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Current time:0:00Total duration:10:12

Video transcript

let's do a couple of problems where the roots of the characteristic equation are complex and just as a little bit of a review and we'll put here this up here in the corner so that it's useful for us we learned that if the roots of our characteristic equation are R is equal to lambda plus or minus mu I that the general solution for our differential equation is y is equal to e to the lambda X times c1 or some constant cosine of MU X plus c2 times sine of MU X and with that said let's do some problems so let's see this first one it's that our differential equation I'll do this one in blue our differential equation is the second derivative Y prime prime plus 4 for y prime plus 5y is equal to 0 and they actually give us some initial conditions so they say Y of 0 is equal to 1 and Y prime of 0 Y prime of 0 Y prime of 0 is equal to 0 so now we'll actually be able to figure out a particular solution or the particular solution for this differential equation so let's write down the characteristic equation so it's R squared plus 4 R plus 5 is equal to 0 let's break out our quadratic formula the roots of this are going to be negative B so minus 4 plus or minus the square root of B squared so that's 16 minus 4 times a well that's 1 times 1 times C times 5 let's go us out there all of that over 2 times a a is 1 so head over 2 and see this simplifies this equals - 4 plus or minus C 16 this is 20 right 4 times 1 times 5 is 20 so 16 minus 20 is minus 4 minus 4 over 2 and that equals let's see this equals minus 4 plus or minus 2i right square root of minus 4 is 2i all of that over 2 and so our roots - the characteristic equation R - 2 just dividing both by 2 minus 2 plus or minus we could say I or 1 i right so if we wanted to do some pattern match if we just want to do it really fast what's our lambda where lambda is just minus 2 let me write that down that's our lambda and what's our mu well mu is the coefficient on the I so mu is 1 mu is equal to 1 and now we're ready to write down our general solution so the general solution to this differential equation is y is equal to e to the lambda X well lambdas minus 2 minus 2 X times c1 cosine of MU X but Mews just 1 so c1 cosine of X plus c2 sine of MU X when mu is 1 so sine of X fair enough now let's use our initial conditions to find the particular solution or a particular solution so when x is 0 Y is equal to 1 so Y is equal to 1 when x is 0 so 1 is equal to let's substitute X is 0 here so e to the minus 2 times 0 that's just that's just 1 so this whole thing becomes 1 so we could just ignore it's just 1 times this thing so I'll write that down e to the 0 is 1 times C 1 times cosine of 0 plus C 2 times sine of 0 now what's sine of 0 sine of 0 is 0 so this is this whole term that's going to be 0 cosine of 0 is 1 so there we already solved for C 1 we get this this is 1 so 1 times C 1 times 1 is equal to 1 so we get our first coefficient C 1 is equal to 1 now let's take the derivative of our general solution and we can even substitute C 1 in here just so that we have to stop writing C 1 all the time and we can solve for C 2 so right now we know that our general solution is y we could call this our pseudo general solution because we already solved for C 1 y is equal to e to the minus 2x times C 1 but we know that C 1 is 1 so I'll write times cosine of X plus C 2 times sine of X now let's take the derivative of this so that we can use the second initial condition so Y prime is equal to we're going to do a little bit of chain a product rule here so what's the derivative of the first expression it is e it's minus 2 e to the minus 2x and we multiply that times the second expression cosine of X plus C 2 sine of X and then we add that to just the regular first expression so plus e to the minus 2x times the derivative of the second expression so what's the derivative of cosine of X it's minus sine of X minus sine of X and then what's the derivative of C 2 sine of X well it's plus C 2 cosine of X cosine of X and let's see if we can do any kind of any kind of simplification here well actually the easiest way instead of trying to simplify it algebraic and everything let's just use our initial condition our initial condition is y prime of 0 is equal to 0 let me write that down the second initial condition was Y prime of 0 is equal to 0 so Y prime when X is equal to 0 is equal to 0 and let's just let's substitute X is equal to zero into this thing we could have simplified this more but let's not worry about that right now so what if X is zero this is going to be one right e to the zero e to zero is one so we're left with just with minus 2 right minus two times e to the 0 times cosine of 0 that's 1 plus C 2 times sine of 0 sine of 0 is 0 so that's just 1 plus 0 plus e to the minus 2 times 0 that's just 1 times minus sine of 0 sine of 0 is just 0 plus C 2 times cosine of 0 cosine of 0 is 1 so plus C 2 that simplified things didn't it so let's see we get 0 is equal to this is just 1 minus 2 plus C 2 or we get C 2 is equal to 2 add to add 2 to both sides C 2 is equal to 2 and then we have our particular solution we have our particular solution I know it's C 2 is equal to 2 C 1 is equal to 1 so actually let me erase some of this just so that we can go from our general solution to a particular solution let me erase some of this so we had figured out you can remember C 1 is 1 and C 2 is 2 that's easy to memorize so I'll just delete all of this I'll write it nice and big so our particular solution given these initial conditions were or are or is y of X is equal to this was a general solution e to the minus 2x times we solved for C 1 we got it's equal to 1 so we can just write cosine of X and then we solved for C 2 we figured out that that was 2 plus 2 sine of X and there you go we have our our particular solution to this sorry where did I write it to this differential equation with this and these initial conditions and what's neat is when we originally kind of proved this formula when he originally showed this form we had all of these eyes and we simplify we said see to it was a combination of some other constants times some eyes and we said oh we don't know whether they're imaginary or not so let's just merge them into some constant but what's interesting is this particular solution has no eyes anywhere in it and so well that tells us a couple of neat things one that it's you know you can if we had kept this c2 in terms of some multiple of eyes and our constants actually would have had eyes and they would have canceled out etc and it also tells us that that this formula is useful beyond formulas that just involve imaginary numbers for example this differential equation I don't see an eye anywhere here I don't see an eye anywhere here and I don't see an eye anywhere here but given this differential equation to get to this solution we have to use imaginary numbers in between and I think this is the first time if I if I'm remembering all my playlists correctly this is the first time that we used imaginary numbers for something useful we use it as an intermediary tool where we got a real a real a non imaginary solution to a real problem a non imaginary problem but we used imaginary numbers as a tool to solve it so hopefully you found that slightly interesting and I'll see you in the next video