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Lecture Comments (4)

1 answer

Last reply by: Dr. William Murray
Mon Dec 8, 2014 5:33 PM

Post by Harjinder Singh on December 8, 2014

Hi Dr. Murray,

I am working on a similar problem. However, the problem asks you to find two linearly independent solutions of the DE and compute the Wronskian. How would I proceed with that after finding the general equation?

Find two linearly independent solutions of (2x^2)y" + 3xy' - y = 0 for x > 0 and compute their Wronskian.

I was able to get the following general solution:

y = (c1)(x^1/2) + (c2)(x^-1)

Can I just plugin some arbitrary values for c1 and c2 to get two separate solutions of y and then test them out for linear independence using the Wronskian? Or are the individual terms [c1*x^(1/2)] and [c2*x^-1] considered two separate solutions of the DE and that's all I need?

1 answer

Last reply by: Dr. William Murray
Thu May 1, 2014 1:29 PM

Post by Khanh Nguyen on April 29, 2014

Hi mr. I could not open you video . That is on the Euler equation
Is there anything go wrong with the website?

Euler Equations

Euler Equations (PDF)

Euler 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 0:11
    • Euler Equation
    • Real, Distinct Roots
    • Real, Repeated Roots
    • Complex Roots
  • Example 1 3:51
  • Example 2 6:20
  • Example 3 8:27
  • Example 4 13:04
  • Example 5 15:31
  • Example 6 18:31

Transcription: Euler Equations

Hi welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and today we are going study Euler equations, let us jump right in.0000

Euler equations have a very special form, they are the form x2 y″ + a constant which I’m calling α × XY′ + β (another constant) × y=0.0012

The idea there is that there is this pattern of descending powers on the x, this is very important.0027

We got an x2 here, we got an x here, and you want to think this is x 1, you want to think that this is x0 because it is just a constant.0036

You really have to have that exact pattern of the powers, x2, x and x0 in order to make the order of the equations work.0045

If you do have that pattern then there is a very easy solution it is basically xr.0054

Let me show you how to figure out what r is, what you are going to do is you are going to solve the characteristic equation for r and there is a little subtlety here.0060

Basically you take the coefficients from the order equation but there is a small change which is just instead taking α and β, you take α – 1 and β.0071

You solve this characteristic equation r2 + α – 1r + β=0.0083

That is just a quadratic polynomial, you can solve that by factoring, by quadratic formula, and whatever works for you to solve that and you will get 2 roots for r, you get an r1 and r2.0090

This is not really hard, this is just algebra, you could get 2 real roots, so you could get r1=3, r2=5 or something like that.0106

You could get 1 repeated root, meaning you will get a double root or you could get 2 roots that are complex conjugates.0118

Something like a + π and a – π, that is if you are on a quadratic formula and you get a negative number under the square root sign, then you are going to get 2 complex conjugate roots.0125

Let me tell you what to do in each one of those situations and how you will write down the solution to the differential equation.0136

If you have real distinct roots, meaning 2 different real roots the general solution to the equation is just xr1 × constant and xr2 × another constant.0144

It is very easy there if you have repeated roots, same copy of root appears you have xr1 and xr1 × natural log of x.0155

Again, multiplying each one by a constant and if you have complex roots then it will always occur in conjugate pairs.0166

Remember that the quadratic formula is something (+ or -) the square root of something.0173

If you get a negative number under a square root then you will end up getting a conjugate pair, something like a (+ or -) π.0182

It is a little more complicated if you do that but it is just a formula to remember.0191

It is xa × cos(x) of logx to the b and xa × sin of natural log of xb.0195

Those are your 2 fundamental solutions and you just multiply each one by an arbitrary constant and you will get your general solution.0208

That is pretty much all you need to know for Euler equations, you solve the characteristic equation, get the 2 roots.0215

Just depending on which of these 3 situations it is, you just drop it in to one of these 3 formulas, let us try that out with some examples.0223

In our first example, I want to find the general solution to x2 y″ – 3xy′ + 3y=0.0231

We got this Euler pattern on the coefficient x2, x and then a constant here, so it is an Euler equation.0240

The α here is -3 because that is the coefficient here, the β is 3 and remember the characteristic equation is always r2 + α- 1r + β=0.0249

That is the equation that you always solve for r and in this case we have r2,0272

Now α is -3, so α – 1 is -4, this is actually -4r and β is 3, +3 =0.0280

That equation factors quite nicely that is r-1 × r-3=0.0292

Our 2 values of r are 1 and 3, r=1 and r=3, now we can drop those right into the general solution.0303

The general solution is c1 × x + c2 x3, that is the solution to the differential equation.0317

It is a really nice relief that this Euler’s equations are much easier than some of the other things that we have been doing recently like series solution.0336

Let me recap quickly how we worked that out, we noticed that this was an Euler equation because it has that descending pattern of powers x2, x, and constant.0345

We read off our values of α and β, just the coefficients there -3 and 3.0356

Drop them into this characteristic equation, the key point there is to remember to subtract 1 from the alphas.0362

We got r2 -4 +3 =0, factor it out, get our roots and those become the powers and we get our general solution.0367

Let us go ahead and try another one, we still have an Euler equation here because the powers on x is descending.0380

We have x2, x, and a constant, our α here is -7, β is 16.0389

We are going to use our generic characteristic equation r2 + α -1r + β=0.0402

Our α is -7 so α – 1 is -8, -8r + 16 = 0.0412

(r) factors into r-42=0, we get the double root at r =4 or 4.0422

We had a different formula for double root.0433

The general solution c1 x4 but we can not have c2 x4 because that would just be a copy of our first solution.0439

C2 x4 natural log of x, we got that straight from the lesson overview at the beginning.0449

Whenever you have that double root, that second solution you just multiply it up by natural log of x.0459

That is the end of that one but just to recap we noticed our descending powers of x.0467

X2, x, and a constant, that is what triggers you that it is an Euler equation.0473

Write down your α is the coefficient of first, β is the coefficient of the second one.0478

Plot those into our generic characteristic equation where you have an α -1 so that -7 turns into -8, factor it out.0483

Once you get a double root for r you know one of your solutions is x4, but the other solution, in order to get an independent solution you got to multiply it by natural log of x.0493

That is our second independent solution there.0503

Our next example here, again we have Euler equation, we can check that by noticing the descending powers of x, x2, x, and a constant.0508

We got the descending powers of x, we got an Euler equation.0519

α here is -1, β is 5, those are the values we are going to use in the characteristic equation.0522

R2 + α – 1r + β= 0, if α is -1, then α -1 is 2, β is still 5.0533

Now that thing does not factor so we are going to jump in the quadratic formula to solve that.0551

Remember the quadratic formula says –b (+ or -) square root of b2 - 4ac/2a.0557

Here our b is -2 so our- b is +2 (+ or -) b2 is -22 so that is 4.0573

4ac is 4 × 1 × 5=20, over 2a is 2, that is 2 (+ or -) 4 – 20 is -16.0584

The square root of -16 is 4i, I’m going to get complex roots here, that simplifies down 1 (+ or-) 2i.0599

Remember we have a generic solution formula when you have complex roots.0614

The generic solution formula, let me remind you what that was from the lesson overview, the beginning of class.0620

C1 xa × cos natural log xb + c2 xa × sin of natural log of xb.0627

You could also use your natural log rules and write this as b natural log x inside, that will be ok if you look that better.0648

The a and b here, they are not the a and b from the original equation so be careful about that.0655

The a and b come from the a (+ or -) π here, it is not the same as these a’s and b’s that we used in the quadratic formula, be very careful about that.0661

What we have here is our general solution, c1 xa, our a is 1, I just write x × cos of natural log of xb, b is 2 here,0676

So natural log of x2 + c2 x × sin of natural log of x2 and that is our solution.0690

Let us take a look and see how that worked out, we had our Euler equation because of those descending powers of x, x2, x, and a constant.0712

We read off our coefficients, α is -1 the coefficient of the x term, β is 5, we plug those into the characteristic equation.0721

Remember the subtlety where the characteristic equation you drop the α down by 1.0732

α -1 becomes -2 and β is 5, to solve that we have to go to our quadratic formula.0737

That simplifies down to 1 (+ or-) 2i and we use that as our a and b, not the a and b from the quadratic formula but the 1 and the 2 are a and b,0747

That we are going to plug into our generic solution for Euler equation when you have complex roots.0760

We plug in x1 here and we get (x cos log of x2) and (x sin log of x2) as our 2 independent solutions.0768

We put them together with constants to make our general solution.0779

For example 4, we got x2 y″ – 6xy′ – 12 y=00785

We check out first that this satisfies the properties of being Euler equation, those descending powers on x, (x2, x, and a constant), (x2, x1, x0).0793

Remember it got to fit that format exactly to be an Euler equation, if it does not fit that format then you are really out of luck, you got to use some other techniques which certainly be more complicated.0805

It is worth checking if something is an Euler equation if so, these things are real pretty quick to solve.0817

If not then you got to have to do something that is probably be much more lengthy.0823

Since this one does work, we read off our α as -6, our β is 12, and we set up our characteristic equation.0827

R2 + α – 1, r + β= 0, that is our squared.0837

α -1 is -7, r + 12=0, that is another easy one to factor, that is r-3 × r-4=0.0846

R is 3 or 4, so 2 distinct real roots that is the easiest situation for an Euler equation.0861

Right away I can write down my general solution it is just c1 x3 form that first root + c2 x4 from that second root.0871

Just like that we are done with that one.0884

Let me remind you of how that one worked out, we first check the powers x2, x1, x0.0893

That looks like an Euler equation so that means I can write down the coefficients.0900

α=-6, β=12, drop them right into my characteristic equation which has that shift to the α down by 1.0904

We get r2 - 7r that is α-1 + 12=0, factors nicely and gives me couple of roots.0914

Those roots become the exponents on x for the general solution.0923

On example 5 here, again we have an Euler equation because we recognized this descending power of x’s.0932

X2, x1,and x0 here, no x at all means x0 or you think of it as being a constant.0939

I can read off my coefficients, α is 5, β is 4, I’m going to set up my characteristic equation.0948

It always has a form r2 + α -1r + β=0.0958

Remember to shift that α by 1, that is the subtlety of Euler equations that we have not really seen anywhere else.0967

In this case the α is 5, we get r2 + 4 r + 4 =0, that first 4 here, this one right here, that comes from 5-1, it has nothing to do with β being 4 there.0973

That is easy to solve, that is r + 22=0, r=-2 and that is a double root.0988

Remember we have a special formula for one you have a double root of an Euler equation and that formula is, first you have x to that power, c1x to that power.1001

But then you can not have c2 × x to that power again because that would just be a copy of your first solution.1013

The way you get around that is you multiply on a natural log of x.1021

That is our general solution to that one.1028

Let us recap how that problem worked out, we recognized the descending powers x2, x1, x0, ok that means it is an Euler equation.1039

We identify our α as a coefficient of the x term, β is our coefficient of the constant term.1051

Then we go to our generic characteristic equation with an α -1 in it and we get r2 + 4r + 4, that is using α -1 to get that 4 right there.1059

That factors easily into r + 22 = 0 that turns into r=-2, that is a double root.1075

In case of a double root, our solution is just like with 2 distinct roots, we form x to that power, x-2.1086

For the second one, we can not just form x-2 because that would be a copy of our first solution.1096

We go x-2 natural log of x to get ourselves a distinct solution and we put a constant on each one.1101

On example 6, we got the differential equation x2 y″ – 3xy′ + 29y=0.1112

As before, look at this we got x2, x1, x0, that means it is an Euler equation so we can use what we have been learning in this lecture.1122

We are going to identify our α and our β, α is -3 here, make sure you include the negative, sometimes people just see the 3 here and say ok α is 3.1135

Of course that is going to lead you to disaster later on, do not forget those negatives if they are part of it.1145

β is 29, we go to our generic characteristic equation r2 + α – 1 r + β =0.1152

α is -3, α -1 is -4, r + 29 =0 and it will be great if that would factor but it does not factor easily so we are going to the quadratic formula for that.1166

R=-b (+or -) the square root of b2 - 4ac / 2a.1185

In this case our b is -4 so –b is +4, (+ or -), b2 is 16 and 4ac is 4 × 1 × 29 which is 116, that is all divided by 2a but a is just 1 so that is 2.1196

That is 4 (+ or -) the square root of -100/2, now -100 the square root of that is 10i.1224

This is 4 (+ or -) 10i/2, 2 (+ or -) 5i and we have a generic solution formula for when we get complex roots to the characteristic equation.1237

It is c1 xa × cos of natural log of xb + c2 xa × sin of natural log of xb.1256

You could also write that as b natural log of x, that would be correct.1273

The a and b, they are not the a and b from back here in the quadratic formula, they are the a and b that you got in the end.1279

Let me go ahead and plug those in, the general solution here is (c1 × x2),( cos r of natural log5 + c2 x 2 × sin of natural log of x5.1290

You could write that as 5 natural log of x if you like.1312

That is our general solution to this Euler equation.1319

Let us go back and see how that worked out, first thing to do is recognize the powers, x2, x1,and x0.1329

That is what you have to have for an Euler equation to work out as those descending powers of x’s.1337

If you do its going to work out easily, if you do not you have to abandon the Euler methods completely and you have to try something completely different.1343

We found that it worked out for this example, we identify our α is the coefficient of the x term, our β is the coefficient of the constant term.1355

We go back to the generic characteristic equation r2 + α – 1r + β = 0.1372

Our α is -3 so subtracting 1 from that is where this -4 came from, we got this quadratic equation that really did not factor nicely.1381

So we use the quadratic formula, we got a –number under the square root so we get complex numbers 2 (+ or -) 5i.1391

Then we drop those as our a and b, not the a and b from the quadratic formula but our a and b from the complex number we found.1400

We drop those in to our generic solution formula for Euler equation for complex roots, which is xa cos log b + xa sin log of xb.1409

We get (c1 x2 cos log of x5) (c2 x2 sin of log of x5).1423

That is our general solution to the differential equation.1431

That is the end of our lecture on Euler equations, by the way you might have come here looking for Euler methods which is actually something quite different.1436

That is something we are going to get in to a later lecture here on the differential equations lectures.1446

If you are looking for Euler methods here and this did not look at all like what you are expecting, just scroll through and look for a later lecture on Euler methods.1451

That is a numerical solution to differential equation and you will find that here on www.educator.com.1459

It is just that this was a specific lecture about Euler equations.1465

That wraps everything up here, this has been the Euler equations lecture in the differential equations series here on www.educator.com.1470

My name is Will Murray and I thank you very much for watching, bye.1479