WEBVTT mathematics/differential-equations/murray
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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.
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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.
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The idea there is that there is this pattern of descending powers on the x, this is very important.
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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.
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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.
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If you do have that pattern then there is a very easy solution it is basically x^r.
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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.
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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.
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You solve this characteristic equation r² + α – 1r + β=0.
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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.
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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.
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You could get 1 repeated root, meaning you will get a double root or you could get 2 roots that are complex conjugates.
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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.
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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.
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If you have real distinct roots, meaning 2 different real roots the general solution to the equation is just x^r1 × constant and x^r2 × another constant.
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It is very easy there if you have repeated roots, same copy of root appears you have x^r1 and x^r1 × natural log of x.
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Again, multiplying each one by a constant and if you have complex roots then it will always occur in conjugate pairs.
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Remember that the quadratic formula is something (+ or -) the square root of something.
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If you get a negative number under a square root then you will end up getting a conjugate pair, something like a (+ or -) π.
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It is a little more complicated if you do that but it is just a formula to remember.
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It is x^a × cos(x) of log^x to the b and x^a × sin of natural log of x^b.
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Those are your 2 fundamental solutions and you just multiply each one by an arbitrary constant and you will get your general solution.
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That is pretty much all you need to know for Euler equations, you solve the characteristic equation, get the 2 roots.
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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.
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In our first example, I want to find the general solution to x² y″ – 3xy′ + 3y=0.
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We got this Euler pattern on the coefficient x², x and then a constant here, so it is an Euler equation.
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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.
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That is the equation that you always solve for r and in this case we have r²,
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Now α is -3, so α – 1 is -4, this is actually -4r and β is 3, +3 =0.
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That equation factors quite nicely that is r-1 × r-3=0.
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Our 2 values of r are 1 and 3, r=1 and r=3, now we can drop those right into the general solution.
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The general solution is c1 × x + c2 x³, that is the solution to the differential equation.
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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.
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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.
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We read off our values of α and β, just the coefficients there -3 and 3.
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Drop them into this characteristic equation, the key point there is to remember to subtract 1 from the alphas.
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We got r² -4 +3 =0, factor it out, get our roots and those become the powers and we get our general solution.
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Let us go ahead and try another one, we still have an Euler equation here because the powers on x is descending.
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We have x², x, and a constant, our α here is -7, β is 16.
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We are going to use our generic characteristic equation r² + α -1r + β=0.
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Our α is -7 so α – 1 is -8, -8r + 16 = 0.
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(r) factors into r-4²=0, we get the double root at r =4 or 4.
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We had a different formula for double root.
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The general solution c1 x⁴ but we can not have c2 x⁴ because that would just be a copy of our first solution.
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C2 x⁴ natural log of x, we got that straight from the lesson overview at the beginning.
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Whenever you have that double root, that second solution you just multiply it up by natural log of x.
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That is the end of that one but just to recap we noticed our descending powers of x.
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X², x, and a constant, that is what triggers you that it is an Euler equation.
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Write down your α is the coefficient of first, β is the coefficient of the second one.
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Plot those into our generic characteristic equation where you have an α -1 so that -7 turns into -8, factor it out.
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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.
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That is our second independent solution there.
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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.
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We got the descending powers of x, we got an Euler equation.
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α here is -1, β is 5, those are the values we are going to use in the characteristic equation.
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R² + α – 1r + β= 0, if α is -1, then α -1 is 2, β is still 5.
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Now that thing does not factor so we are going to jump in the quadratic formula to solve that.
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Remember the quadratic formula says –b (+ or -) square root of b² - 4ac/2a.
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Here our b is -2 so our- b is +2 (+ or -) b² is -2² so that is 4.
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4ac is 4 × 1 × 5=20, over 2a is 2, that is 2 (+ or -) 4 – 20 is -16.
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The square root of -16 is 4i, I’m going to get complex roots here, that simplifies down 1 (+ or-) 2i.
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Remember we have a generic solution formula when you have complex roots.
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The generic solution formula, let me remind you what that was from the lesson overview, the beginning of class.
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C1 x^a × cos natural log x^b + c2 x^a × sin of natural log of x^b.
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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.
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The a and b here, they are not the a and b from the original equation so be careful about that.
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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.
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What we have here is our general solution, c1 x^a, our a is 1, I just write x × cos of natural log of x^b, b is 2 here,
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So natural log of x² + c2 x × sin of natural log of x² and that is our solution.
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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.
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We read off our coefficients, α is -1 the coefficient of the x term, β is 5, we plug those into the characteristic equation.
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Remember the subtlety where the characteristic equation you drop the α down by 1.
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α -1 becomes -2 and β is 5, to solve that we have to go to our quadratic formula.
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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,
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That we are going to plug into our generic solution for Euler equation when you have complex roots.
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We plug in x¹ here and we get (x cos log of x²) and (x sin log of x²) as our 2 independent solutions.
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We put them together with constants to make our general solution.
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For example 4, we got x² y″ – 6xy′ – 12 y=0
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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).
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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.
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It is worth checking if something is an Euler equation if so, these things are real pretty quick to solve.
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If not then you got to have to do something that is probably be much more lengthy.
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Since this one does work, we read off our α as -6, our β is 12, and we set up our characteristic equation.
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R² + α – 1, r + β= 0, that is our squared.
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α -1 is -7, r + 12=0, that is another easy one to factor, that is r-3 × r-4=0.
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R is 3 or 4, so 2 distinct real roots that is the easiest situation for an Euler equation.
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Right away I can write down my general solution it is just c1 x³ form that first root + c2 x⁴ from that second root.
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Just like that we are done with that one.
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Let me remind you of how that one worked out, we first check the powers x², x¹, x⁰.
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That looks like an Euler equation so that means I can write down the coefficients.
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α=-6, β=12, drop them right into my characteristic equation which has that shift to the α down by 1.
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We get r² - 7r that is α-1 + 12=0, factors nicely and gives me couple of roots.
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Those roots become the exponents on x for the general solution.
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On example 5 here, again we have an Euler equation because we recognized this descending power of x’s.
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X², x¹,and x⁰ here, no x at all means x⁰ or you think of it as being a constant.
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I can read off my coefficients, α is 5, β is 4, I’m going to set up my characteristic equation.
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It always has a form r² + α -1r + β=0.
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Remember to shift that α by 1, that is the subtlety of Euler equations that we have not really seen anywhere else.
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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.
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That is easy to solve, that is r + 2²=0, r=-2 and that is a double root.
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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.
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But then you can not have c2 × x to that power again because that would just be a copy of your first solution.
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The way you get around that is you multiply on a natural log of x.
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That is our general solution to that one.
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Let us recap how that problem worked out, we recognized the descending powers x², x¹, x⁰, ok that means it is an Euler equation.
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We identify our α as a coefficient of the x term, β is our coefficient of the constant term.
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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.
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That factors easily into r + 2² = 0 that turns into r=-2, that is a double root.
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In case of a double root, our solution is just like with 2 distinct roots, we form x to that power, x^-2.
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For the second one, we can not just form x^-2 because that would be a copy of our first solution.
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We go x^-2 natural log of x to get ourselves a distinct solution and we put a constant on each one.
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On example 6, we got the differential equation x² y″ – 3xy′ + 29y=0.
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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.
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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.
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Of course that is going to lead you to disaster later on, do not forget those negatives if they are part of it.
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β is 29, we go to our generic characteristic equation r² + α – 1 r + β =0.
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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.
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R=-b (+or -) the square root of b² - 4ac / 2a.
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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.
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That is 4 (+ or -) the square root of -100/2, now -100 the square root of that is 10i.
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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.
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It is c1 x^a × cos of natural log of x^b + c2 x^a × sin of natural log of x^b.
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You could also write that as b natural log of x, that would be correct.
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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.
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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.
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You could write that as 5 natural log of x if you like.
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That is our general solution to this Euler equation.
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Let us go back and see how that worked out, first thing to do is recognize the powers, x², x¹,and x⁰.
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That is what you have to have for an Euler equation to work out as those descending powers of x’s.
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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.
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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.
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We go back to the generic characteristic equation r² + α – 1r + β = 0.
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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.
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So we use the quadratic formula, we got a –number under the square root so we get complex numbers 2 (+ or -) 5i.
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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.
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We drop those in to our generic solution formula for Euler equation for complex roots, which is x^a cos log b + x^a sin log of x^b.
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We get (c1 x² cos log of x⁵) (c2 x² sin of log of x⁵).
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That is our general solution to the differential equation.
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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.
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That is something we are going to get in to a later lecture here on the differential equations lectures.
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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.
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That is a numerical solution to differential equation and you will find that here on www.educator.com.
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It is just that this was a specific lecture about Euler equations.
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That wraps everything up here, this has been the Euler equations lecture in the differential equations series here on www.educator.com.
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My name is Will Murray and I thank you very much for watching, bye.