Repeated roots of the characteristic equation

let's say we have the following second-order differential equation we have the second derivative of y plus four times the first derivative plus 4y is equal to zero and we're asked to find the general solution to this differential equation so the first thing we do like we've done in the last several videos we'll get the characteristic equation that's R squared plus 4 R plus 4 is equal to 0 this one's fairly easy to factor we don't need the quadratic equation here this is R plus 2 times R plus 2 and now something interesting happens something that we haven't seen before the two roots of our characteristic equation are actually the same number R is equal to minus 2 so you can say we only have one solution or one root or a repeated root however you want to say it we only have one R that satisfies the characteristic equation so you might say well that's fine maybe my general solution is just Y is equal to some constant times e to the minus 2x using my one solution and my reply to you is this is a solution then and if you don't believe me you can test it out but it's not the general solution and why do I say that because this is a second-order differential equation and if someone wanted a particular solution they would have to give you two initial conditions the two initial conditions we've been using so far are what Y of 0 equals and what Y prime of 0 equals you could have you know but you could they could give you what Y of 5 equals who knows but in general when you're having a set when you have a second order differential equation they have to give you two initial conditions now the problem with this solution and why it's not the general solution is if you use one of these initial conditions you can solve for C right you could you'll get an answer you'll solve for that C but then there's nothing to do with the second initial condition and in fact except for only in one particular case whatever C you get for the first initial condition it won't be that this equation won't be true for the second initial condition and you could try it out I mean if we said Y of zero is equal to a and y prime of 0 is equal to I don't know five a let's see if these work if Y is 0 is equal to a what's so that tells us that a is equal to C times e to the minus 2 times 0 so e to the 0 or C is equal to a right so if you just had this first initial condition say fine my particular solution is y is equal to a times e to the minus 2x but let's see if this particular solution satisfies the second initial condition so what's the derivative of this Y prime is equal to minus 2a e to the minus 2x and it says that 5 5 a this initial condition says that 5 a is equal to minus 2 a times e to the minus 2 times 0 so e to the 0 or another way of saying that E is 0 is just 1 it says that 5 a is equal to minus 2 a which we know is not true so no note when we only have this general silora pseudo general solution it can only satisfy generally one of the initial conditions and if we're really lucky both initial conditions so that at least gives you an intuitive feel of why this isn't the general solution so let me clean that up a little bit so that I could have a feeling off to use this real estate so what do we do so we can use a technique called reduction of order and it really just says well let's just guess a second solution you know in general when we first thought about these linear constant-coefficient differential equations we said well e to you know our X might be a good guess and why is that because all of the derivatives of e are kind of you know multiples of the original function and that's why we used it so if we're looking for a second solution it doesn't hurt to kind of make the same guess in order to be a little bit more general let's make our guest for our second solution I'll call this G for guess say it's some function of x times our first solution e to the minus 2x I could say some function of x times C times e to the minus 2x but the C is kind of encapsulated it could be part of this some random function of X so let's be as general as possible so let's assume that there is this is a solution and and then substitute it back into our original differential equation and see if we can actually solve for V that makes it all work so before we do that let's get its first and second derivatives so the first derivative of G is equal to well this is product rule and I'll drop the V of X we know that V is a function and not a constant so product rule derivative of the first V prime times the second expression e to the minus 2x plus the first function or expression times the derivative of the second so minus 2 times e to the minus 2x so x minus 2 e to the minus 2x or just to write a little bit neater G prime is equal to V prime e to the minus 2x minus 2v e to the minus 2x now we have to get the second derivative I'll do that in a different color just to fight the monotony of it so the second derivative we're going to do the product rule twice derivative of this first expression it's going to be V prime prime e to the minus 2x minus 2 V prime e to the minus 2x that was just the product rule again and then the derivative of the second expression is going to be and see the derivative the first one is V prime so it's going to be minus 2 V prime e to the minus 2x times or pono plus I'm just probably rule 2 minus 2 times minus 2 plus 4v e to the minus 2x I hope I haven't made a careless mistake and we can simplify this a little bit so we get the second Rivet of G which is our guests solution is equal to the second derivative of V prime e to the minus 2x minus 2 V Prime no minus 4 sorry I could we have minus 2 minus 2 minus 4 V prime e to the minus 2x plus 4v e to the minus 2x and now before we substitute it into this we can just make one observation that'll just make the algebra a little bit simpler notice that G is something times e to the minus 2x G prime is we could factor out an e to the minus 2x and G prime prime we can factor out an e to the minus 2x so let's factor them out essentially so we can when we write this we can write so the second derivative is G prime prime which we can write as and I'm gonna try to do this it's e to the minus 2x times the second derivative so now we can get rid of the e to the minus 2x terms so that's V prime prime minus 4v prime plus 4v right if I just distribute this out I get the second derivative which is this plus 4 times the first derivative and I'm also going to factor out the e to the minus 2x so plus 4 times this so it's going to be plus 4v prime minus 8v right minus 8 V once again I factored out the e to the minus 2x right plus 4 times y so plus 4 times we factored out the e to the minus 2x so plus 4 times V I did that because if I didn't do that I'd be writing e to the minus 2x and I'd probably make a careless mistake and I'd run out of space etc but anyway I essentially to get this I just substituted the second derivative the first derivative and G back into the differential equation and we know that that has to equal 0 let's see if we can simplify this a little bit more and then hopefully hopefully solve for V so let's see some things are popping out at me so I have plus 4 V plus 4 v plus 4 V that's plus 8 V minus HV right so plus 4 minus 8 plus 4 those cancel out it's plus 8 minus 8 those cancel out and then I also have minus 4 V prime plus 4 V prime so those cancel out and lo and behold we've done some serious simplification it ends up being e to the minus 2x times V prime prime we could call it V prime prime of X now that we've saved so much space is equal to 0 we know this could never equal to 0 so essentially we have now established that this that this expression has to be equal to 0 and we get a separable second-order differential equation we get that the second derivative of V with respect to X or it's a function of X is equal to 0 so now we just have to differentiate both sides of this equation twice you differentiate once you get what V V prime V prime of X is equal to let's call it c1 and if we were to take the antiderivative both sides again we get V of X is equal to c1 x c1 X plus some other c2 right now remember what was our guess our guess was thought that our solution was going to be or our general solution was going to be some arbitrary function V times kind of that first solution solution we found e to the minus 2x and when we actually took that guess and we substitute it in we actually were able to solve for that V and we got the V is equal to this so this is interesting so what is G or what is our guest function equal it's no longer and guess we've kind of established that it works G which we can call our solution is equal to V V of x times e to the minus 2x well that equals this C 1 X plus C 2 e to the minus 2x that equals C 1 X e to the minus 2x plus C 2 e to the minus 2x this is an e and now we have a truly general solution we have two constants so we can satisfy two initial conditions and if you are looking for pattern this is the pattern when you have a repeated root of your characteristic equation the general solution is going to be is going you're going to use that e to the that whatever root is twice but one time you're going to have an X in front of it and this works every time for second order homogeneous constant coefficient linear equations I will see you in the next video