WEBVTT mathematics/differential-equations/murray
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Hello and welcome back to www.educator.com, I'm Will Murray and we are studying differential equations, today we are going to study second order equations and there are several different cases that arise when you study second order equations.
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We are going to study something called the characteristic equation which will have 2 roots and there are different things that can happen if those 2 roots are distinct if it is different from each other or if they are the same.
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Or they turn out to be complex numbers, we are going to have one lecture on each one, today we are going to talk about the case of distinct roots, let us jump right in there.
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We are going to solve a second order linear homogeneous constant coefficient differential equations, that is a quite a mouthful there, let me explain what each one of those words means.
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Linear means you have this form Ay″ + By′ + Cy, that is where the linear differential equation is contrast of that would be if you had something like y x y′ would be a non-linear differential equations.
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Right now we are just studying linear differential equations, second order means that you have y″ appearing, that is what the second order refers to is that you have not just y′ but y″.
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That is what that word means, homogeneous means that the right hand side is 0, the fact that we have a 0 over here makes it a homogeneous equation, later on we are going to study inhomogeneous equations where we have other functions on the right hand side.
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We will learn how to solve them when we do not have 0 on the right hand side but that is more complicated issue, we are not going to get to that today.
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And then constant coefficient means that the a, b, and c are constants, they are just numbers there, that is what constant coefficient means, that is what all those words mean.
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Linear means Ay″ + By′ + Cy, second order means you have y″ in there, homogeneous means the right hand side is 0 and constant coefficient means the a, b, and c are constants they are not functions of x or t or any other variable.
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Let us learn how to solve those, it is actually a pretty straight forward algorithm to solve those, what you have to do is to solve this thing called the characteristic equation.
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And that is basically translating the differential equation and writing this other equation in terms of R, our variables are going to be R and we are going to solve this quadratic equation AR² + BR + C = 0.
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Remember A, B, and C are constants, this is just a quadratic equation, you can solve it using the quadratic formula or completing the square or if you are lucky you can factor it.
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Whatever techniques you have learned from algebra you can use to solve this quadratic equation.
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This is really just an algebra problem to solve the quadratic equation, of course as you remember when you solve quadratic equation and algebra you get 2 roots, r1 and r2, and today we are talking about the case where the roots are distinct.
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Sometimes you have multiple roots or sometimes you have complex roots but we are going to put those off for another lectures, today we are talking about 2 distinct real roots.
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And then the general solution to the differential equation is C1 e⁺r1(T) + C2 e⁺r2(T), basically all of this is prefab except for you put the r1 and r2 in as exponents on the e terms.
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That is the general solution to the differential equation, it has 2 constants in it, the reason it has 2 constants is because it was a second order equation.
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The general solution to a first order equation always has one arbitrary constant in it and the general solution to a second order equation always has 2 arbitrary constants.
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And to find those arbitrary constants it is just like with a first order equations you use initial conditions, before we had one initial condition for a second order equation you need 2 initial conditions.
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They are usually given in the form of y and 0 equals a certain number and y′ of 0 is equals to a certain number, what you will do is you take those initial conditions, plug goes back into the general solution and you get 2 equations and 2 unknowns.
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Each initial condition gives you one equation and the 2 unknowns that I'm talking about here are C1 and C2, you get 2 equations for C1 and C2, and then you will solve those to figure out what C1 and C2 are.
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Then you will plug goes back into the general solution and get a specific solution, that is the way it works, it will make more sense after we work out some examples, let us go ahead and try one out.
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We are asked to find the general solution to the differential equation y″ of T +2y′ of T -8 y of T is equal 0, this is a second order linear homogeneous constant coefficient differential equation.
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And remember the way we solve these things is we write down the characteristic equation which is AR² + BR + C is equal to 0 and A, B, and, C are just the coefficients from the differential equation.
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In this case the A is one, the B is 2 and the C is -8, remember to keep track of that negative sign, that is a part of the coefficient, we got r² + 2r -8 is equal to 0 and that is just a quadratic.
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You can solve using quadratic formula if you like, this one as is actually a pretty easy one, we can factor this one, I can factor this into r + 4 x r - 2 is equal 0 and that means my roots are r=-4 or 2.
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Now I can write down my general solution here, the general solution and I'm following the format the we had at the beginning of the lesson, C1 e⁻4t, putting the roots up in the exponents + C2 e⁺2t.
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I do not have any initial conditions for this one, I'm just going to stop here with the general solution I can not find what the values of the constants are unless I have initial conditions.
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I'm done with the general solution, just to recap what we did there, we took the differential equation, we wrote down the characteristic equation, filled in the constants, factored it, solve for values of r.
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And in those values of r became the exponents in the general solution and we did E to each one of those values of r x TE and then we multiply each one by an arbitrary constant.
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We got our general solution, we cannot figure out what the values of the constants are unless we have initial conditions, in our next example we have to solve the initial value problem y″ +2y′ -8 y =0.
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And y(0)=1 and y′ of 0 is equal to 10, this is actually the same differential equation that we had in the previous problem, I'm going to go ahead and write down the general solution that we had from the previous problem.
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Because that much of the problem is the same, y general is equal to c1e⁻4T + C2e⁺2t, that I came from the previous problem, if you do not remember how we derive that just check back in example 1 and you will see where that comes from.
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That was the solution to the differential equation, we solve that now we have to use the initial conditions to figure out what the values of the constants are, what we are going to do is look at that first initial condition.
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We are going to plug in T equal 0 and y equals one, I will say one is equal to c1 E⁻⁴ x 0 + C2 e² x 0 and ofcourse e⁰ is just one, that tells me that one is C1 + C2 , that is one equation in my unknown C1 and C2.
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To get a second equation I'm going to use this other initial condition y′ of 0 is equal to 10, in order to use that, I got to find the derivative of y, I'm going t go back to the general solution and I'm going to find y′ is equal to.
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Now C1 x is derivative of e⁻4T, that -4 will pop out, I get -4 C1e⁻4T, this is the derivative I'm finding now, +2 C2e⁺2t and then I will plug in n, again T is equal to 0 and y′ is equal to 10.
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I will plug-in 10 for y′ and I will plug in 0 for T, which will give me -4C1 e⁰ is just 1, +2 x C2 and the e⁰ is just one again.
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What I have here is one equation and another equation, 2 equations and 2 unknowns to solve for C1 and C2 and there is a lot of different ways you can solve 2 equations and 2 unknowns.
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You probably learned several different ways in algebra class, you can use substitution, you can use linear combinations, you can use addition and subtraction, you can use matrices.
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There is a lot of different ways to solve this, if you do not remember how do this, you might want to check back at the www.educator.com lectures on algebra.
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The way I'm going to do it is use linear combination, I think I'm going to multiply this equation both sides by 4, the reason I'm doing that is I'm going to add it to this equation and I want those 4's to cancel.
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If I multiply this equation by 4, I will get 4 is equal to 4C1 + 4C2 and then I will write the other equation over here 10 is equal to -4C1 + 2C2 and I'm going to add those equations together and the point of doing that is that I will get 14 on the left is equal to 0C1 + 6C2.
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Now I'm going to solve for C2, I get C2 is equal to 14/6 which I can simplify down into 7/3 and now I plug that back into the first equation and try to figure out what C1 is.
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I got one is equal to C1 + C2 and that is equal to c1 + 7/3 is one, I will subtract 7/3 from both sides and I will C1 is equal to 1 is 3/3, that is -4/3, now I figured out my values of C1 and C2 I will plug those back into the general solution.
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This is the specific solution which satisfies the initial conditions y=-4/3 e⁻4T + 7/3 e⁺2T, that is my specific solution that is supposed to satisfy both initial conditions.
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Let me just recap what we did there, we started with the general solution and we got that from example 1, that is where that came from we did not actually solve that here, but if you check back in example 1, you will see where that general solution came from.
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And then we tried to make it fit the 2 initial conditions, we looked at the initial conditions y(0) equal one, we plug y(0) in, we plug T= 0 and and we got e⁰ on both sides we just got C1 + C2=1.
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That is where we got one equation for C1 and C2 and then we took the derivative of the original equation, we got Y′ =-4 C1 x e⁻4T and 2C2 x ei⁺2T and we plug in 0 there because we are looking at y′ of 0 is equal to 10.
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We got 10 is equal to 4C1 + 2C2 and that gave us a second equation in C1 and C2, we got 2 linear equations and C1 and C2, there are lots of different ways you can solve these.
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Any of them should work, if you have another way of solving 2 equations and 2 unknowns, it is okay if you use a different way but the way I used was to multiply the first equation by 4 and then add it to the second equation.
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The point was that it made the C1 term go away and we got 6 C2 is equal to 14 , C2 is equal to 14(6) or 7/3 and then we plug back in for C2 into the first equation and we got C1 is equal to -4/3 and then we plugged C1 and C2 back into the general solution.
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And we got our specific solution that satisfies both initial conditions, for our next example we have to find the general solution of y″ -4 y′ -5 y is equal to 0.
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Again this is a linear homogeneous constant coefficients second-order differential equation, we are going to use the method of the characteristic equation to solve it, I'm going to write down the characteristic equation.
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It is just the same equation as the differential equation except you use R instead of y and you translated into a polynomial, we have r² from the y″ - 4 r -5 = 0.
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And now we want to solve this to find the roots of this quadratic equation, you can use the quadratic formula, I'm going to factory it because it turns out that this one factors nicely.
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We get r -5 x r + 1 is equal 0, my r is equal to 5 or-1 and what you do with those roots is you put them in the exponents and you multiply them by T, my general solution is y is equal to C1 e⁻1t, I will just put e⁻t + C2 e⁺5T.
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By the way it does not matter which order you put them in, you can switch the order and have it C1 e⁺5T or C2 Ee⁺T, works just as well.
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What we found here is the general solution to that second-order differential equation and we do not have any initial conditions on this, we are going to stop to the general solution, we do not have a way to find the values of the constant C1 and C2.
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That is going to come in the next example, let us go ahead and take a look at that.
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In example 4, we have to solve the initial value problem, it is the same differential equation as before y″ -4 y′ -5 y is equal to 0, but now we had 2 initial conditions, y(0)=4 and y′(0)=2.
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Remember in the previous example we found the general solution to the differential equation y(gen) is equal to C1 e⁻T + C2 e⁺5T and what we are going to doing with that is plug in our initial conditions.
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This says when T is equal 0, y is equal 4, I will say 4 is equal to C1 x e⁰+ C2 e⁵ x 0, that is equal to C1 + C2 because e⁰ is just 1 and then y′ of 0, to use that we have to take the derivative of our general solution.
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y′ is negative C1 e⁻T +5 C2 e⁺5T and again I will plug in T equal 0 and I will plug in 2 for y′, 2 is equal to -C1 e⁰ + 5 C2 E⁵ x 0, that is -C1 +5 C2 because e⁰ just 1 is equal to 2.
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I have 2 equations and 2 unknowns and again there are several different ways you can solve this, you could do a substitution, solve for 1 variable and 1 equation and substitute it to the other equation.
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You could use matrices, you can use linear combinations, what I'm going to do is take these 2 equations and I will just add 1 to the other one because I see that if I do that, the C1's will cancel each other out and I will get 6 C2.
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I'm adding these 2 equations here, 6C2 is equal to 6 and there is no C1's because those cancel each other out, I will get C2 is equal to 1 and then if I plug that back in here to C1 + C2 is equal 4, I will get C1 is equal to 3.
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Now I got my values of C1 and C2, I will plug goes back into my general solution and get the specific solution as C1 was 3, I get 3 e⁻T + C2 is 1, just + e⁺5T.
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That is my specific solution which satisfies both the differential equation and the 2 initial conditions, just to recap there we got our general solution that was from the previous example, example 3.
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If you do not remember how we derive that, just check back in example 3 and you will see where that came from and then we plug in our 2 initial conditions, we plug our first one in and plug in T=0 and y=4.
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We plug in 0 for T and 4 y and that reduce down to C1 + C2=4, for the second initial condition we had to figure out the derivative, we took the derivative of the general solution and we got -C1 e⁻T + 5C2 e⁺5T.
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And then we plug in 2 for y′ and we got 2 is equal to what we get plugging in 0, again that turn into -C1 +5C2=2, each one of those initial conditions gave us an equation and 2 unknowns.
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Our unknowns now are C1 and C2, once we got those 2 equations and 2 unknowns, it turns into an algebra problem of solving for C1 and C2 and this one I was able to solve by adding the 2 equations.
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I got 6c2=6, C2 is equal to 1 then I substituted back to get C1 is equal to 3, if you do not like the way I did that by adding the equations, you can also solve for one variable and substitute it into the other equation.
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You can use matrices and determinants, there are lots of different ways to do and that just depends on what you are most comfortable with from your algebra class.
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Once you figure out the C1 and C2, we plug those into the general solution up here and here and we get our specific solution y=3 e⁻T + e⁺5T, that is our specific solution to the differential equation and both initial conditions.
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Our last example here, we have to find at least one non zero solution to the differential equation y″ + 6y′ + 9y is equal to 0.
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Just like all the others, we start out with the characteristic equation, that means we convert the y into r and the derivatives into exponents, r² + 6r +9 = 0.
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That actually factors as a perfect square, that is r +3² is equal to 0, we get a double root, r=-3 and then our other root is also -3, we really only get one root from that, we had a solution here y is C1 e⁻³ T.
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We can not find a second solution because our other root is also -3, let me write that down.
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But we can not find a second solution that is independent of our first solution that was e⁻3T, left a little space here after the word "can't" because this is actually the subject of a later lectures, I'm going to say we can't, I will put that in red.
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We can't yet find a second solution because a couple lectures later we will figure out what to do with repeated roots but in the meantime we do not know how to solve that.
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I will say a common student mistake would be to write the general solution as well if you just can not blindly look at these 2 roots, you would say C1 e⁻3T + C2 e⁻3T because you just copy your 2 roots into the exponents here.
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If you wrote that, that would be incorrect, that is not the general solution and the reason is because you really have 2 copies of the same solution there, you need to find a second independent solution.
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We have not learned how to do that yet, in the meantime all we can offer as a solution is just the C1 e⁻³ T, there is another way to solve these equations with repeated root and we will learn how to do that, learn how to solve this in a later lecture.
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You have to peek at a couple more lectures down the line and you will see one called repeated roots.
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And that one we will learn how to find a second independent solution to an equation when we do have repeated roots, just to recap here, we started out with a characteristic equation where we converted it into a polynomial in r with exponents instead of derivatives.
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We factored it down just like the others, we found the roots and then the new wrinkle in this example was that both the roots were the same, we were able to form one solution, C1 e⁻3T but we can not find a second independent solution.
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So we cannot tack on a C2 e⁻3T because that would just be a copy of the first solution, in order to find the second solution we will have to come back and study this in more detail on our later lecture on repeated roots.
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That is the end of our lecture on distinct roots, we will come back later and start talking about complex roots and repeated roots, these are the lectures on differential equations, my name is Will Murray and you are watching www.educator.com, thanks.