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# Repeated roots of the characteristic equation: What happens when the characteristic equation only has 1 repeated root?

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- Let's say we have the following second order
- differential equation.
- We have second derivative of y, plus 4 times the first
- derivative, plus 4y is equal to 0.
- And we're asked to find the general solution to this
- differential equation.
- So the first thing we do, like we've done in the last several
- videos, we'll get the characteristic equation.
- That's r squared plus 4r plus 4 is equal to 0.
- This one is fairly easy to factor.
- We don't need the quadratic equation here.
- This is r plus 2 times r plus 2.
- And now something interesting happens, something that we
- haven't seen before.
- The two roots of our characteristic equation are
- actually the same number, r is equal to minus 2.
- So you could say we only have one solution, or one root, or
- a repeated root.
- However you want to say it, we only have one r that satisfies
- the characteristic equation.
- You might say, well that's fine.
- Maybe my general solution is just y is equal to some
- constant times e to the minus 2x, using my one solution.
- And my reply to you is, this is a solution.
- And if you don't believe me you can test it out.
- But it's not the general solution.
- And why do I say that?
- Because this is a second order differential equation.
- And if someone wanted a particular solution, they
- would have to give you two initial conditions.
- The two initial conditions we've been using so far are,
- what y of 0 equals, and what y prime of 0 equals.
- They could give you what y of 5 equals, who knows.
- But in general, when you have a second order differential
- equation, they have to give you two initial conditions.
- Now the problem with this solution, and why it's not the
- general solution, is if you use one of these initial
- conditions, you can solve for a c, right?
- You'll get an answer.
- You'll solve for that c.
- But then there's nothing to do with the
- second initial condition.
- In fact, except for only in one particular case, whatever
- c you get for the first initial condition, it won't be
- that-- this equation won't be true for the
- second initial condition.
- And you could try it out.
- I mean, if we said, y of 0 is equal to A and y prime of 0 is
- equal to 5A.
- Let's see if these work.
- If y of 0 is equal to A, that tells us that A is equal to c
- times e to the minus 2 times 0.
- So e to the 0.
- Or c is equal to A, right?
- So if you just had this first initial condition, say fine,
- my particular solution is y is equal to A times e
- to the minus 2x.
- Let's see if this particular solution satisfies the second
- initial condition.
- So what is the derivative of this?
- y prime is equal to minus 2Ae to the minus 2x.
- And it says that 5A-- this initial condition says that 5A
- is equal to minus 2A times e to the minus 2 times
- 0, so e to the 0.
- Or another way of saying that, e to the 0 is just 1.
- It says that 5A is equal to minus 2A, which
- we know is not true.
- So note, when we only have this general, or
- pseudo-general, solution, it can only satisfy, generally,
- one of the initial conditions.
- And if we're really lucky, both initial conditions.
- So that at least gives you an intuitive feel of why this
- isn't the general solution.
- So let me clean that up a little bit so that I-- I have
- a feeling I'll have to use this real estate.
- So what do we do?
- We can use a technique called reduction of order.
- And it really just says, well let's just
- guess a second solution.
- In general when we first thought about these linear
- constant coefficient differential equations, we
- said, well e to rx might be a good guess.
- And why is that?
- Because all of the derivatives of e are kind of multiples of
- the original function, and that's why we used it.
- So if we're looking for a second solution, it doesn't
- hurt to kind of make the same guess.
- And in order be a little bit more general, let's make our
- guess for our second solution-- I'll call this g
- for guess-- let's say it's some function of x times our
- first solution, e to the minus 2x.
- I could say some function of x times c times e to the minus
- 2x, but the c is kind of encapsulated.
- It could be part of this some random function of x.
- So let's be as general as possible.
- So let's assume that this is a solution and then substitute
- it back into our original differential equation, and see
- if we can actually solve for v that makes it all work.
- So before we do that, let's get its first and second
- derivatives.
- So the first derivative of g is equal to-- well this is
- product rule.
- And I'll drop the v of x, we know that v is a function and
- not a constant.
- So, product rule, derivative of the first, v prime times
- the second expression, e to the minus 2x, plus the first
- function, or expression, times the derivative of the second.
- So minus 2 times e to the minus 2x.
- Or, just to write it a little bit neater, g prime is equal
- to v prime e to the minus 2x minus 2ve to the minus 2x.
- Now we have to get the second derivative.
- I'll do it in a different color, just to fight the
- monotony of it.
- So the second derivative-- we're going to have to do the
- product rule twice-- derivative of this first
- expression.
- It's going to be v prime prime e to the minus 2x, minus 2v
- prime e to the minus 2x.
- That was just the product rule again.
- And then the derivative of the second expression is going to
- be-- let's see, derivative of the first one is v prime-- so
- it's going to be minus 2v prime e to the minus 2x, plus
- 4ve to the minus 2x.
- I hope I haven't made a careless mistake.
- And we can simplify this a little bit.
- So we get the second derivative of g, which is our
- guess solution, is equal to the second derivative of v
- prime, e to the minus 2x minus 2v prime-- no, minus 4, sorry,
- because we have minus 2, minus 2-- minus 4v prime e to the
- minus 2x, plus 4ve to the minus 2x.
- And now, before we substitute it into this, we can just make
- one observation.
- That will just make the algebra a little bit simpler.
- Notice that g is something times e to the minus 2x.
- G prime is-- we could factor out an e to the minus 2x.
- And g prime prime, we can factor out an e
- to the minus 2x.
- So let's factor them out, essentially.
- So when we write this, we can write-- so the second
- derivative is g prime prime, which we can write as-- and
- I'm going to try to do this-- it's e to the minus 2x times
- the second derivative.
- So now we can get rid of the e to the minus 2x terms. So
- that's v prime prime minus 4v prime plus 4v, right?
- If I just distribute this out I get the second derivative,
- which is this.
- Plus 4 times the first derivative.
- And I'm also going to factor out the e to the minus 2x.
- So, plus 4 times this.
- So it's going to be plus 4v prime minus 8v, right?
- Once again, I factored out the e to the minus 2x, right?
- Plus 4 times y.
- We factored out the e to the minus 2x-- so plus 4 times v.
- I did that, because if I didn't do that I'd be writing
- e to the minus 2x, and I'd probably make a careless
- mistake, and I'd run out of space, et cetera.
- But anyway, I essentially-- to get this, I just substituted
- the second derivative, the first derivative, and g back
- into the differential equation.
- And we know that that has to equal 0.
- And let's see if we can simplify this
- a little bit more.
- And then hopefully solve for v.
- So let's see, some things are popping out at me.
- So I have plus 4v plus 4v, that's plus
- 8v, minus 8v, right?
- So plus 4 minus 8 plus 4, those cancel out.
- It's plus 8 minus 8, those cancel out.
- And I also have minus 4v prime plus 4v prime.
- So those cancel out.
- And lo and behold, we've done some serious simplification.
- It ends up being e to the minus 2x times v prime prime--
- we could call that v prime prime of x, now that we've
- saved so much space-- is equal to 0.
- We know this could never equal to 0.
- So, essentially, we have now established that this
- expression has to be equal to 0.
- And we get a separable second order differential equation.
- We get that the second derivative of v with respect
- to x-- or it's a function of x-- is equal to 0.
- So now we just have to differentiate both sides of
- this equation twice.
- You differentiate once, you get what?
- v prime of x is equal to, let's call it c1.
- And if we were to take the anti-derivative of both sides
- again, we get v of x is equal to c1 x plus
- some other c2, right?
- Now remember, what was our guess?
- Our guess was that our general solution was going to be some
- arbitrary function v times that first solution we found,
- e to the minus 2x.
- And when we actually took that guess and we substituted it
- in, we actually were able to solve for that v.
- And we got that v is equal to this.
- So this is interesting.
- So what is g, or what does our guess function equal?
- And it's no longer a guess.
- We've kind of established that it works.
- g, which we can call our solution, is equal to v of x,
- times e to the minus 2x.
- Well, that equals this, c1 x plus c2 e to the minus 2x.
- That equals c1 xe to the minus 2x, plus c2 e to the minus 2x.
- And now we have a truly general solution.
- We have two constants, so we can satisfy two initial
- conditions.
- And if you were looking for a pattern, this is the pattern.
- When you have a repeated root of your characteristic
- equation, the general solution is going to be-- you're going
- to use that e to the, that whatever root is, twice.
- But one time you're going to have an x in front of it.
- And this works every time for second order homogeneous
- constant coefficient linear equations.
- I will see you in the next video.

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