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# Repeated roots of the characterisitic equations part 2: An example where we use initial conditions to solve a repeated-roots differential equation.

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- Let's do another problem with repeated roots.
- So let's say our differential equation is the second
- derivative of y minus the first derivative plus 0.25--
- that's what's written here-- 0.25y is equal to 0.
- And they've actually given us some initial conditions.
- They said that y of 0 is equal to 2, and y prime of 0 is
- equal to 1/3.
- So like we've done in every one of these constant
- coefficient linear second order homogeneous differential
- equations, let's get the characteristic equation.
- So that's r squared minus r plus 0.25-- or we can even
- call it plus 1/4.
- So let's see, when I just inspected this, it always
- confuses me when I have fractions.
- So it becomes very hard to factor.
- So let's just do the quadratic formula.
- So the roots of this are going to be r is
- equal to negative b.
- Well, b is negative 1.
- So negative b is going to be 1.
- Plus or minus the square root of b squared.
- b is negative 1.
- So that squared is 1.
- Minus 4 times a, which is 1, times c.
- Well, 4 times 1 times 0.25, that's 1.
- Ah-ha.
- So notice that when you have a repeated root, this under the
- square root becomes 0.
- And that makes sense, because it's this plus or minus in the
- quadratic formula that gives you two roots, whether they be
- real or complex.
- But if the square root is 0, you're adding plus or minus 0
- and you're only left with one root.
- Anyway, we're not done yet.
- What's the denominator of a quadratic equation?
- 2a.
- So a is 1, so over 2.
- So our one repeated root is 1 plus or minus 0 over 2, or it
- equals 1/2.
- And like we learned in the last video, you might just
- say, oh well, maybe the solution is just y is equal to
- ce to the 1/2 x.
- But like we pointed out last time, you have two initial
- conditions.
- And this solution is not general enough for two initial
- conditions.
- And then last time, we said, OK, if this isn't general
- enough, maybe some solution that was some function of x
- times e to the 1/2 x, maybe that would be our solution.
- And we said, it turns out it is.
- And so that more general solution that we found, that
- we figured out that v of x is actually equal to some
- constant plus x times some other constant.
- So our more general solution is y is equal to c1 times e to
- the 1/2 x soon. plus c2 times xe to the 1/2 x.
- I forgot the x here.
- Let me draw a line here so you don't get confused.
- Anyway, that's the reasoning.
- That's how we came up with this thing.
- And it is good to know.
- Because later on when you want to know more theory of
- differential equations-- and that's really the whole point
- about learning this if your whole goal isn't just to pass
- an exam-- it's good to know.
- But when you're actually solving these you could just
- kind of know the template.
- If I have a repeated root, well I just put that repeated
- root twice, and one of them gets an x in front of it, and
- they have two constants.
- Anyway, this is our general solution and now we can use
- our initial conditions to solve for c1 and c2.
- So let's just figure out the derivative of this first. So
- it becomes easy to substitute in for c2.
- So y prime is equal to 1/2 c1 e to the 1/2 x, plus-- now
- this becomes a little bit more complicated, we're going to
- have to use the product rule here-- so plus c2 times--
- derivative of x is 1-- times e to the 1/2 x, that's the
- product rule.
- Plus the derivative of e to the 1/2 x times x.
- So that's 1/2 xe to the 1/2 x.
- Or we can write-- I don't want to lose this stuff up here--
- we can write that it equals-- let's see, I have 1/2-- so I
- have c2 times e to the 1/2 x and I have 1/2 times c1
- e to the 1/2 x.
- So I could say, it's equal to e to the 1/2 x
- times c1 over 2.
- That's that.
- Plus c2.
- That takes care of these two terms. Plus c2 over 2
- xe to the 1/2 x.
- And now let's use our initial conditions.
- And let me actually clear up some space, because I think
- it's nice to have our initial conditions up here where we
- can see them.
- So let me delete all this stuff here.
- That, hopefully, makes sense to you by now.
- You know the characteristic equation.
- We figured out the general solu-- I don't want to erase
- our initial conditions-- we figured out the general
- solution was this.
- I'll keep our general solution there.
- And so, now we just substitute our initial conditions into
- our general solution and the derivative of the general
- solution, and hopefully we can get meaningful answers.
- So substituting into our general solution, y of 0 is
- equal to 2.
- So y is equal to 2 when x is equal to 0.
- So c1-- when x is equal to 0, all the e terms
- you become 1, right?
- This one will become 1.
- And then notice, we have an xe to the 0.
- So now this x is 0.
- So this whole term is going to be equal to 0.
- So we're done. c1 is equal to 2.
- That was pretty straightforward.
- This x actually made it a lot easier.
- So c1 is equal to 2.
- And now we can use the derivative.
- So let's see, this is the first derivative.
- And I'll substitute c1 in there so we can
- just solve for c2.
- So our first derivative is y prime is equal to-- let's see
- c1-- 1/2 plus c2-- so it's-- well I'll write this first--
- it's equal to 2 over 2.
- So it's 1 plus c2 times e to the 1/2 x plus c2 over 2 times
- xe to the 1/2 x.
- There was an x here.
- So when x is equal to 0, y prime is equal to 1/3.
- So 1/3 is equal to-- well, x is equal to 0, this'll be 1--
- so it's equal to 1 plus c2.
- And then this term, when x is equal to 0, this whole thing
- becomes 0, right?
- Because this x just cancels out the whole thing.
- You multiply by 0 you get 0.
- So then we get 1/3 is equal to 1 plus c2, or that c2 is equal
- to 1/3 of minus 1 is equal to minus 2/3.
- And now we have our particular solution.
- Let me write it down and put a box around it.
- So this is our general solution.
- Our particular solution, given these initial conditions for
- this repeated root problem, is y is equal to c1-- we figured
- that out to be 2 fairly quickly-- 2e to the 1/2 x plus
- c2. c2 is minus 2/3.
- So minus 2/3 xe to the 1/2 x.
- And we are done.
- There is our particular solution.
- So once again, kind of the proof of how
- do you get to this.
- Why is there this x in there?
- And it wasn't a proof, it was really more of just to show
- you the intuition of where that came from.
- And it did introduce you to a method called, reduction of
- order, to figure out what that function v was, which ended up
- just being c1 plus c2 times x.
- But all that can be pretty complicated.
- But you see that once you know the pattern, or once you know
- that this is going to be the general solution, they're
- pretty easy to solve.
- Characteristic equation.
- Get your general solution.
- Figure out the derivative of the general solution.
- And then substitute your initial conditions to solve
- for your constants.
- And you're done.
- Anyway, I'll see you in the next video.
- And actually, we're going to start solving non-homogeneous
- differential equations.
- See

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