'll stay in the light blue the general solution is going to be y is equal to c1 times e to the first root let's make that the plus version so lambda plus mu I all of that times X plus c2 times e to the second root so that's going to be lambda minus mu I times X let's see if we can do some simplification here because that I they're really kind of makes things kind of crazy so let's see if we can do anything to either get rid of it or simplify it etc so I don't let's multiply the X out just do some algebraic manipulation I'm trying to use as much space as possible so we get Y is equal to c1 e to the what lambda X distributing at X plus mu X I plus c2 times e to the lambda X minus mu X I just distributed the X's in both to both of the terms and let's see what we can do well these when you add exponents this is the exact same thing as y is equal to c1 e to the lambda X times e to the MU X I right if you have the same base and you're multiplying you can just add exponents so this is the same thing as that plus c2 times e to the lambda x times e to the minus mu X I now let's see we have an e to the lambda X in both of these terms so we can factor it out so we get Y is equal to why is equal to let me draw a line here when you get confused with all this quadratic equation stuff y is equal to e to the lambda x times c1 e to the MU X I that's an i plus c2 times e to the minus mu X I now what we can we do and this is where it gets fun if you watched the calculus playlist especially when I talked about approximating functions with series we came up with what I thought was the most amazing result in calculus just from a or in mathematics just from a metaphysical point of view and now we will actually use it for something that you'll hopefully see is vaguely useful so here we have two 2x2 terms that have something times you know e to the something times I and we learned before Euler's formula and what was utilize that in the special I'll write it in purple that e to the I theta or we could write e to the i-x is equal to cosine of X plus I sine of X and what's amazing about that is if you put negative 1 in here then you get you get you get e to the or don't actually if you put I if you put Pi in here so e to the I pi is equal to negative 1 right if you sup to do this because sine of pi is 0 so I thought that was amazing or you could write e to the you know I 2 pi is equal to 1 that's pretty amazing as well and you can so you know in one equation you have all of the fundamental numbers of mathematics why that's amazing but let's let's get back down to earth and get practical so let's see if we can use this to simplify Euler's this is actually a definition and the definition makes a lot of sense because when you do the power series approximation or the Maclaurin series approximation of e to the X it really is it really looks like or e to the I write e to the X it really does look like cosine of X plus I times the power series approximation of of of X anyway we won't go into that now there I have like six or seven videos on it but let's use this to simplify this up here so we can rewrite that as Y is equal to e to the lambda e to the lambda x times this to the first one see one it's e to the MU X I so that can be Ritu the MU X I so instead of an X we have a mu X that will be equal to that will be equal to cosine of whatever is in front of the I so cosine of MU X plus I sine of MU X and then plus c2 plus c2 times what times cosine of minus minus mu X plus I sine of minus mu X and let's see if we can simplify this further so one thing that you might you might want what's the distribute the C's so now we get I'll do in a different color actually I'm running out of time so I'll continue this in the next video see you soon