Sign In | Subscribe

Enter your Sign on user name and password.

Forgot password?
  • Follow us on:
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Differential Equations
  • Discussion

  • Study Guides

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Lecture Comments (10)

1 answer

Last reply by: Dr. William Murray
Thu Dec 22, 2016 11:22 AM

Post by Marwa Alzubaidi on December 20, 2016

I have problem I want solve by matlab Could you helpe me?

1 answer

Last reply by: Dr. William Murray
Mon Mar 30, 2015 1:05 PM

Post by Ulysses Hickey on March 29, 2015

Where does the "i" in the imaginary root end up going ?

- In the process demonstrated it seems as if the i is dropped completely.
But further study in my own text indicates that the i is essential to the solution.

    Please Explain.

1 answer

Last reply by: Dr. William Murray
Fri Feb 6, 2015 12:28 PM

Post by Salah Issa on February 4, 2015

Thank you Dr.Murray,

I am really thankful

Salah Issa

3 answers

Last reply by: Dr. William Murray
Fri Feb 6, 2015 12:29 PM

Post by Salah Issa on January 17, 2015

Dear Dr.Muray,

Kindly, can you explain what is the reason behind sine and cosine in the general solution?

Complex Roots of Second Order Equations

Complex Roots of Second Order Equations (PDF)

Complex Roots of Second Order Equations

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 0:15
    • Sometimes The Characteristic Equation Has Complex Roots
  • Example 1 3:21
  • Example 2 7:42
  • Example 3 15:25
  • Example 4 18:59
  • Example 5 27:52

Transcription: Complex Roots of Second Order Equations

Hi these are the lectures on differential equations, my name is Will Murray and today we are talking about second-order equations and the case where you have complex roots to the characteristic equation.0000

Let us go ahead and take a look at those, remember to solve second-order equations at least the linear constant coefficient differential equations, what you do is you take the differential equation and you translate it into a characteristic equation AR^2 + BR + C=0.0013

And when you solve that as a polynomial, you try to factor it when you complete a square maybe use the quadratic equation to solve that, if you are lucky you get 2 distinct real roots and we talked about those in the previous lecture.0033

If you have not check that out, yet might want to check out the previous lecture about what to do when you get 2 distinct real roots.0048

Today we are going to talk about what happens when you get complex roots to the characteristic equation, when you're factoring this down, when you're using the quadratic formula and you end up with complex roots we are going to figure out how that works out today.0054

Let us take a look at that, when you get complex roots they will always be in this form α + β(i) and α - β(i), those are called conjugate pairs because you have an α + and α - β(i).0072

The always common conjugate pairs you will always get that form assuming you had real coefficients to the original differential equation which is generally what you get in most differential equations classes.0089

When you have those conjugate α + β(i) and α - β(i), we have a special form from the solution that always works to get the general form of the solution is C1 e^α(t) x cos β(t) and C2 e^α(T) x sin(β t).0105

You just look at that α and you look at that β and drop them into this formula E^α(t) cos β(T), E^α(t) sin β(t) and that is general solution.0131

You still have C1 and C2, just like before the previous lecture on real distinct roots and when you find C1 and C2 is you have to be given an initial conditions.0146

Usually those are given as y(0) equal some number and y′ 0 is equal to some number, you plug each one of those in and each one of those we give you one equation and 2 unknowns.0155

The 2 unknowns are C1 and C2, the 2 unknowns that you are trying to solve for are C1 and C2 and each one of theses Y of 0, each one of these initial conditions y(0) and y′ 0 we give you an equation.0168

And you can end up with 2 equations and 2 unknowns C1 and C2 that you can solve for using whatever algebraic technique is most comfortable for you.0189

Let us take a look at some examples and see how that plays out, first example we are asked to find a general solution to the differential equation Y″ -4y′ + 13y equal 0.0198

The way you solve that is you use the characteristic equation just like we did in the previous lecture, we go r ^2 - 4r +13 is equal to 0, now that is a polynomial equation.0212

You can try to factor this one but this will be hard to factor so instead we are going to use the quadratic formula, I will just remind you the quadratic formula says that r=-b + or -the square root of b^2 - 4AC all over to A.0230

That is the quadratic formula that we are going be using a lot today to solve these characteristic equations, in this case b is -4, -B is +4 plus or minus b^2 will be 16 - 4AC, A is 1 and C is 13 so -4× 1× 13 all over 2 x a that is 2.0252

4 x 13 is 52, we get 4 plus or -16 - 52 is -36 still divided by 2 and -36 the square root of -36 is 6(i), that is the imaginary number (i) there over 2, we get 2 plus or minus 3(i).0282

Now what that tells me is that that 2 is the α, that 3 is the β and we are going to drop that α and that β into our general solution.0314

Our general solution using our formula from before is the general solution C1 e^α(T), that is e^2t x cos(β T), our β is 3, plus C2 e^α(t) so 2t x sin(β t), sin(3t).0327

That is our general solution, we do not have any initial conditions given to us in this problem, we stopped with the general solution, that means we just leave the C1 and C2 the way they are.0359

To recap what we did there, we started out by translating the differential equation into the characteristic equation for r, that characteristic equation for r did not factor nicely, we had to go the quadratic equation in to solve it.0382

The quadratic formula here is what I worked out using the quadratic formula, we got a negative number under the square root which means we got an imaginary numbers coming up for the roots.0399

Simplify down to 2 + 3(i) and 2 -3(i), that means the α is 2 and the β is 3 and then we drop those into our formula for the general solution e^2T x cos(3T) e^2T x sin(3t) and we attach constants C1 and C2 to them.0413

Since we do not have initial conditions we just leave it that way, we just get the general solution, if we had initial conditions we will be using them at this point to solve for the C1 and C2 in the general solution.0436

And that is actually what is going to happen in the next example, we are going to keep going with the same differential equation and then introduce initial conditions to find the C1 and C2.0452

In example 2, we got the initial value problem Y″ -4y′ + 13y equal to 0, now that is the same differential equation we had before, we already know the general solution to that.0464

I will just copy it here, the general solution that we figured out last time was C1e^2T x cos(3T) + C2 e^2T x sin(3T) and then we now have the initial conditions y(0) equal 2 and y′ of 0 is equal to 1.0480

We are going to use those initial conditions, we are going to plug them in and get 2 equations to solve from the 2 unknowns C1 and C2, y(0) means we plug in 0 for t.0508

Now ei^2T would be e^2 × 0, C1 x 1 because e^0 is 1, cos(3 × 0) is 1 + C2 x 1 sin(0) is 0, I will just put 0 here, it just simplifies down to C1 and that is equal 2, I got that from up here.0525

Now the second initial condition up with the derivative, I need to say the derivative of the general solution, y′ now this is going to get a little bit messy because I have to take the derivative of a product of C1 e^2T x cos(3T).0558

I have to use a product rule, I will take the derivative of e^ 2T first, that is 2, C1 e^2T x cos(3T) + the derivative of cos(3T) which is -3 sin(3T),-3 C1 e^2T sin(3T).0578

I have to take the derivative of the C2 part, again derivative of e^2T is 2e^2T, 2C2 e^2T sin(3T), now the derivative of sin(3T) is 3 x cos(3T), +3 C2 e^2T cos 3T, I got that by expanding out the product rule on each one of these 2 terms.0608

It gets a little messy y′ of 0, I'm plugging in t equal 0 everywhere here so I get 2 x C1 x 1 x cos 3T is 1 - 3C1 x 1e^0, sin(3T) sin(0) is 0+ 2 C2 x 1 sin(0) is still 0+ 3C2 x 1 because e^2 × 0 x cos(3) x 0 is cos(0) which is 1.0646

That all simplifies down to 2C1 + 3C2 and that is equal 1, that 1 is coming from that initial condition right there, we know that y′ of 0 is equal 1.0688

Let us keep solving that, fortunately I already figured out that C1=2, 2 x 2 + 3C2 =1 and 4 + 3C2=1, I get C2 = -1.0708

And if I take the values for C1 and for C2 and I plug those back into the general solution up here, I will get y is equal to C1 was 2.0728

2e^2t x cos(3T), C2 is -1, so minus e^2t x sin(3T), that is my specific solution that satisfies both of the initial conditions there.0748

Let us go back over there and make sure we still understand everything, we started out with a general solution that was from example 1, that is where that came from.0775

If you do not remember where that came from just check back and rewatch example 1, you will see where we got the general solution and we are incorporating the 2 initial conditions.0784

Here is our first initial condition y(0) equals 2, that means we plug in T=0 here, I plug in T=0 everywhere across the board.0797

Now e^0 is just 1 and then cos(0) is also 1 e^0 again is 1 and sin(0) is 0, this simplifies down just to C1 and then it is equal 2 because I was given that in my initial condition there.0808

For the second initial condition we have to take y′ and that gets a little messy because we have to expand out the derivatives of all of this using the product rule.0830

This long expression here came from using the product rule to take y′ and it expanded out into this long expression and then we plug in T=0 everywhere and all the E^2t give us e^ 0 , those turn into 1.0843

All the cosines, cos of 0 turns into 1, all the sin of 0 terms turn into 0, that dropped out a couple more terms here and it just simplifies down to 2C1 + 3C2 is equal to 1.0866

Then we had 2 equations and 2 unknowns but really one of them told us immediately what the value of C1 is so we can plug that into the second equation and we got C2 is equal -1.0881

And finally we took C2 and C1 and plug them back into our general solution for that C1 right there and that C2 and we got y equal 2^2T x cos 3T + -1 or just -e^2 2T x sin 3T so that is our specific solution to the differential equation and also to the initial conditions.0893

For example 3, we are going to find the general solution of Y″ -10 y′ + 29 y is equal 0, again we are going to our characteristic equation that means you convert the y's and r's, they will go r^2 -10r +29 is equal to 0.0926

And again that is not an equation that solves easily by factoring so I'm going to use the quadratic formula, I will do r is equal to -B is 10, plus or minus the square root of B^2, 10 ^2 is 100-4,a is 1, c is 29 all divided by 2a, since a is 1 that is divided by 2.0950

This is 10 plus or -4× 29 is 116, 100 -116 is -16 still divided by 2, this is 10 plus or minus now the square of -16 since it is negative we get a complex number here 4i/2 which is just 5 + -2i.0984

They always occurring conjugate pairs, it comes straight out of the quadratic formula that the complex roots always occur in conjugate pairs assuming you start with real numbers in the original equation.1015

What that means is that 5 is the α and that 2 is the β so we can write down our general solution, we just drop our α and β into the general solution that we had at the beginning of the lecture which was C1 e^α T x cos(β T) + C2 e^α T x sin (β T).1030

If you have initial conditions now is when you would plug them into solve for C1 and C2, but we do not have initial conditions in this one, I will just stop here and present that as our general solution.1067

Just a recap here we started out with the differential equation with constant coefficients, translated that into the characteristic equation, solved it using the quadratic formula, when we solve that we got complex numbers as roots and they came in conjugate pairs.1085

We always write in the form α plus or minus β(i) and then we take the α and β and just drop them into the formula we had at the beginning of lecture there, there is α and there is β.1104

It is always e^α T cos β T + e^α T sin β T and of course you put constants on each one if you have initial conditions you would use those to solve for the constants but we do not in this one so were just going to present the general solution.1117

For example 4, we have an initial value problem Y″ + 4y is equal to 0 and a couple of initial conditions y0 equal 5 and y′ 0 equal 6, we start out with the differential equation, we will not worry about the initial conditions yet.1141

We are going to write down the characteristic equation and there is a very common mistake that my students often make, you might make it to when you are first learning about these things, is you will see something like this.1160

And when you write down the characteristic equation you will say r ^2 + 4r is equal to 0 and you will go off and you will solve r, that is incorrect and what incorrect about it is your thinking that y should translate into an r and that is not really the right way to think about it.1171

Instead Y″ is r ^2, now what you want think about this is that is like saying we taken 0 derivatives,we have r^ to the 0 power which is just 1, it is just r ^2 + 4=0.1197

If we had a y′ that is the term that would translate into some multiple of r, but we do not have that which is why people often end up making this mistakes in this kind of equation.1216

y′ would translate into an r, y by itself you want to think of that as being 0 derivatives of y, we have r^0 in other words we just have a constant 4 here.1228

We are solving r ^2 + 4 is equal 0 that is an easy equation to solve, you get r^2=-4, r= to the square root of -4, plus or minus that, plus or -2i.1240

We want to think about that is 0 as 0 + or -2i because our general solution was I always had the form of α plus or minus β(i) and then our general solution from that was e^α T cos β T C1 + C2 e^α T sin β T.1261

That is what you do with the α and β, in this case our α is 0 and our β is 2, I will write down our general solution is C1 e^α T would be e^0T which would just be 1.1292

I'm going to dropout that term and I'm just going to write cos of β T cos of 2T + C2, again I have e^α T but T is 0, it is e^0 is just 1, I will drop that term out and I get C2 sin of 2T.1312

That is my general solution, the E terms dropout because they are e^0T because that α was 0, now we are going use the initial conditions to solve for the constants.1333

We use y 0 equal to 5, y of 0 means we plug in 0 for T, C1 x cos of 0 + C2 sin of 0, now cos of 0 is just 1, sin of 0 is 0 and that was supposed to be equal to 5, I'm using this 5 right here and I find out right away that C1 is equal to 5.1346

Now my second initial condition, I have to look at y′, I will take the derivative of this, y′ is equal to now derivative of cos of 2T is - 2 x sin of 2T, -2C1 should been a subscript there, C1 x sin of 2T.1377

Now the derivative of sin of 2T is 2 x cos of 2T, I'm using the chain rule there, +2 C2 x cos of 2T and I'm going to plug in 0T equal 0 there, y′ of 0 plugging in T=0, -2C1 sin of 0 + 2C2 x cos of 0 and sin of 0 is 0.1402

That term disappears, I get 2C2 and that is supposed to be equal 6, I'm getting that from here, I get C2 is equal to 3 and now I figured out what my C1 is, I figure out what my C2 is, I'm going to plug those back into the general solution.1440

We get y equal to my C1 was 5, cos 2T + my C2 is 3 sin of 2T and that is my specific solution both of the differential equation and initial conditions.1465

Let us go back and see how that worked, we started off with the differential equation, we translated that into the characteristic equation and that is where a lot of students make that common mistake of translating the y into r but that is wrong.1493

You really want to translate the y just into the constant 4, we get r ^2 + 4 equal 0 and that solves quickly into r is equal plus or -2(i) and we think of that 0 as being the α there and we have α plus or minus β(i).1510

And we plug those back into our general solution, there is our generic form for the general solution, we are going to plug α and β in there but then e^α T since α is 0 just turns into e^0.1533

The E terms just turns to 1 and effectively they disappear, we are just left with the the sin and cos terms, C1 cos of 2T + C2 sin of 2T.1548

By the way it does not really matter if you switch the sin and cos, if you put the sin first with the C1 an the cos with a C2, does not really matter at this point, you could switch them, it will matter later on because you have to figure out specifically what is C1 and C2 are.1561

There is only one that goes with sin and one that goes with cos, but if you switch them at this point it would have been okay, we would have just switch the C1 C2 later on.1578

That was our general solution and then to find the specific solution we plug in 0 for T, that corresponded to T is equal to 0, we plug that in here T is equal 0, this turns out be very nice set of equations for C1 and C2.1588

Because cos 0 is 1 and sin of 0 is 0, we just get C1 + 0 is equal to 5 so immediately we get C1, to use the second initial condition, the derivative condition we have to take y′ of 0.1606

We have to take the derivative here so I was using the chain rule to take the derivative y′ here that is why we got these -2 sin 2T, that is the derivative of cos of 2T and then 2 cos of 2T came from the derivative of sin of 2T using the chain rule.1623

I will plug in T equal 0 to each one of those and we get y′ of 0 is 2C2 because the sin term dropped out, that 6 came down and became that 6, we figure out what C2 was and then we plugged then our C1 and C2 back into the general solution here.1642

And we got our specific solution that satisfies both the differential equation and both initial conditions, for our last example we are going to find the general solution of y″ -8y′ + 20y equal 0.1664

As usual when you have constant coefficients we are going to set up the characteristic equation r ^2 -8r + 20 is equal to 0, it is the same equation as the differential equation except we change the y to r and we change the derivatives to exponents.1683

And now we would try to factor this but it turns out it does not factor, instead we are going to the quadratic formula, we get r is equal to -B is +8 plus or minus.1705

B ^2 is -8 ^2 so 64 - 4AC, A is 1, C is 20, all over 2A so that is 2, now 4 × 20 is 80, we get 64 - 80 there, 80 plus or - 64 -80 is-16, we get -16 under the radical over 2.1718

Now the square root of -16 is imaginary, that is 8 plus or -4(i)/2 and that simplifies down to 4 + 2(i), as usual the complex roots come in conjugate pairs.1750

B4 is my α, the 2 was my β and I'm going to drop that the α and β back into the generic form for the general solution that we had back at the beginning of the lecture.1769

My general solution is C1 e^α T α is 4 x cos of β T, β is 2 + C2 e^α T x sin of β T and since I do not have any initial conditions on this one, I'm going to stop there and just say my general solution is the last word for this example.1783

To highlight what we did there, we took the original differential equation, translated it into the characteristic equation which did not factor nicely, we went to the quadratic formula to solve for that.1822

We ran it through the quadratic formula, turned out we got 4 plus or -2(i), that gives me my α and β that I can plug back into the generic form for the general solution.1836

There is α, there is β, it is always this form the e^α T cos β T, e^α sin β T and then you tack constants on each one.1848

At this point we have no initial conditions, we are done this example, if we been given initial conditions we would have use them to figure out what the constants are.1864

We would have got a couple of equations and a couple of unknowns, we do not have that so at this point we are done with this example with the general solution.1875

That is the end of our lecture on complex roots to the characteristic equation when you have second-order equations we will move on in a later lecture and study what happens when you have multiple roots, repeated roots to the characteristic equation.1885

That will be the next in our series of lectures on differential equations, you are watching and my names Will Murray, thanks for joining us.1901