Second order homogeneous linear differential equations with constant coefficients

in this video I'll show you guys how to approach to solve a second order linear differential equation with constant coefficients and in this case we want to focus on the right hand side to be 0 and when the right hand side is equal to 0 this is called a homogeneous situation and once again a B and C they are constants and in this case we don't want a to be 0 otherwise this cannot be done right so it's not taking water anymore alright as we can see we have y y prime which is the first derivative and this is y double Prime the second derivative and it seems like this is just a constant multiples of the original first and the second derivative right do we know a function so that its derivative is just a constant of its original I think so we do right on the side let me just show you if I start with y equal to well if apps that say e to some power times X or T in this case we just evaluate for second water your T let's say well e to the 3t well if you differentiate that you get Y Prime and you know this is going to give you e to the 3t the function prostate exactly the same but the tune will set I will have to multiply by 3 right so let me put it down right here right and you see these three stays and this is just the original so 3 times y right and likewise I can do it again y double prime is going to be we will keep this 3 times e to the 3t but then we multiply this 3 rifle at the review by the chain rule 3 times 3 is 9 + e to the 3t is the original and you see forever you have to eat with some power the first derivative the second derivative we are just going to be the constant multiple of the original right so what this is telling me is that this was suggest let me just put this down this is just that for the function we should have the form y equal to e to some power times t I don't know what this number should be early I just use sweet right so in general we put R times T and once again for the second water situation we usually use T because there are a lot of applications that enforced time and sin theta anyways this is my starting first of all we begin by saying Y is equal to e to the RT power and the idea is that I'm just going to go ahead differentiate this twice and then plug in home only we can squeeze out some conditions that will help us to solve for this kind of differential equation all right y prime is going to be e to the RT times R let me just put R in the front and then my double prime it's going to be this right here repeat that so our e RT but then we multiply by another arc which will be awesome job which is R squared and now let me put all this into their curricula T here we have a right double prime is odd we're into the R T and then we add it with B so we put on plus B Y prime is our e to the RT and then we continue plus C Y is e to the RT and this right here is equal to zero and now can we kind of reach our condition that people can we do every term has EGR key to the facility of course that e to the RT and that will give us a R squared plus B R plus C equals to 0 all right this is a exponential part right exponential function e to the something this right here we know is never 0 isn't it never certain just audio much better so when we have this quantity comes back on the tea-things e to the RT is never 0 so you can either D by D out what's going to forget about it because we just want to focus on this so that means we must have this part we must have a limit is really doubt we must have the situation that a R square plus B are plus C equal to zero in fact this right here is the condition that we need because from here this is pretty much just a quadratic equation isn't it quadratic equation in comes apart from here we can so far and we can just plug in 2 we are here and we can generate the building blocks of the solution and I'll show you get slamming by that and before I show you guys an example let me tell you guys that this equation here has a name this is called the characteristic equation and this is also called the a-pillar in equation and now let's go ahead solve this for y double prime minus bi y prime minus 6y is equal to 0 the first step is we have to take this and change you to its corresponding closed through state equation and check this out earlier when we start with a y double prime we will end up a R squared right so Y double prime corresponds to R squared that means right here the first term is going to give me for R squared and then we just continue minus 5 and the y prime correspond with our to change that to our right here and the minus 6 if you have why we didn't have any our right so it's just minus 6 this is it and now we just have to solve this quadratic equation and do whichever way that you would like and our faculty stuff for you guys and I'll show you guess with the tick titled factoring and to do so I want to ask myself what comes forgive me who are Square and let me tell gets a color combination which is 4 times R and what time so give me negative 6 and once again déméter gets the correct combination which is negative 2 past mystery in this box it's like this in this order well I'll convince you guys this is correct to do so you cross multiply who are times negative 2 which is negative they are and then you take 3 times aqua cheers to our is correct because negative AR plus 3 R is negative PI R so this is correct isn't it right okay so that's good and now we have to read the answers correctly strong on the boxes here to really answer if you go across so the first factor is going to be 4 plus 3 and the second one is r minus 2 and we still have equal to 0 of course alright for the first one we know R is equal to negative 3 over 4 and for the second one we know R is equal to 2 you see we end up with two different our values in this case we will end up with two different building blocks for the solution e 21st R which is this and the time T the second one is going to be e to the 2t right so let me just write down the putting plus for you guys first this right here it's going to give me e to the negative 3 over 4 T and the second one is going to give me e to the 2t and I'm going to explain to you guys what do I mean by building blocks first of all let me just show you guys that both of them will satisfy the original differential equation and just for simplicity purpose let me just check this one for you guys only ok and you can do this angle so let me just put down this right here let me just say this is the second one I guess so let me put on Y 2 y 2 equal to e to the 2t identify I have to do is differentiate is which is going to give me 2 e 2 T right and then 3 again Y 2 double prime which is for e 2 T right and now I will plug in all this into the original and you see it will give me 4 5 and y double prime which is that so we multiplied by 4 e 2 T and the minus 5 y prime which is that which is x 2 e 2 T and the minus 6 Y which is that e to the 2t like this do we enter 0 yes we do this is 16 minus 10 which is 6 and then minus 6 or support everything and the series equals zero so check this right here definitely works isn't it and you can do exactly the same thing for this but you can also take my work word you will work out nicely as well alright now this are just a function part what we'll be doing is that we'll remember when were solving differential equations we have that constant right and keep in mind whenever we have the second order differential equation we'll have two different constants we also put on plus C word divided by kya know the truth is we must apply the function part by c1 and c2 so this right here t1 and this right here - so let me just demonstrate this right here on the side for you guys real quick because now you see 4/9 multiply this function by C so let me just put on scene black edges guess what I will just have to multiply everything by C Ethernet C and C and then pretty much this part will now have a speed right here and this function here this part here has a see here as well this right here is just x feet right just plugging everything accordingly guess what when they have this C here I was to end up with zero on the left hand side of course T or equal to zero the point of that is to show you these are the places where tgop you wanna see two are just the multiple of the function part you want to see to adjust the multiple of the building blocks of the structures at the end the overall solution is that you are just going to add them together and this is it this is the solution for that why it's equal to c1 e to the negative Z of F of T plus c2 e to the 2t that's it and the truth is that we can just add in together and this is aa prostitution it's because for you differentiate this right the emergence I'm differentiating the Y we just have to differentiate the first and a second to differentiate this and depression that do again to again plugging plugging plugging you get zero is equal to zero once again so this is the idea just to summarize this real quick all you have to do to keep everything simple we have the courage to escape equation solve the quadratic equation R if you get two different our values but e to the first R times P e to the second our party and be sure your multiplies u 1 and C 2 course one IDI and then add them together this is the general solution this is the first situation if you watch my next video I will show you what happens when I have these two our values being the same

Channel: blackpenredpen

Views: 155,849

Rating: 4.9182363 out of 5

Keywords: Second order linear differential equations, 2nd order linear differential equations with constant coefficients, second order homogeneous linear differential equations, Homogeneous second order linear differential equations with constant coefficients, intro to second order differential equations, blackpenredpen, characteristic equations, auxiliary equations, auxiliary equations with distinct roots

Id: iQoAbO_xlFs

Length: 11min 43sec (703 seconds)

Published: Thu Mar 09 2017