Repeated roots of the characterisitic equations part 2

An example where we use initial conditions to solve a repeated-roots differential equation.

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