Repeated roots of the characteristic equations part 2

let's do another problem with repeated roots so let's say our differential equation is the second derivative of Y minus the first derivative plus 0.25 that's what's written here 0.25 y is equal to zero and they've actually given us some initial conditions they said that y 0 is equal to 2 and Y prime of 0 is equal to 1/3 so like we've done in every one of these constant coefficient linear second order homogeneous differential equations let's get the characteristic equation so that's R squared minus minus R plus 0.25 or we could even call it plus point 1/4 so let's see when I just inspecting this it always confuses me when I have fractions so it becomes very hard to factor so let's just do the quadratic formula so the roots of this are going to be R is equal to negative B well B is negative 1 so negative B is going to be 1 plus or minus the square root of B squared B is negative 1 so that squared is 1 minus 4 times a which is 1 times C well 4 times 1 times 0.25 that's that's one uh-huh so notice then when you have a repeated root this under the square root becomes 0 that makes sense because it's this plus or minus in the quadratic formula that gives you two roots whether they be real or complex but if the square root is 0 you're adding plus or minus 0 and you're only left with one root anyway we're not done yet and what's the denominator of the quadratic equation to a so a is 1 so over 2 so our one repeated root is 1 plus or minus 0 over 2 or it equals 1/2 and like we learned in the in the last in the last video you might just say oh well maybe the solution is just Y is equal to C e to the one-half Exce but like we pointed out last time you have to initial conditions and this solution is not general enough for two initial conditions and then last time we said okay this isn't general enough maybe some solution that was you know some function of x times e to the one-half X maybe that would be our solution we said it turns out it is and so that more general solution that we found that we figured out that V of X is actually equal to some constant plus x times some other constant so our more general solution is y is equal to c1 times e to the 1/2 X plus c2 times X e to the one-half X forgot the X here they draw a line here so you don't get confused anyway that's the reasoning that's how we came up with this thing but and and it is good to know because later on when you want to know more theory of differential equations and that's really the whole point about learning this if your whole goal isn't just to pass an exam it's good to know but when you're actually solving these you could just kind of know the template if I have a repeated root well I just put that repeated root twice and one of them gets an X in front of it and they have two constants but anyway this is our general solution and now we can use our initial conditions to solve for c1 and c2 so let's just figure out the derivative of this first so it becomes easy to substitute in for c2 so y Y prime is equal to let's see it's equal to 1/2 C 1 e to the 1/2 X plus now this becomes a little bit more comfortable here so let's see plus c2 times derivative of X is 1 times e to the 1/2 X that's the product rule plus the derivative of e to the 1/2 x times X so that's 1/2 X e to the 1/2 X or we can write I don't want to lose this stuff up here we can write that the it equals let's see I have see if I can I have one half so I have C two times e to the one-half X and I have one half times C one e to the one-half X so I could say it's equal to e to the one half x times C 1 over 2 that's that plus C 2 that takes care of these two terms plus C 2 over 2x e to the one-half X and now let's use our initial conditions let me actually clear up some space because I think it's nice to have our initial conditions up here so we can see them so let me delete all this stuff here that hopefully make sense to you by now we know the characteristic equation we figured out the general so look I don't want to erase our initial conditions we figured out the general solution was this I'll keep our general solution there and so now we just substitute our initial conditions into our into our general solution and the derivative of the general solution and and hopefully we can get meaningful answers so substituting into our general solution Y of 0 is equal to 2 so Y is equal to 2 when X is equal to 0 so C 1 when X is equal to 0 all the e terms become 1 right this one will become 1 and then notice we have an x e to the 0 so now this X is 0 so this whole term is going to be equal to 0 so we're done C 1 is equal to 2 that was pretty straightforward this X actually made it a lot easier so C 1 is equal to 2 and now we can use the derivative so let's see this is this is the first derivative so let's see and I'll substitute C 2 C 1 in there so we can just solve for C 2 so our first derivative is Y prime is equal to let's see see C 1 1 half plus C 2 so it's it's well I'll write this first it's equal to 2 over 2 so it's 1 plus C 2 times e to the 1/2 X plus C 2 over 2 times X e to the 1/2 X there was an X here so when we when X is equal to 0 Y prime is equal to 1/3 so 1/3 is equal to well X is equal to 0 this will be 1 so it's equal to 1 plus C 2 and then this term when X is equal 0 this whole thing becomes 0 right because this X just cancels out the whole thing it multiplied by 0 you get 0 so then we get 1/3 is equal to 1 plus C 2 or that C 2 is equal to 1/3 minus 1 is equal to minus 2/3 and now we have our particular solution and let me write it down and put a box around it so this was our general solution our particular solution given these initial conditions for this repeated root problem is y is equal to c1 we figure that out to be 2 fairly quickly 2 e to the 1/2 X plus C 2 C 2 is minus 2/3 so minus 2/3 X e to the 1/2 X and we are done there is our particular solution so once again kind of the proof of how do you get to this you know why is there this X in there and it wasn't a proof it was really more of just to show you the intuition of where that came from and it did introduce you to a method called reduction of order to figure out what that function V was which ended up just being you know c1 plus c2 times X but all of that can be pretty complicated but you see that once you know the pattern once you you know that this is going to be the general solution they're pretty easy to solve characteristic equation get your general solution figure out the derivative of the general solution and then substitute your initial conditions to solve for your constants and you're done anyway I'll see you in the next video and actually we're going to start solving non-homogeneous differential equations see you soon