you
okay cool so yeah we're going to start
out with number one so we're going to
solve y prime plus 2xy squared equals
zero by separation of variables so yeah
our first step is basically going to be
to subtract two XY squared from both
sides so then we're going to get Y prime
equals negative two x y squared and so
we can actually rewrite this Y prime
piece as dy/dx and say that that still
equals negative 2xy squared
so then with separation of variables you
basically just want to get all the X
variables on one side and all the Y
variables on the other side so what
we're going to do is multiply over the
DX piece and divide over the Y squared
piece so then we're going to get dy over
y squared
poppin down the Y squared and moving
over the DX and that's an equal negative
two X DX
okay
so then we can integrate both sides so
we're going to take the integral and I'm
going to rewrite this one over y squared
as Y to the negative 2nd power because
that's the one of
we have you can flip things from the
numerator to the denominator or vice
versa just by changing the sign of the
exponent and so then that's also we have
to integrate the right hand side so
that's going to be the integral of
negative 2x DX
okay so then we're gonna get Y to the
negative first adding 1 and dividing by
our new exponent so we divide by
negative 1 equals negative x squared
adding 1 and dividing by 2 and then we
should toss on a plus C so then we get Y
inverse equals x squared minus C popping
over this negative into each piece and
so then we get 1 over y equals x squared
minus C so then we can take the
reciprocal of both sides and get that y
equals 1 over x squared minus C okay is
that okay with everyone does anyone have
any questions about that
okay cool
see ya that's number number one done and
I think in Pinsky solution he writes
plus C there can anyone tell me why
that's okay
the
concentrate yeah exactly so if you have
a constant you make it negative it's
still a constant yeah exactly
cool so let's move on to number two so
we have this differential equation y
prime equals y squared minus y cubed and
you want to sketch the YT plane find the
equilibrium points and then describe the
stability of each equilibrium point okay
so what is our first step
you
you
you
you
you
and then you guys can also feel free to
use the chat too if you don't want to
talk on your mic I'll keep an eye on the
chat but what is our first step right
here what do we do if we're looking for
the equilibrium point
you
you
you
yep exactly so we want to find the zeros
so that basically means we take Y prime
and we set it equal to zero and so all
that means is that you're going to have
a slope of zero so we're going to set
zero for our y Prime
and we're going to set that equal to y
squared minus y cubed so that becomes
zero equals we can pull out a y squared
and we're left with 1 minus y so that
tells us y equals zero or Y has to be
one so these Y values are going to be
our equilibrium points and so if we try
to sketch this out
let's say we have
why equals one right here so that's
actually going to be a solution so if
you take a y-value of one your slope
this setup right here is always going to
be zero
we specifically solve this equation so
our slope is zero whenever our height of
Y is 1 so that means you don't have a
change in Y and you just keep moving
along a constant over time so that's
what this would solution would look like
ok and we also know we have the same
thing at y equals 0
okay and so then can anyone tell me our
next step
stability yeah exactly
stability slash plug in values perfect
so then what we can do is we can make
this little number line and so we can
analyze what happens between our zeros
so we know that when we have a y-value
of 0 we get 0 and we know that when we
have a y-value of 1 we also get 0 but we
want to figure out what happens in
between so if we plug in a number say
like 1/2 between these two so that would
go right in there so we would have 1/2
squared which is 1/4 and then we'd have
1 minus 1/2 which is 1/2 so that comes
out to be positive so we know it's
positive in between there in between 0 &
1
ok and if we look at say negative 2
we're going to get a positive and then 1
minus negative 2 is another positive so
this is also positive
you
normally when I have it factored like
this I don't figure out the exact number
I just kind of say like positive times
positive and positive times negative
type of thing and then lastly if we plug
in three or something like that we're
gonna get positive and then 1 minus 3 is
negative 2 which is negative so that's
positive times a negative which is a
negative okay
so yeah that's basically
our setup for y prime so that tells us
so let's consider this piece first so
for it we have a y-value between 0 & 1
so that means we're in here then our
slope is positive so that means we need
to be gaining Y so we need to be gaining
and Y so basically we're gonna have to
come up it's positive it's going upwards
but then we have to get bounded by this
y equals 1 piece so it starts to come
down a little bit like that ok and so
that's like one solution if we look at
stuff below zero again it has positive
slope so again it has to be when we move
this out of the way so again we need to
be moving upwards so we're moving
upwards it's all fine but then we kind
of start to run into this bound and we
need to taper it down a bit
so that we can kind of coincide with
this solution so that's another kind of
solution if we consider this negative
piece for our Y values above one we need
to come down but then we need to start
agreeing with this other solution so we
kind of go like that okay so now we
would say that this equilibrium point is
stable or an attractor because it's like
pulling everything in so even if you
start a little bit away from it you're
good because you're still going to
arrive basically at y equals 1 so we
would call this stable
or a tractor
and then this this is what we would call
semi-stable because if you choose a
number slightly below y equals zero
you're fine you're still going to come
back to y equals zero but if you choose
a number above y equals zero you're not
fine it goes somewhere else so this we
would call semi stable
okay any questions about that
you
okay cool so yeah to summarize your
first step is to set y equal
to set Y prime equals to zero and then
your second step is to do this kind of
number line idea to figure out the
behavior of y Prime in between your
equilibrium points okay
cool and yeah feel free to ask questions
too in the chat yeah let's keep going
so go on to number three so here we have
whoops my bad here we have y prime is
equal to two times XY okay so some
differential equation why evaluate it at
1 equals 1 so our starting kind of point
is 1 for an x value of 1 and we want to
compute 2 steps of Euler's method with
our Delta X being 0.01 ok cool so we
know Y of 1 equals 1 so then we want to
find out what Y prime is at 1 because
then we can use that Y prime which is
kind of like how much height and Y you
get for how much change in X you give
and since we know how much change in X
we're going to be using we can find out
how much change in Y we get so we know Y
prime of 1 is going to be 2 times X
X is going to be one and Y is also one
okay so this becomes two so that means
for every step we take in the
x-direction we gain two steps in the
y-direction so for us we're only going
to take point one of a step in the
x-direction which means we're only going
to gain
point two steps in the y-direction so
then what should we do from here any
ideas now that we know how much of a
change in Y we are going to get
you
you
you
okay so what we do is we gain this point
- and why so we're going to say like our
next y-value maybe I'll call it Y
that's why one because it's like our
first iteration through the process this
would have been y 0 so y 1 is going to
be our previous y plus 0.2 our previous
y was 1 so we're going to take one and
add 0.2 so that's 1.2 so that's one
iteration through Euler's method so
that's one step basically so if we will
only need to do one step that would be
our answer but we need to do two steps
so then we just kind of repeat the
process again so now we need to find y
prime this Y height so Y prime is going
to be
two times X so our new x value remember
is we took one we started at one and
then we added point one so our new x
value is one point one
okay and then we need to plug in our
current y-value and our current y-value
is 1.2 1.2 so what does that come out to
be 11 times 12 is that's gonna be 132
I'm doing my math right yeah 132 so
that's going to be two times one point
three two which is two point six four so
now this is our Y prime again we're only
taking a step of 0.1 in the X direction
we're only taking 1/10 of a step so
we're only going to gain one tenth of
that so delta y is going to be a tenth
of that which just means move the
decimal to the left so that means our
delta y is going to be 0.26 four and so
then our next y value maybe we'll call
it y sub two is going to be our previous
y-value which was y sub one plus this
Delta Y okay and our previous Y value
was one point two
and we just found out Delta Y was zero
point two six four so that means it's
going to be one point four six four okay
so basically what we did was
we have some sort of Y Prime and it's
changing and so we took like a first
approximation here and said it kind of
looks like this we kind of took one step
and then we took like this slope and put
it there and we're just kind of
estimating so it's an approximation okay
and yeah so this should be approximately
the value of y of one point two because
we took two steps of 0.1 in the
x-direction okay does anyone have any
questions about Oilers method
you
you
okay cool
sweet so let's keep keep moving on
so number four we have
system of differential equations so I
think this should actually be a capital
y to represent kind of like the idea of
a vector yeah anyway we know that dy DT
the rate at which this Y vector changes
with respect to time is basically this
linear transformation of itself okay so
this Y vector is changing as a linear
transformation applied to itself but
anyway what's our first step when we
want to solve these systems of
differential equations
you
you
you
they're like lambda times the identity
matrix yeah exactly so we want to find
what's called the eigenvalues of this
linear transformation or this matrix and
the way you find those eigenvalues is
you take the determinant
if we call this matrix a you take the
determinant of a minus lambda I so
that's going to be the determinant of a
which we know is one negative eight one
negative three we want to subtract
lambda I and all that lambda I is is
lambdas on the diagonal and the zeros
everywhere else
okay so this becomes the determinant of
and so the way you subtract two matrices
is you take this top left corner and you
subtract this top left corner top right
- top right bottom left - bottom left
bottom right - bottom right so you just
kind of subtract it position wise so
we're going to get the determinant of 1
minus lambda 1 minus lambda negative 8 1
and the negative 3 minus lambda so now
we want to find this determinant okay
let me make sure it did that step right
1 minus lambda negative 8 1 and negative
3 - I'm okay cool sweet so can anyone
tell me how to find the determinant of a
2 by 2 matrix what do we do
so drink and yeah exactly
cool so what we're gonna do is we're
gonna take whatever is here and multiply
it by this and then we're going to
subtract whatever is here and multiply
it by this so that's going to be one
minus lambda times negative three minus
lambda minus now we do this product one
times negative eight okay
and so then once you have this
polynomial what you want to do is you
want to factor it completely and so if
you want to factor completely you kind
of actually need to expand it first and
then factor it so that you get all the
like terms fully combined so this is
going to be we're going to have a lambda
squared piece
we're going to have
positive lambda positive three lambda
and then we're gonna have negative
lambda so that means positive three
lambda minus lambda which is just two
lambdas and then we're also going to get
a minus three piece
okay yeah I lambda squared plus 3 lambda
minus lambda and then minus 3 yeah and
then this simplifies to plus 8 so that
becomes lambda squared plus 2 lambda
plus 5 and so then we think what two
numbers multiply to 5 and add to 2 so 5
is prime so our only options are 1 times
5 which adds to 6 or negative 1 times
negative 5 which adds to negative 6 so
then we realize that this doesn't really
factor nicely in terms of real numbers
and so if you're kind of stuck when it
comes to factoring and it's not natural
or obvious you can always resort to the
quadratic formula which is given by
negative b plus or minus the square root
of b squared minus 4ac
all / qà okay so then we're going to get
negative 2 because this is a B let's see
so we're going to get negative 2 plus or
minus the square root of B squared so
that's 2 squared which is 4 minus 4
times a which is 1
times C which is five
/ 2 times a a is 1 so that's 2 times 1
which is just 2 so this becomes negative
2 plus or minus the square root of so
we're gonna have 4 minus 24 minus 20 is
negative 16 and the square root of
negative 16 is 4 I and then we're still
dividing by 2 so this becomes negative 1
plus or minus 2i okay
our first step is to find the
eigenvalues of this matrix which we just
figured out are these two and they're
complex okay so does anybody have any
questions about finding those
eigenvalues is everyone okay with we're
at so far
you
yes you need all right sweet so then our
next step is to find the eigenvector
associated to our I can value so I
actually went ahead and wrote down the
formulas we're going to need so let me
talk about what they are so this is
going to be our first solution and this
is going to be our second solution and
our final solution is going to be c1 x1
plus c2 x2 because you can take linear
combinations of solutions to
differential combination two
differential equations and still get us
our solution so yeah our first solution
is what's called x1 equals beta b1 cows
beta t minus b2 sine beta T e to the
Alpha T and so beta is just the piece
attached to I alpha is the real piece
and b1 and b2 are vectors and the
vectors that they are they're the real
component of k1 that's B 1 and B 2 is
the imaginary component of k1 so now we
need to know what k1 is and k1 is just
the eigenvector associated to just the
positive 1 so k1 is the eigenvector
associated to negative 1 plus 2i ok so
that's kind of a lot of information but
basically we already figured out most of
this stuff we figured out the alphas and
the beta's we still need to figure out
the B 1 and the B 2 and the way you
figure those out is you have to figure
out k1 which is an eigen vector
associated to the positive eigen value
ok so yeah now our job is to find out
what this k1 is let's I can rector so
now we want to find the eigen vector
associated to lambda equals negative 1
plus 2i
okay so how do we find an eigenvector
that's associated to an eigenvalue what
do we do
I'll go back into the matrix for Landon
yeah exactly so we're going to take
lambda which is this and plug it back in
here so we know a minus lambda is 1
minus lambda negative 8 and then 1
and then negative three minus lambda
cool and so now we can plug in what
lambda actually is so this is the same
thing as 1 minus
negative one plus 2i and make sure you
have parentheses on it because you need
to distribute that negative so negative
eight and then one on the negative three
and - I'm gonna get a 1 plus 2i
so that's just lambda right there okay
so that's going to become we're gonna
have one minus negative one which is the
same as 1 plus 1 which is 2 and then
we're gonna have 1 - all right now then
we're gonna have negative 2i we're gonna
have - yeah because we get this
simplifies to 1 minus 2i and so then you
would get 2 minus 2i so then we get
negative 8 and then we get a 1 and then
here we're gonna get negative 3 plus 1
minus 2i by distributing that negative
into each piece so negative 3 plus 1 is
negative 2 and then we have plus 2i
and so then what we can do is if we say
that this K k1 vector
did you have a question but 2-2 lie
right on the bottom right corner
yeah great catch nice thanks yeah sweet
that's why I need you guys yeah a lot of
mistakes I'm cool so K 1 is going to be
K 1 K 2 so if we write k 1 vector as
like this piece and then this piece then
this translates to 2 minus 2i times this
K 1 piece
plus negative eight times this k2 piece
and we're gonna actually set this to 0
because we set this matrix
equal to the zero vector okay and you
actually don't really need to worry too
much about this bottom row because it
ends up kind of being the same
information as the top row okay
so then what we can do is we can say 2
minus 2i k1 equals positive 8 k2
yeah and so then what we can do is we
can say all right let's make this even
simpler let's kind of take out all the
twos so divide by two on each side to
get 1 minus I k1 equals 4 k2
okay and so then what I would do is I
would probably set
k1 equal to 4 so that means this is 4
so on the left hand side we get 4 minus
4i equals on the right hand side we
just still get that 4 k2 so dividing
by 4 we get k2 equals 1 minus i so
that's what I would probably do I notice
that Pinsky chose a different one so I'm
gonna actually convert like what I would
naturally do to it he would like to do
so this means k1 is going to be based
on what I would do it before 1 minus i
but what he wants to do is he doesn't
really want the complex in there he
wants the complex there so what we can
do is we can scale this by any complex
number
so an eigenvector
has the property that you can still scale it
and it'll still be an eigenvector so
we're going to do that now so we're
going to get 4 plus 4i and then we need
to figure out what 1 plus i times 1
minus i is so that's a difference of
squares so that becomes 1 minus i
squared which is the same as 1 minus
negative 1 which is 2 so then we get a 2
there well we had kind of a more twos
than we need so we can scale this by 1/2
to get 2 plus 2i, 1 and I believe that's
the eigenvector that he wanted to use
yeah it is cool so that's the
eigenvector that he wanted to use I'm
gonna switch it to that just so we stay
consistent with the solution key but
yeah that's how you can find an
eigenvector so kind of a lot of steps
there but basically the
gist of it is you plug in what your
lambda is what your eigenvalue is and
then for us we only need to focus on
this top equation because it turns out
these two equations are actually
conveying the same exact information and
so once we know (2 - 2i) k1 plus
negative 8 k2 equal to 0 you can solve
basically for how k1 and k2 relate to
each other and then I want to emphasize
you can put whatever you want in for k1
and k2 to make it easier for yourself
except you never want to have an
eigenvector of 0 0 so this should never
be 0 0 so be a little bit careful you
don't want it to be that that's not
allowed as an eigenvector but yeah so i
chose k1 to be 4 so that i could just
divide over the 4 nicely and that gave
me basically this eigenvector but then I
figured I'd stay consistent with the
solution key and this is still an
eigenvector because you can scale
eigenvectors but yeah so that is K 1 ok
any questions about that step
yeah we're making pretty fast progress
so feel free to to ask questions
okay so that's that step so now we can
start to figure out what b1 and b2 are
now that we know what k1 is so b1 is the
real part of k1 so if we write k1 as
let's write it as 2 times 1 plus i times
2 0 is that ok with everyone because 1
is the same thing as 1 plus 0 i so you
can kind of split up k1 that way so that
means this is the real part B1 and this
is the imaginary part B2 okay cool so
now we're ready to plug it into our
solutions so I'm just going to copy what
they are
which is kind of the important
information
okay let me get cleaned it up a little
bit yeah so here's what our solution is
and here's what B1 and B2 are up here so
now we're ready to plug it in so we know
x1 is going to be this this is B1 is
what we just found to be 2 1 so it's
going to be 2 1
cosine and then beta what's beta
what's beta if this is our eigenvalue that
we're using
okay so this is going to be the same
thing that's alpha plus or minus beta i
so your beta is the thing attached to
the 2 your alpha is the real piece okay
so that's how you find alpha and beta so
in our case beta is 2 so we're going to
have B1 cosine 2 t minus B2. What's B2?
it's 2 0
and then we have sine beta t which beta
is 2
and then we have e to the alpha t and
okay yeah cool someone put in the chat 2
then we have e to the Alpha t and we
found alpha is negative one
alpha is negative one so we're going to
put e to the negative one t which is the
same as e to the negative t ok cool so
that's that solution we have another
solution which is going to be B2 cosine
ß t. B2 remember is this so 2 0
sine of beta t is the same as sine of 2t, my bad cosine and then we have plus B1
B1 is 2 1
all right this is B1 and then sine beta
t so this is sine 2t and then we have e
to the negative t ok and so yeah we're
pretty much done technically all you
need to do is
pop on a c1 there do you plus and pop
on a c2 there and that's your
final answer but a lot of the times
they'd like you to simplify it a little
bit so if it's a multiple choice you may
have to clean it up a little bit so
we're gonna clean our solution up a
little bit so our final solution is
going to be I'm sorry these should
really be y's it's not too important
but
since we did technically start out with
y it's a good idea to stick with that
not really that important but anyway so
we're going to get y equals c1
times x1 which is all of this and then
I'm just going to go ahead and clean up
this inside a little bit so if we clean
up this inside we can actually pop this
cosine 2 t into each top and bottom same
with the sine and then we can add and
subtract the top so up top we're going
to have to cosine t minus 2 sine 2t so up
top we're going to get
actually yeah this is already a vector
So up top we're going to get 2 cosine t
and then to sine 2t okay is that okay
with everyone
on the bottom we're going to get one
cosine t minus zero sine 2t which just disappears because it's zero so on the bottom we're
gonna get one x cosine 2 t which is just
cosine
2 t okay and then we have our second
solution
and we need to clean up this inside so
up top we're going to have to cosine 2 t
plus two sine 2t
2 cos 2t plus two sine 2t and on the
bottom we're going to get 0 cos 2t + 1
sine 2t so that's just sine 2t ok
and yeah that is the correct answer and that's the most
kind of simplified version to get it in
so if it's a multiple-choice just
heads up like this is totally correct
but they may write it in an alternate
form to look like that even though your
answers right okay any questions about
that
can you repeat how you obtained b1 and
b2 yeah for sure so the way you find b1
and b2 is you find this eigenvector and
this eigenvector Jake are you cool with
how we got this this eigenvector right
here?
okay cool so then the way you do it
basically when you move this over a
little bit
so then we can write this is the same
thing as this
we're plus 2i yeah so this is what we
want to rewrite this step I was just
multiplying by 1/2 to make it a little
bit nicer
so yeah this this was our eigenvector
this was k1 so we want to find find out
how to get B 1 and B 2 from this k1 so
we know we can write k 1 like this there
was a secret 0 i write here so we really
had just the one but you want to have
everything with an i so we can call that
1 plus 0 i and so then what we can do is
we can say ok this is 2 1 plus 2i 0i
because that's how you add vectors the
way you add vectors you add the tops and
then add the bottoms
and then what we did was we factored out
an i from this vector and put it right
in there
does that make sense Jake
okay cool yeah good question so that's
how you kind of split up the eigenvector
in terms of its real part and
its imaginary part okay so yeah this was
our eigenvector and then we had a secret
0i right there and so you can split it
into two pieces and then we factor out
that i to get b1 and b2 any other
questions
okay cool and if you want you guys can
also
use your mics - and feel free to talk
don't worry about interrupting me or
anything like that but yes and now let's
move on to repeated eigenvalues I went
ahead and wrote down some formulas we'll
need but yeah it's gonna be the same
kind of set up you know actually let's
break it up a little bit let's not just
do matrices we'll go do something else
give ourselves a break yeah I already
solved this one
ignore that we'll come back to five
let's just mix it up a little bit and
not get too absorbed in matrix stuff and
keep it fresh we'll go back but yeah for
now we're going to solve y double prime
plus two y prime equals Delta of t minus
one where y 0 is zero so our starting
point is zero zero and our derivative at
our initial time zero is one okay so and
they tell us to do this by using Laplace
transforms
and yeah so basically what's our first
step going to be if we want to solve
this differential equation using
Laplace
yeah exactly we're gonna want to take
the Laplace of both sides so in order to
do that we need to know what the Laplace
of Y double prime is we also need to
know what the Laplace of Y prime is and
we also need to know what the Laplace of
Delta t minus 1 is so I wrote down some
formulas to help us out so the Laplace
of Y double prime is s squared Y of s
minus s Y of 0 minus y prime of 0 if you
guys have a formula sheet I would write
all the Laplace's I could probably
but with enough practice you'll start to
have them memorized I personally haven't
taken diff eq in a little bit so I'm
kind of rusty on them so I always have
to use the formulas a little bit yeah
there's not like a fantastic way to
memorize Y double prime in my opinion
Laplace of Y prime is s YS minus y
of 0 so we kind of see a pattern that
this power of 2 drops every time and
this power of 2 drops every time and
then I wrote down the Laplace of Delta t
minus C which is kind of the more
general version of Delta t minus 1 and
that's e to the negative CS and then I
also wrote down some inverse Laplace's
that we'll need a little bit later so
yeah exactly we're gonna take the Laplace
of both sides so we just figured
out that Laplace of y double prime
is s squared Y of s minus s y 0 minus y
prime of 0 yep and then the Laplace is
linear so you can actually just kind of
pop in the Laplace onto just this piece
and multiply by 2 so we're gonna have 2
times a Laplace of Y Prime and the
Laplace of Y prime should be s Y of s
minus y of 0
sY of s minus 0 perfect so that's the left hand
side done and then we want to take the
Laplace of Delta t minus one and we
know the little plus of Delta t minus C
is e to the negative CS so this is going
to be e to the negative one s which is
just e to the negative s perfect so now
what should we do
now that we were finished taking the
Laplace of both sides
plug in the value given in a problem
combine like terms yeah both of those
are correct
so we can make it easier on ourselves
and you guys are both right whatever
order you want to do it it's fine but
yeah we're gonna plug in what these are
and then combine like terms so y 0 is 0
so that vanishes that also vanishes and
we're gonna get s squared Y of s we know
Y prime of 0 is 1 so minus 1 because
that's just 1 and then plus 2 s Y of s
and that still equals e to the negative
s
cool yeah so s squared YS minus one plus
two s YS equals e to the negative s cool
so then we're going to combine like
terms so maybe we'll add over this one
first so on the right hand side we're
going to get e to the negative s plus
one and then over here we'll just have
these two left over and we'll factor out
a Y to the S piece and we'll be left
with Y s times s squared plus 2 s add
one factor out the Y s, s squared plus 2 s, 
yeah that looks good and so then we want
to get Y of s by itself
and the reason we want to get Y of s by
itself is because when we take the
inverse Laplace of capital Y of s that
gives us our solution lowercase y of T
so yeah you want to get y of s by itself
so then to do that we divide over a
squared plus two s underneath so that's
going to be e to the negative s plus 1
all divided by s squared plus 2s all
right so this is the part where the
inverse Laplace comes into play so now
that we have capital Y by itself we
can take the inverse Laplace and inverse
Laplacess are always kind of like a
puzzle you know they're like integrals
there's not necessarily a set way you
just kind of have to toy around with
them and break them into things that
look more familiar so for us this is our
puzzle okay so we need to think to
ourselves how are we gonna inverse Laplace
this so I'm gonna open up to you
guys what do you guys think any ideas on
how to inverse Laplace this
it doesn't really look familiar right
now at all so we want to make it look
more familiar sine of X multiplied by
step function yeah that's definitely a
piece of it looking at this yeah very
good so it's helpful to kind of know
things that you are familiar with so for
us these are some inverse Laplace's that
are probably going to be helpful so if
you have e to the negative CS times
anything
that is how it transforms so it
transforms into a step function times
what y would normally be but instead of
plugging in t you plug in t minus C okay
so that's really general and abstract
but we'll see how that helps us later it
would also be helpful to know how that
transforms and how that transforms so
let's say okay I know how to do
something when there's an e to the
negative s attached so I'm going to
divide this into each piece okay and
that's allowed if you have a plus B over
a fraction you can pop it into each
piece
but you can't really do it the other way
so you can't do C over a plus B you
can't like pop that into each piece
that's the common mistake but you can't
so for us we can take this whole thing
and divide it into each piece so this is
going to become e to the negative s
times 1 over s squared plus 2 s
also gonna have plus 1 over s squared
plus 2 s so that's my best advice if you
ever see an e and a plus I would say try
to get the e by itself from the plus by
itself because if we know how to inverse
Laplace this we're done with that part
if we know how to inverse Laplace this
then we can actually can do the
whole thing because we have a formula
for the inverse Laplace of something we
know times the e piece so really if we can
figure out how to do this piece we're
done if we can figure out how to inverse
Laplace that does anyone have any ideas
on how to inverse Laplace that
partial fraction, yeah partial fraction
decomposition perfect
so probably work through the partial
fraction decomposition a little bit
quickly so this is the same thing as 1
over if we pull out an SS times s plus 2
and the reason we're doing partial
fraction is because we know how to do
this and we know how to do that so this
is going to be we're going to say a over
s plus B over s plus 2 that's how
partial fraction decomposition, we just want to figure out that
and why do we want to figure out that? Like I said, we know how to do these, so you can just toss on an a in front of them
and so then our next step is to multiply
this to both sides
so we get 1 equals a s s plus 2 over s
so the S is canceled and then we're also
going to get plus B s s plus 2 over s
plus 2 B s plus 2's cancel so we're left
with that if you plug in s equals
negative 2 that thing goes to 0 and
we're going to get B equals negative 1/2
taking this negative 2 and dividing it
over if that's a little bit fast about
you to kind of check it out on your own
and work it out for yourself I did too
many steps but keep going if we plug in
s equals 0 that term vanishes this is
just the 2 we divide over the 2 and a
equals 1/2 so a would be 1/2 so again
it's a lot of steps
but yeah feel free to plug it in for
yourself so then we get this equals a is
1/2 so we can write the 1/2 up here and
divide by us but I'm gonna write it like
this 1/2 times 1/s and then B is
negative 1/2 times 1 over s plus 2 okay
cool so now it's our job to inverse
Laplace this which actually seems
reasonable because we know the inverse
Laplace of 1 over s is 1 and we know how
the inverse Laplace of 1 over s minus a
transforms so this is going to inverse
Laplace to become 1/2 times 1 minus 1/2
so this is secretly s minus negative 2
so according to this formula you take
negative a and you put the positive
version so we have minus negative 2 so
we're gonna have eat to the negative 2t
great so now we know that this becomes
1/2
- 1/2 e to the negative 2t so let me
kind of move this scratch work out of
the way
so we know Y of s or y of t is the
inverse Laplace of
this whole thing and so yeah we know
we're gonna have a plus we figured out
what that piece was we figured out it's
1/2
- 1/2 e to the negative 2t
okay
so that's that piece we still need to
take care of this piece but we have a
formula for how to do that so what we do
our first step is we consider this e to
the negative CS and we plug in the step
function or the Heaviside function
evaluated at t minus C so for us
this is negative one so the first thing
we're gonna do is we're going to
multiply by u t minus one okay and
then what our formula says to do is it
to take what you know regular thing
should be but instead of plugging in
just a regular t you have to plug in a
new weird t you have to plug in t
minus C okay so that means plug in what
this would regularly be and this would
regularly be we already figured it out
that so it's gonna be one half minus one
half e to the two t but you have to fix
it a little bit you can't pop in an
a regular t you have to plug in
a weird t which is t minus C
t minus one
okay so that kind of gives you an idea
of how to inverse Laplace anything with
an e attached to it I know that can be
kind of challenging to think about like
plugging in a weird value for t but yeah
that's kinda how you do it does that
make sense to everybody does anyone have
any questions about this
okay cool yeah that's how you do the e
piece and yeah cool and that's our final
answer right there it's here so that is
the answer let me double check that
on the solution key
perfect cool oh wait this should
actually be
it should be e to the negative 2t I
forgot the negative
yeah I had it here but I dropped it
negative two negative two cool all right
that's our final answer
so we have another Laplace one and we
have undetermined coefficients in linear
why don't we go back to the matrix one
so we don't get too sick of Laplace so
we'll come back to this matrix one cool
yeah so here we have a repeated
eigenvalue matrix question and so dy/dt
is going to be negative 6 5 negative 5
4 y and so our first step is to find
the eigenvalues so we want to find the
determinant of negative 6 minus lambda 5
negative 5 4 minus lambda
and these straight bars are a notation
for determinant okay so that's going to
become product product minus product
product so that's gonna become negative
six minus lambda times four minus lambda
minus negative twenty five
so that's just +25 so this becomes
lambda squared then we get negative 4
lambda plus 6 lambda and negative 4
lambda plus 6 lambda is just 2 lambda
and then we also have a negative 24
attached so we have minus 24 plus 25
which is just plus 1
okay can anyone tell me how that factors
lambda plus one squared yeah exactly
so this becomes lambda plus 1 squared
cool all right so that means our
eigenvalue it's going to be lambda
equals negative 1 with multiplicity of 2
and so I went ahead and wrote down some
formulas we're going to need so this is
in the case of repeated eigenvalues the
formulas are going to want
so you have x1 equals K e to the lambda
t can anyone tell me what that K means
in this notation
it's going to be the eigenvector
associated to our eigenvalue that's what K
this K vector is
then our second solution is going to be
this K that we find T e to the lambda T
that's fine we know what lambda is and
then we have to add a P e to the lambda
T and then we have to say that way
what's P well P is a solution to this
equation it's a solution to you take a
minus lambda I which is some matrix you
set it equal to K and you solve for P
okay
step one was to find lambda step two is
to find K so the way we find K is we
take a minus lambda I and we set it
equal to zero so we're going to set zero
equal to a minus lambda I a minus lambda
I is going to be negative six minus
lambda so that's negative six plus one
because if we subtract negative one then
we have five negative five and then we
have four minus negative one which is 4
plus 1 which is 5 negative 6 plus 1 is
the same thing as negative 5
so then we see that these two are
redundant so this equation is the same
exact thing of this equation so we can
translate this to saying okay the only
thing this is telling me is that zero
equals negative five K 1 plus 5 k2 where
again we're writing this capital K as
two pieces K 1 and K 2 so then we can
pick whatever we want to make this easy
on ourselves so let's stick with what he
wants to pick
so he chose K 1 to be one so if we set K
1 equal to 1 then we know 0 has to equal
negative 5 plus 5 K 2 if we add over the
5 and divide by 5 we get 1 so K 2 is
also 1
but yeah you could have picked whatever
you wanted as long as it this didn't
come out to be zero zero for us it came
out to be one one which is fine so that
is our eigenvector k associated to our
eigenvalue lambda equals negative one
so now we already know one solution but
we can do better we can find another so
now we want to find P and the way we do
that is we set a minus lambda I which is
negative 5 5 negative 5 5 and we set it
equal to K which is 1 1
okay so this is basically shorthand for
a minus lambda I times P equals K cool
then again these are redundant there are
same information so this tells us
negative 5 p1 plus 5 p2 equals 1
again we're writing capital P as P 1 P 2
so then we can plug in whatever we want
for
P 1 and P 2
he chooses p1 to be zero so if we set p1
equal to zero that means this term
vanishes and 5 p2 is equal 1 which tells
us p2 is 1/5 so that means P1 P2 is 0
1/5
and that is okay as an eigenvector
because it's not zero zero so now we
know what P is and we know what K is
then we know what lambda is so we can
plug in everything into here to get x1
or actually let's do this all is one
step so this is going to be C one times
K e to the lambda T but we know what K
is K is just one one
we know what lambda is lambda is just
negative one so that's going to be
negative t
that's c1 x1 right there and then we
want to do plus c2 times this whole
thing
okay but we know lambda is negative one
and we know K is one one
and we know P is zero one fifth and that
should be our final answer
let me double check yeah perfect cool so that
would be our final answer for the case
of repeated eigenvalues okay does anyone
have any questions about that one
okay so let's see what all do we have to
do left we still have we did all of that
did 6 so we only have seven eight and
nine why don't we take like a
five-minute break just so everyone's
brain can kind of get a break and you
can like get some water or whatever and
yeah we'll meet back here at at 10:15
okay
so feel free to take a break and just
relax for a second we have plenty of
time
can everyone see my screen okay
yeah okay cool
sweet so let's keep going
where were we so we have number eight
so we already do both Laplace's, no
this is my old work
we actually have number seven so we
haven't done this one yet so we want to
find the solution to Y double prime plus
4y equals one minus U of T minus one
which is like the step function or the
Heaviside function and we know Y of zero
is zero and y prime of zero is negative
one and we want to use the Laplace so yeah like
we said earlier our first step is to
take the Laplace of both sides so s
squared Y of s minus s y of 0 minus y
prime of 0 so that's Y double prime and
then we're going to have plus 4 and
we're going to have s y of s minus y of
0
so that's how why oh my bad actually I
was thinking of Y Prime
this should actually just be Y of s
cool yeah so the way Y transforms is
it's kind of like by definition the
notation for capital Y of s and so
then we have Laplace of 1 which is 1
over s then we have minus u of t minus 1
which I believe is e to the negative T
yeah it is Oh actually it's gonna be e
to the negative s over s yeah because
for example oh I wrote it down Laplace
of u t minus C is yeah okay so that's how
Laplace of U of T minus one goes is you
take e to the negative s and then you
divide by s cool
so now we'll plug in what we know so we
know Y of 0 is 0 and Y prime of 0 is 1
so we get s squared Y of s minus 1 plus
4 y of s equals 1 over s minus e to the
negative s times 1 over s
okay so then we can add one to both
sides so we get
let's pull out a Y of s over here so
we're going to add one so on the right
hand side we're gonna get one over s
plus one actually let me write it this
way one plus one over s plus e to the
negative s times 1 over s and then over
here we're going to be left with
s squared plus 4 because we got rid of
that one
okay so then we know Y of s is going to
be 1 over s squared plus 4 plus 1 over s
squared plus 4 plus e to the negative x
times 1 over X s squared plus 4 does
that okay with everyone how I kind of
distributed this into each piece we're
dividing through by it is that cool
you
okay sweet so then we're good we know
that this is very close to what we want
this we should be thinking partial
fraction decomposition we do know how to
do that okay so let's see what did I
write that would be helpful so I wrote
down some helpful inverse laplace --is
so if you have an S here and an S
squared there and an a squared there
that's Coast a T if instead of an S you
have an A then that's sign a T so for us
looking at this we're like ah that's
just really close to what we want we
would want a to to be there but we don't
have it to there so we're kind of like
dang there's anyone know how we could
fix that
you
any ideas how we can kind of force a two
to be right there
when really we have a one yeah perfect
multiply by 2 over 2 so we can multiply
by 2 over 2 to get that this is the same
thing as 1/2 times 2 over s squared plus
4 so that way we can kind of force it
but we didn't alter anything because
this still simplifies to 1 so that's
good we got what we wanted without
changing it too much and then this we're
gonna want to do partial fraction
decomposition decomp or do we want to
just kind of see what it is and not go
through all the steps what do you guys
think
you
you
I don't think it's too bad I just take a
decent amount of time
you
you guys have any preferences if you
want to want to do the partial fraction
decomposition into a calculator for us
it's your call whether you feel
comfortable with it or not
preference anyone else
yeah okay cool so it sounds like
everybody's cool with just figuring it
out what it is so let me do that real
quick
save ourselves some time
you
so let's plug in we really want to know
one over s times s squared plus 4 1 over
s s squared
plus four okay so that comes out to be 1
over 4 s minus s over 4 s squared plus 4
so this piece comes out to be 1/4 times
1 over s minus 1/4 times s over s
squared plus 4 ok so that's what that
piece is and yeah maybe I'll just say
like what it would have how you would
have at least set it up so you would
have set it up as a over S Plus be s
plus C over s squared plus 4 because you
can't reduce this any further
that doesn't factor over the reals any
further and so you need to have this be
s plus C linear term to do it and so you
would have solved that but we just
plugged it in to save ourselves some
time or hassle so yeah that's what that
piece is so maybe I'll actually pull out
the quarter
you
have it written nicely like that okay
cool so yeah one fourth one over s minus
one fourth s s squared perfect so that
means this next piece is going to be
plus one-fourth e to the negative s and
then we figured out how this can be
rewritten that can be rewritten as 1
over s minus
one over minus s over s squared plus 4
where it has a quarter in each slot but
I just put it right there
hey Darren we can't see what you're
writing
is it your screen okay let me let me
refresh it real quick
you
yeah it for some reason it does that
sometimes
how's that good yeah okay where did I
lose you guys that I know we had this
was it kind of here
right after you started writing
one-quarter on the first side of old
quarter okay gotcha yeah so basically we
uh we have the same exact thing so this
is this so this is going to be the same
exact thing but with an e to the
negative s attached
okay because this piece is still that
and that just like it was over here but
the only difference is now we have an e
to the negative F does that make sense
to everyone
okay cool yeah yeah so that's how that
transforms sweet and then now what we
can do is this is good so we know this
is going to be we pretty much know what
all of these are now so we're ready to
take the inverse Laplace so this is
going to be one half and if you have
constants I believe that's yeah that's
sign so this is going to be one half
sine of 2t because that's a two squared
right there is that okay with everyone
cool so they're going to have plus
one-fourth and we want to find the
inverse Laplace of that piece that's
just one then we have - we want to find
the inverse Laplace of that piece that
should be Coast so Coast is when you
have an S so that's going to be - Coast
cutie because that's two squared and
then we're gonna have plus 1/4 okay and
so now we have to remember how things
that we know but attached with an e how
those transform so what we do is we take
the the U of T - C take what Y would
normally be and plug in something
different T - C so for us this is going
to be U of T minus one that's kind of
what the e does and then we would say
okay well it's pretty much the same
thing 1 - we know this is Coast - T but
then we have to remember we don't get to
plug in regular tubes anymore instead we
have to plug in this which is t minus 1
okay cool
and yeah that should be good any
questions about that that's kind of our
full answer right there cool so that's
in rissalah plusses yeah we did
battle-ax plush did that one okay cool
so now we just have eight and nine to do
so for number eight we are going to do y
double prime plus y prime minus 6y
equals 2x using the method of
undetermined coefficients and so there's
kind of two steps to doing the method of
undetermined coefficients can anyone
kind of tell me what they are
you
you
you can just find me um characteristic
right here somewhere yeah yeah so what
we do is we find the characteristic
equation so we're going to set this this
translates to M Squared this translates
to plus M and this translates to minus 6
and we set that equal to 0 ok and so
this the reason we're doing this is to
find our
complementary solution so the
complementary solution you get by
basically solving the easy version of
this differential equation which is when
I equals zero and so the way you do that
is you take M Squared that's what this
becomes M and negative 6 and you said
that equal to 0 then you solve that
and then later on we're gonna find the
particular solution we'll talk about
that later but this is how you view the
complementary solution and so then we
factor this
so what multiplies the negative six and
adds to one 3 and negative 2 M plus 3 M
minus 2 so that means M equals negative
3 or 2
why complimentary it's going to be c1 e
to the negative 3 T plus c2 e to the 2t
is that okay with everybody
cool
sweet yeah that's kind of the general
general form for when you have two
solutions to this equation
and now we want to find why particular
so the way you find why particular is
you take a guess at what this kind of
right-hand side looks like and so we
think that looks kind of like a linear
equation so we write ax plus B I
personally would probably just do the ax
but if you have plus B that gives you an
even greater degree of accuracy so
whenever you have a 2x even though
there's only one piece you do you want
to give it the option a but plus B there
and what do we do now that we have a
guess at what Y particular should look
like I should actually call this a T now
I should call these axes
yeah what do we do take yeah exactly
and you also want to make sure that why
particular does not look at all similar
to why complementary so there's an issue
you can run into if they're the same but
we're fine
ours ours is different so we don't need
to worry about that so you guys want I
can talk about that that later what like
an example would be in how to fix it but
for us these two things are completely
different so we don't need to adjust
this at all and yeah now we want to take
its derivative and second derivative so
that we can plug it into this equation
so the first derivative it's going to be
Y P prime equals a second derivative is
going to be yd double prime which is
zero and then we can plug that in so
we're going to get zero right there
that's a zero
why prime y prime is just a minus six
times y but Y we know is ax plus B and
that still has to equal to X so then
what we want to do is we want to
distribute so we get Y a minus 6a X - 6
B equals 2x so that's the same thing as
negative 6a X plus a minus 6 B equals 2x
I do that math right minus 6 X minus 6 e
yeah that looks good so then we can say
this is the same thing as 2x plus 0 so
then we can say this piece has to match
up with this piece things attached to X
have to match up with things attached to
X and this piece has to match up with
this piece is that cool with everyone
whenever two polynomials are equal the
the coefficients has to be exactly the
same so then we know negative 6a that's
equal to we know a minus 6b is equal to
0 so then a has to be negative 1/3
and then we get that negative 1/3 minus
6b is equal to zero so then we get
negative six B equals one-third and B
equals negative one over 18 so seems
kind of like a weird number
you make sure that's good no that's good
so like if I were taking a test and I
saw kind of not nice numbers like this I
kind of be suspicious and probably
double check my work but yeah we did it
right this is good so then we know now
that we know what a and B are we know
what YP is so we know YP is ax plus B so
it's negative 1/3 X minus 118 okay
so then we can say step 3 our final
solution is going to be
our first solution that we found our
complementary solution plus our
particular solution so that's going to
be why complementary which is this
oops
which is that that's why complimentary
plus why particular which is this and so
we get that okay cool let me switch my
my headphones real quick I think my ear
pods are about I'll be right back
you
you guys hear me okay
yeah okay and then somebody asked isn't
there supposed to be a plus C somewhere
um actually the way this works out there
doesn't end up being a plus C we have
these c1s and c2s maybe what you're
thinking of is a plus C that comes from
like when you have to solve separable
differential equations separable
differential equations you get a plus C
you'll also get them when you have to
solve linear differential equations so
this is a linear ode ee will get a plus
C in here as well but for this one you
actually don't get a plus C the only the
constants you get are the c1 and this
seats you and then these don't have any
constants
does that make sense okay good question
let's keep going
homestretch number nine we have a nice
linear OTE so we kind of peaked in
difficulty with those systems of
differential equations and Laplace is
out saying now we have a little bit
nicer one so for this you want to make
sure that the you have a one right here
which is good we do so then we want to
find our integrating factor which is
called me and that's going to be the e
to the integral of this whatever's
attached to your Y so it's going to be e
to the 2x DX which is going to become e
to the x squared and so then what do we
do now now that we've found our
integrating factor
you
what we integrate both side rail yeah so
we want to multiply both sides by e to
the x squared and then integrate both
sides so we're going to get e to the x
squared Y prime plus 2xy e to the x
squared equals x cubed times e to the x
squared and so the way the integrating
factor is set up is so that this always
becomes e to the x squared times y
Prime so it's always set up so that it's
the derivative of that some
mathematicians like however long ago
figure out a way to solve these by
whenever you pick mu in the special way
you'll always get this left-hand side of
simplify nicely like that
so yeah that's the prime right there or
we could also call this DDX does that
make sense to everyone okay and so then
we're going to integrate both sides so
if we integrate both sides
this essentially can
and we get e x squared y equals this
integral so for this we can do lui x
squared minus 1/2 yeah it doesn't look
too nice in terms of au substitution
yeah we may have to do integration by
parts so integration by parts says let's
set u equal to X cubed D U is going to
be 3x squared um DV is whatever you
didn't use we didn't use this e x
squared DX and V should be that integral
but actually that integral is not very
nice either so that means we want to
adjust our U and V a little bit so maybe
we don't want e to the x squared right
here maybe instead you want something
nice like a 2x it'd be nice if we had a
2x right there because integrating that
that would just become e to the x
squared does everyone agree with that it
would be nice for out of 2x because then
this perfectly derives to that
okay so how can we force the 2x there
well we can write X cubed as 1/2 times x
squared times 2x e to the x squared DX
okay so there's that piece that we
wanted and notice that this is still X
cubed so I didn't cheat and so this is
going to be our u right there what we
haven't used yet U is going to be 1/2 x
squared and D U is going to be 2x so
unfortunately we didn't get like a super
nice integral to do but it's definitely
possible
does anyone have any questions about
kind of what I just walked through right
there I know it's not really a typical
integral so feel free to ask anything ok
so then what we can do is we can say
integration by parts is going to be UV
so that 1/2 x squared e to the x squared
minus BD u minus the integral of VDU so
that's 2x e to the x squared DX is that
ok with everyone be you okay so then
that becomes 1/2 x squared e to the x
squared and then this should be minus e
to the x squared plus C cool so then we
can kind of summarize what we just found
we know e x squared y equals this
hm
yeah when your d you found ocean x just
regular x oh yeah good catch yeah yeah
yep just X all right catch
so that should become UV so that's still
good and then it should be minus the
integral of VDU so instead of a 2x we
have a regular x so instead of a 1 we
have a 1/2 yeah and if that is too much
of a step you can do a u sub of u equals
x squared to double check if you don't
if you want to do it all the details but
that should be minus 1/2 e to the x
squared plus C yeah thanks so then we
can did I fix it over here minus 1/2 e
to the x squared plus C and so then we
can divide by e to the x squared so Y is
going to be the e to the x squared
cancels when we divide it there so we're
gonna have 1/2 x squared and then we're
gonna have minus 1/2 and then we're
going to plus C I'll write it like this
e e to the negative x squared because
the negative exponent is the same thing
as dividing and yeah that is the final
answer is that okay with everybody
okay cool so that completes the review
yeah any questions I
didn't talk about one of those scenarios
when why particular does look like that
I can talk about that if you guys want
but other than that yeah do you guys
have any questions
okay cool yeah so hopefully that helped
you guys out yes my best advice for
diffic you is definitely just do a lot
of practice problems with the laplace's
just look at what they look like and
what you want them to be so there's kind
of a recurring theme when we solve some
of these things like we want it to look
a certain way cuz I'd be nice but it
doesn't but you can just force it we saw
that with some inverse laplace ism and
with that integral we just did and yeah
good luck you guys hopefully that helped
out and you're more than welcome to come
into the math Center and ask other
questions about difficu or these
questions in more detail or anything
like that but yeah so we'll end the
review there thanks
