For more information, please see full course syllabus of Differential Equations

For more information, please see full course syllabus of Differential Equations

### Repeated Roots & Reduction of Order

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Lesson Overview 0:23
- If the Characteristic Equation Has a Double Root
- Reduction of Order
- Example 1 7:23
- Example 2 9:20
- Example 3 14:12
- Example 4 31:49
- Example 5 33:21

### Differential Equations Online Course

### Transcription: Repeated Roots & Reduction of Order

*Hi, you are watching www.educator.com and I am Will Murray, we are learning about differential equations and today we are going to study second-order equations when the characteristic equation has repeated roots.*0000

*We are also going to learn a method called reduction of order for finding a second solution to a differential equation if you already know the first solution., let us jump right in.*0013

*The equations we are going to be studying are linear second-order homogeneous constant coefficient differential equations, we have learned what all those words mean before.*0024

*But I will reminder quickly linear means you just have terms of Y′ Y, Y″ and Y′ and you do not have the multiplied together, you do not have the raised powers, you just have nice of nice linear configuration of those terms.*0034

*Homogeneous means the right hand side is = 0 constant coefficients that refers to the A, B, C see here, those are all constants, those are not functions of T, R, X or anything else.*0052

*Those are the kinds of differential equations we are going to be studying and we learned that to solve those equations you can set up the characteristic polynomial which is this polynomial AR ^2 + BR + C is = 0.*0064

*And that is just a quadratic equation so you can solve that using whatever techniques you learn in algebra, if you are lucky you can factor it if not you might have to go to a quadratic equation to find the roots.*0080

*But in any case generally you get 2 roots R1 and R2 and what we are studying today is the case when those 2 roots are actually equal to each other and it turns out that changes the form of the solution.*0091

*If you do have a double root then we are just going to call it R since there is only one of them, about the general equation to the differential equation is C1 e^RT, now that is the same as we have been studying before.*0108

*But you do not just tack on a C2 e^ RT because that would be a copy of the first solution what you do is find C2 TE^RT, that is the sort of new element here when you do have a double root your first solution is e^RT.*0124

*The second solution is Te^RT and then from that point on you try to solve for the 2 coefficients using any initial conditions which are given, the coefficients are C1 and C2 and you try to use initial conditions.*0144

*They are usually given as y0 and Y′ 0 and you get 2 equations and 2 unknowns of course the 2 unknowns are C1 and C2, you will try to solve for those 2 coefficients and then you will be finished with solving the initial value problem.*0161

*We will see some examples of that before we get started I want to mention this other technique called reduction of order which was actually how we found the second solution.*0180

*The idea of reduction of order is that by one way or another you know one solution to a differential equation, maybe somebody is giving you one solution to a differential equation and you are trying to find a second solution to the differential equation.*0191

*That is what reduction of order is all about, the ideas you try to write your differential equation in this form Y″ of T + P1 of T x Y′ + P2 T x yT is = 0 and the important thing here is that there is no coefficients in front of the Y″.*0208

*If you have an equation that does have a coefficient in front of the Y″ then you want to divide that away, you just get Y″ by itself.*0231

*We will see some examples of that and it is a place where students often make mistakes, you want to be very careful to divide away that coefficient in front of the Y″.*0242

*And then here is what you do with that, we are going to find a second solution y2T and I have written it in the form of the vT x 1T x Y1 of T, the Y 1 of T is the known first solution and VT is a function of T that we are trying to find.*0253

*How do we find this VT well we are going to solve this first order equation down here, this is a first-order differential equation for new variable W, the equation that we get is Y 1 that is unknown Y1 is known.*0286

*And then we have W′ we are solving for W + 2Y 1′ + P1 Y1 that P1 here that is also known because that comes from the differential equation that comes from right here, this P1 of T becomes that P1 is down here .*0306

*P1 Y1 x W is = 0, this is a differential equation for W, we are going to solve this for W(T), the key thing here is that this is a first order equation for W.*0328

*It is easier to solve we can use some of our old techniques to solve it such as separation of variables or linear equations to solve that first-order equation.*0348

*That is usually not too hard to solve for W and then what we will do is we will integrate W to get the V, V is the integral of WT DT, then we will know V and then in turn we all use V to form Y2(T) by multiplying V x Y1.*0361

*It is a several step process here, we will set up this first-order equation, we will solve it for W and then we will integrate W to get V and then our second solution will be V x Y 1.*0386

*We will work through this with some examples and you will get the hang of it but that this is originally where that second solution came from when we are talking about e^RT and Te^RT.*0402

*Remember we set our generic solution when you have a double root is Ygen =C1 e^RT + C2 TE^RT, this reduction of order technique is how we found that second solution there, that is that is the origin of that second solution.*0413

*And we will see how we can use the same method to solve other differential equations as well, let us get started here with our first example, we are going to find the general solution to the differential equation Y″ -4Y′ + 4 Y = 0.*0437

*This is a pretty straightforward second-order constant coefficient linear differential equation, we solve this by setting up the characteristic equation r^2 -4r +4 is = 0 and then we try to solve that for R you can use a quadratic formula or you can factor.*0455

*This one is pretty easy to factor, we get r -2 ^2 = 0, our roots are 2 and then we had 2 again it is a double root, our general solution to the differential equation is C1 e^2T and then instead of just having a second e^2T we are going to go with TE^2T.*0475

*That is our general solution to the differential equation and that is where we stopped because we do not have any initial conditions, we just stop with the general solution and we do not go through and try to find the constants.*0506

*Just to remind you how that worked here, we set up the characteristic equation, we factored it to get the roots since we found a double root we have C1 e^RT and C2 x TE^RT and then we stop there because we do not have an initial condition that will let us solve for the constants.*0520

*We are going with this same equation in the next example, you might want to keep this in mind Y″ -4Y′ + 4Y = 0 and our general solution we are going to take this a step further in the next example.*0545

*On our second example we got that same equation again Y″ -4Y′ + 4Y = 0 and then we got a couple of initial conditions now y(0) =3 NY′ 0 is = 8.*0561

*Let us remember that we already solve this one before, we got the general solution is equal to C1 e^2t + C2 x TE ^2T, that is from example 1, you can go back and check that out if you have not just worked through it.*0577

*Now we are going to use the initial conditions to figure out what the constants should be, first initial condition y(0) = 3, y(0) if I plug in 0 here that is C1 e^0 + C2 x 0 e^ 0 which of course is just C1, e^0 is 1.*0605

*And then C2 x 0 is just 0, that was supposed to be equal to 3, that tells me right away that C1 is equal to 3 and then I have to look at Y′(0) when you take the derivative of my general solution Y′ is 2C1e^T because the 2 pops out.*0629

*Now C2 I have to take the derivative of TE^2T, I have to use the product rule to take that, first I will do T x derivative of e^2T which is 2e^2T + e^2T x derivative of T which is just 1, that is e^2T.*0653

*Then I'm going to plug in T = 0 Y′ 0 is = 2C1, now e^2T, when T = 0 is just 1, that turns into 1 + C2 0 × 2e^ 0 + e^0 and of course this just cancels away the 0 and we got 2C1 + e^ 0 is still 1, C2 = 8.*0680

* But I have already figured out that C1 = 3, that is 3 and I have got 6 + C2 is = 8, C2 = 2, I'm going to plug those back into my general solution and I will get my specific solution to the initial value problem Y= C1 was 3, 3e^2T + C2 is 2 x e^2T.*0715

*That is my specific solution to the initial value problem, just to recap what happened there, my general solution that was something we figured out in example 1, we already had the general solution, we got that by using the characteristic equation and factoring it.*0755

*We got e^2T + C2 TE^2T and then we plugged in y(0) our first initial condition in there, we plugged in 0 for T and we figured out that C1 =3 and then to use the second initial condition Y′ 0, I had to take the derivative of the general solution.*0777

*And that was easy on the first part, the derivative of e^2T is 2e^2T, a little harder on the second part because I had to use a product rule to find the derivative of TE ^2T, I used the product rule to expand that out into 2 terms.*0802

*And then I plugged in T =0 to both parts and I got 2C1 + C2 after I simplified, I already knew what C1 was and I got C2 = 2 and I plugged in my values of C1 and C2 into the general solution and that is how I got Y =3 e^2T + 2Te^2T.*0821

*That is my solution to the initial value problem, in example 3 we are going to start with a more complicated differential equation TY″ - 2T +1Y′ + 4Y is = 0 and there is 2 requirements here.*0847

*We are going to check that Y1 is = e^2T is a solution to the equation and we are going to use reduction of order to find a second independent solution to the equation, we will start off with part A here.*0869

*We are just checking that Y1 is = e^2T, we are going to check that that is a solution to the equation, I'm going to plug that in, I will plug into the DE the differential equation.*0883

*We are going to use Y1 is e^2T and we will need to know its derivatives Y1′ is 2e^2T and its second derivative is 4e^2T, I'm going to take each one of these and plug them into the differential equation.*0904

*T x Y″ is 4e^2T -2 x T +1 Y′ is 2e^2T +4 x Y is e^2T and we are really hoping that this will come out to be equal to 0, I have not confirmed that yet but we have been given that it is a solution, we are just kind of checking that now.*0926

*And first thing I notice right away is that I have e^2T everywhere, I'm going to divide that away e^2T I can divide both sides by e^2T else to 0 on the right-hand side, I get 4T - this 2 and this 2 together give me a 4 - 4 x T + 1 + 4.*0958

*And we are checking to see whether that comes out to be equal to 0 and I get 4T - 4T - 4 + 4 and I see that the 4T cancel each other and the 4 cancel each other and I get 0=0.*0983

*That does indeed check which means that Y1 =e^2T is indeed a solution to the differential equations, I have one solution it was given to us externally we just checked that it was a solution.*1005

*We are going to use reduction of order to find a second independent solution, let us keep going on the next page with that, our equation is still TY″ - 2T + 1Y′ + 4Y = 0.*1020

*And our first solution given to us is Y1 =e^2T and let me remind you of the way reduction of order works, it was the equation I gave you back in the lesson overview a few minutes ago.*1039

*What we do is we solve this first-order differential equation Y1 x W′ + 2Y1′ + T1 Y1 x W = 0 and remember that comes from looking at the original differential equation Y″ + P1 1Y′ + P2 Y = 0.*1054

*Now a key difference between that equation and what we have is that here we have no coefficient in front of the Y″ but here we do have this T in front of the Y″.*1092

*We have to do is divide that T away before we can use our formula to implement reduction of order, what I'm going to do is take this equation and I'm going to divide both sides by T, I get Y″ - 2 x T + 1/T Y′ + 4/T = 0.*1107

*The point is that is it now its agrees with this form that we are going to use to develop our reduction of order and in particular all of this stuff right here is the P1, P1 is -2 x T +1/T and if I expand that out that is -2 x T/T is just 2.*1139

*And 2×1/T is 2/T, -2-2/T that is the P1 that we are going to be using here in our formula for reduction of order, let me go ahead and set that out. *1165

*Y1 is our e^2T, e^2T x W′ + 2 x Y1′, Y1′ is 2e^2T, 2 x Y1′ is 2 x 2e^2T + P1 Y1 now we figured out what P1 was over here, I'm going to make that a - -2- 2/T.*1182

*I'm going to factor out that - and make it a + here, because we R multiplying the whole thing by Y1 which is e^2T, all of this x W is = 0, the point here as we got a first order differential equation that we R going to solve for W.*1230

*And one thing I notice here, I'm going to try to clean this up and simplify it, it simplifies pretty well I noticed that I have got an e^2T everywhere here, e^2T in every term.*1248

*The first thing I'm going to do is divide out all those e^2T and I will get W′ + 4 just 4 now, because we divided away the e^2T -2, I'm going to distribute the negative side again -2/T.*1259

*All that x W is = 0 all the e^2T R gone now, that is quite convenient 4 - 2 is 2 - 2/T, that x W is = 0 we had W′ + all that and now this is actually a separable differential equation, I'm going to move these terms over to the other side.*1282

*W′ is equal to, if I move it over to the other side that is going to reverse their sign so it will be 2/T - 2W, I change which one was positive and which one was negative because it moved over to the other side.*1309

*And I will divide both sides by W′/W is equal to 2/T -2, I'm going to right this W′ as DW by DT and then multiply DT by both sides, I will get the DW/W is = 2/T -2 DT.*1329

*What I'm doing right now is kind of following the steps to solve a separable differential equation, if this seems a little fuzzy to you, if you are not quite sure what is going on here, you do not remember too well.*1361

*There is a earlier lecture in this series on differential equations specifically on separable differential equations, if you just go back to manual look for separable differential equations you get a whole lecture on these techniques and you can review them.*1372

*What I'm doing right now make more sense and I'm going to integrate both sides, I'm going to give myself a little more room for an integral sign here and left hand side DW/W is the natural log of W.*1386

*On the right-hand side I have to integrate 2/T -2, that is 2 x natural log of T - 2T and when you R solving at this point we do not need to add a constant.*1403

*Actually really to do that anywhere in this procedure, we do not need to add this constant because we R just going for a single second solution to this differential equation.*1421

*Now I need to solve for W, to solve for W I will raised e to both sides and get W is = e^2 natural log of T x E ^-2T that is using the laws of Exponents there remember exponents convert addition or subtraction into multiplication.*1431

*And e^2 natural log of T that is E^natural log of T ^2 and that is just T^2 e^-2T, be careful with that. While students will say e^2 natural log of T, they will say that the e in the natural log cancel each other out so it is just 2T, that is not correct at all.*1462

*You have to make sure that you distribute that 2 up into an exponent before you cancel the e in the natural law, what we just did there was we found W but remember what we are actually shooting for is this the V, which is the integral of W(T) DT.*1490

*I have to integrate T^2 e^-2T and I'm going to do that integral by parts and there is little tabular trick that we learned back in the calculus lectures, T ^2 e^-2T.*1510

*I 'm going to do quickly and again if you do not remember how to do this you might want check out the calculus 2 lectures also here on www.educator.com and there is a lecture on integration by parts and you can review how to do something like this really quickly*1524

*I'm going to write down derivatives down the side 2T and 2 and 0 and I'm in going to write integrals of e^-2T so -1/2 e^-2T and positive 1/4 e^-2T and -1/8 e^-2T make little diagonal lines here.*1540

*And on each diagonal line up in a + - + and what I get is my integral, I'm going to multiply along the diagonal lines, I get -1/2 T^2, e^-2T, - 2T/4 -1/2 T e^-2T on the second diagonal line, on the third diagonal line I get -2/8 -1/4 e^-2T.*1566

*That is my function V and then Y2 is V x Y1 and Y1 was e^2T, we are going to be multiplying V by e^2T that just cancels out the e^-2T's, it is -1/2 T^2 -1/2 T -1/4 and finally what we have here is a second solution to our differential equation.*1601

*Now that is a little bit messy and we can actually clean it up if we want since this is a linear differential equation, the linear homogeneous differential equation, homogeneous means the right-hand side is equal to 0.*1640

*It means you can multiply solutions by constants if you like and so we could clean this up and multiply by constant just to get rid of some of the fractions of a negative signs, it is okay here we want to multiply.*1658

*Now I'm looking at those fractions and I'm looking at those negative signs and I'm thinking a way I can get rid of all of them would be -4 and we would get Y = -1/2 x - 4 would be 2T ^2 -1/2 x -4 = 2, -1/4 x -4 will be +1.*1675

*That would be another way we could offer a second solution 2T^2 + 2T +1 will also be a second solution to that differential equation and that one has the advantage that has nice positive whole number coefficients and no negatives no fractions.*1699

*We are done with example 3 here with finding our second solution here but let me just kind of quickly remind you of the steps we followed there, first thing we did was we manipulated the differential equation into the form we needed.*1717

*In particular we had to get rid of that coefficient in front of the Y″, that is why we divided away this T and we got a version of the equation that had no coefficient in front of the Y″.*1730

*Then we got for our P1 was everything in front of Y′s our P1 was -2T + 1/T, then we went back to this formula from the lesson overview, you will find this formula a few pages back in the lesson overview.*1745

*And we plugged everything in y1 was e^2T, remember that was given to us up here, it is not obvious where that came from but it was given as one of the parameters of the problem.*1763

*We plugged in Y1, we plugged in Y1′, we plugged in what P1 was and Y1 again and then we got this first order differential equation for W reduction of order.*1775

*Remember you means you go from a second-order differential equation down to a first order differential equation which is usually easier to solve.*1788

*We got this first order differential equation, we simplified it down, we separated the variables and then we integrated both sides this is following the earlier lecture on our differential equations series on separable equations.*1796

*We integrated both sides and then to solve for W, we raised e to both sides that is we are doing at this step, we had to be careful here , e^2 natural log of t is not 2T, it is T^2.*1814

*And then that gave us our W and to get V we have to integrate W and that is what I was doing here I was the integrating by parts and this is a little shorthand technique for integration by parts that we learned back in the calculus 2 lectures here on www.educator.com.*1830

*We got a fairly nasty expression for V, our second solution is V x Y1, again that is coming from the lesson overview that you will find this back a few pages back just in this lecture.*1851

*Y1 was e^2T, we multiply e^2T by everything here, which is very conveniently cancels off all the -2T's and we got our second solution an then we noticed that since it is a linear homogeneous equation, we can multiply by a constant to clean it up if we want.*1863

*We multiplied it by -4 just to give ourselves slightly nicer coefficients, let me emphasize that is not a necessary step, this would certainly be a valid second solution.*1882

*It is correct it is a solution to the differential equation but if you want something that looks a little nicer, then this is also a solution and we got that just by multiplying our original second solution by -4.*1892

*Let us keep moving with another example here, we want to find a general solution to the differential equation Y″ + 2Y′ + Y = 0 , we are going to do that is set up the characteristic equation as usual R^2 + 2r + 1=0.*1908

*And then we will factor that, that one factors very easily into r +1 ^2 = 0, we get the value of our root here as r -1 and that is a double root, we just see -1 again when we go for the second root.*1931

*We are going to invoke our general formula, Y general is = C1 e^-1T and then C2, I can not just put e^-T again because that is a copy of my first solution, I might go with Te^-T.*1947

*Since we do not have any initial conditions this time, we just stop there with the general solution, we are done with that example, just a recap we set up the characteristic equation, factor it, find a double root.*1968

*The root becomes the coefficient of T in the exponent there but because we have a double root we have to introduce this factor of T to get the second solution there.*1988

*Example 5 we have a more complicated differential equation T^2 Y″ - T x T+ 2Y′ + T + 2Y = 0 and for the first step we are going to check that y 1 =T is a solution to the differential equation.*2003

*We are going to use reduction of order to find a second independent solution, to solve that for part A we are given that Y1 =T might be a solution, if Y1 = T Y1′ would be 1 and Y1″ derivative of 1 is 0.*2022

*We are going to plug all those into the differential equation and see if it checks out to get T ^2 x 0 - T x T +2 x Y′ would be 1 + C + 2 x Y, our YT and we are checking to see if that is equal to 0.*2052

*The 0 term drops out here, we got - T x T + 2 + T2 x T and we are checking to see if that is equal to 0, in each term here we got T x T + 2 1 -1 positive and it just simplifies down to 0 = 0, that checks.*2079

*What we did there was we just found the derivatives of this solution, that hypothetical solution that we are given, plugged them into the differential equation, plugged them in for Y″ for Y′ for Y, simplify it down and check that we did indeed get 0 = 0.*2107

*It is a solution to the differential equation, that was sort of the easy part there, the harder part is going to come in part B where we are going to use reduction of order to find a second solution to the differential equation.*2128

*Let me remind you that with reduction of order we need to have an equation of the form Y″ + P1 Y′ + P2 Y = 0 and we do not have that yet because we have this coefficient T ^2 in front of the Y″.*2145

*Again I'm going to divide that coefficient away that we will get the right form to invoke our formula for our reduction of order, I will get Y″ - T x T + 2/T ^2 Y′ + T x R + T + 2/ T ^2 Y is = 0.*2165

*If I divide the right-hand side by T^2 is still 0, now I have got it into this form and my P1 I can actually simplify that is negative, if I cancel that T with that T ^2 is -T + 2/T and I could even simplify that some more.*2193

*That is -1 + 2/T, I'm going to work with that and I'm going to use the formula that we had for reduction of order, let me remind you what that was.*2220

*We went over this in the lesson overview but that was quite a few minutes ago so I will remind you what that formula was, it is a Y1 x W′ + the quantity 2 Y1′ + P1 Y1 x W = 0.*2250

*The idea here is that we do not know what W is but we know everything else we can fill it in and then solve a first order differential equation for W, I'm going to fill in everything I know Y1 is T, that is T x W′ + now 2 x Y1′.*2271

*Y1′ is just the derivative of T, that is 2 × 1 + P, okay P1 is negative, I will do -1 + 2/T and Y1 is T x W = 0, I get TW′ + let us simplify things in the parentheses here.*2293

*2-, now I'm going to distribute this T and I will get 2 - T -2/T x T -2, all this x W= 0 and simplify things to further, TW′ + now the 2's cancel and I will just get -TW = 0.*2321

*And I can divide by T I will get W′ - W= 0, I'm going to write W′ as DW by DT and maybe bring the W over the other side, equals W and now I will multiply and divide.*2348

*I'm separating the variables here, I'm setting up a separable differential equation, if you do not remember exactly how to solve a separable differential equation you might check back, there is an earlier lecture here on www.educator.com in the differential equation series.*2366

*You will see one titled separable differential equations, you can look that up and practice some examples of these if this is a little fuzzy for you.*2381

*If I multiplied both sides by DT, I will get DT on the right, on the left if I divide by W, DW/W I separate it I got all the T's on one side, all the W's on the other, take the integral both of sides, get natural log of W is equal to the integral of DT is just T.*2390

*W is equal to e^T and remember from there we try to find V, which is the integral of W, which is the integral of e^T DT, which in this case is just e^T itself and finally our Y2 remember is V x Y 1 that is always the formula.*2411

*You can look that up in the lesson overview at the beginning of this lecture, V x Y 1 is e^T x T and I will write that a little nicer as Te^ T that is our second solution to the differential equation.*2437

*We are done with that example but let me recap and just remind you how all the steps worked out there, first thing we had to do was make it fit our generic form to use reduction of order.*2469

*And since there is nothing in front of Y″ in the generic form that meant we had to get rid of this coefficient of the Y″ here.*2471

*We got rid of it by dividing it away which is where we got that T^2 in the denominator here and here and that simplified a little bit our P1, you have to include the negative sign, the T and T^2 cancel down to just T in the denominator.*2481

*And then it simplifies a little more into -1 + 2/T and then we invoke this generic formula for reduction of order, you always use this formula Y1 W′ +2 Y1′ + P1 y1 x W = 0.*2497

*Plug-in what Y1 is, that was given to us, plug-in what Y 1′ is that is 1, plug in the P1 we just figured out, simplify it down, we get to a fairly easy separable differential equation for W, solve that for W we get W.*2514

*To get V we have to integrate W, in this case integrating e^T does not actually change anything, V is also e^T and then to finally to get your second solution, you take that V and multiplied by Y1.*2537

*Our Y1 was T that came from here and finally Y2 is just Te^ T, that is the second solution to our differential equation here, we use reduction of order to start with one given solution and find the second solution.*2543

*That is the end of our lecture on repeated roots and reduction of order, I really appreciate your watching this is the differential equations lecture series and I'm Will Murray for www.educator.com, thanks.*2571

3 answers

Last reply by: Dr. William Murray

Wed Jun 29, 2016 9:50 AM

Post by Silvia Gonzalez on June 22 at 09:33:07 AM

Can we have a general solution when we use the power reduction method? May be mutiplying the two particular solutions by c1 and c2 and adding them?

3 answers

Last reply by: Dr. William Murray

Wed Jun 29, 2016 9:50 AM

Post by Silvia Gonzalez on June 21 at 06:38:45 PM

Thank you for the lectures.I have two questions from this one. Am I right to think that you apply the reduction of order method when the coefficients in the differential equation are not constant? (and that this same method was used to find the general equation for repeated roots so that we do not need to apply it every time?)

The second question is, these two solutions you find when using the reduction of order method, do they correspond to two different initial conditions that we do not know? Thank you.