Applying Newton's Law of Cooling to warm oatmeal

WEBVTT
Kind: captions
Language: en

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- [Voiceover] Let's now actually apply

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Newton's Law of Cooling.

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Just to remind ourselves, if
capitol T is the temperature

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of something in celsius degrees,

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and lower case t is time in minutes,

00:00:10.007 --> 00:00:13.951
we can say that the rate of change,

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the rate of change of our temperature

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with respect to time,

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is going to be proportional

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and I'll write a negative K over here.

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We assumed our concept K is positive,

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then a negative K is going to
proportional to the difference

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between the temperature of our thing

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and the ambient temperature in the room.

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Once again, why do I
have a negative there?

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Well, because if the
temperature of our thing

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is larger than the
temperature of our room,

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we would expect that we would
be decreasing in temperature.

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We would have a negative rate of chance.

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Temperature should be
decreasing over time.

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If, on the other hand,
our temperature is lower

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than the ambient temperature of the room

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then this thing is going to be negative

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and we would want a
positive rate of change.

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Things would be warming up.

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That's why a negative of a negative

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would give you the positive.

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This right over here, this
differential equation,

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we already saw it in a previous video

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on Newton's Law of Cooling.

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We even saw a general solution to that.

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The general solution that I care about,

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because we are now going
to deal with the scenario

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where we are putting
something warm in a...

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Or we are going to put
a warm bowl of oatmeal

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in a room temperature room.

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Given that, we are
going to assume the case

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that we saw in the last video

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where our temperature is
greater than or equal to

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the ambient temperature.

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In that situation, our general
solution boiled down to...

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It boiled down to temperature
as a function of time

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is equal to some constant
times e to the negative KT,

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negative KT, plus our ambient temperature.

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Plus our ambient temperature.

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Once again, we figured
this out in our last video.

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Now, let's actually apply it.

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I said we were dealing with the scenario

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where our temperature is greater than

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or equal to the ambient temperature.

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Let's assume we are in a scenario...

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Let's assume a scenario where

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our ambient temperature
is 20 degrees celsius.

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We assume that doesn't change.

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The room is just large enough that even if

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something that is warmer is put into it

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the ambient temperature does not change.

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Let's say that the thing
that we have put in it,

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our warm bowl of oatmeal,

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let's say it starts off the
moment we put it in the room,

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that time equals zero,

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is 80 degrees celsius.

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Let's say we also know,
just from previous tests,

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that after two minutes, after two minutes,

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it gets to 60 degrees celsius.

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We also know that T of
two is 60 degrees celsius.

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Given all of this
information right over here,

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using Newton's Law of Cooling,

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and using all of this information we know

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about how bowls of oatmeal
that start at this temperature

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have cooled in the past,

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we want to know how long it will take.

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How many minutes will have to pass

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when you put an 80 degree
bowl of oatmeal in the room?

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How many minutes have to pass

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in order for it to get to 40 degrees

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using this model?

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Let me write that down.

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So how long...

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How many minutes for...

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or let me just say to cool
to 40 degrees celsius?

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I encourage you to pause the video now

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and try to figure it out.

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I'm assuming you have paused the video,

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and you have had your go at it

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and the key is to use
all of this information

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right over here to solve
for the constants C and K,

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and once you know that, you essentially

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have described your model.

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Then you can apply it
to solve for the time

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that gets you to a temperature
of 40 degrees celsius.

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So let's do that.

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The first thing we know is

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the ambient temperature
is 20 degrees celsius.

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This right over here

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is 20 degrees.

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The most obvious thing
to solve for or to apply

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is what happens with T of zero.

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What's neat about T of zero,

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when T equals zero, this exponent is zero,

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either the zero power is one,

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and so T of zero is
essentially going to simplify

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to Ce plus 20 degrees.

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Let me actually right that down.

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T of zero, which we
already know is 80 degrees,

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we already know as 80 degrees celsius.

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I'm just going to write 80.

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We will assume it's in degrees celsius.

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That is going to be equal to...

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That is going to be equal to

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when T equals zero, this,

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the e to the zero is just going to be one.

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It's going to be equal
to C plus, C plus 20.

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If you want to solve for
C, you just subtract 20

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from both sides of this equation.

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We are left with...

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We are left with 80 minus
20 is 60, is equal to C.

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60 is equal to C.

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We were able to figure out C.

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Let's figure out what we know right now.

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We know that T, let me
do that in magenta color.

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We know that T of t, that's confusing,

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upper case T of lower case t,

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temperature as a function of time,

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is going to be equal to...

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is going to be equal
to in that same color,

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60 e to the negative
KT, negative KT plus 20,

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plus our ambient temperature.

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Plus our ambient temperature.

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Now, we need to solve for K.

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We can use this information
right over here to solve for K.

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T of two is equal to 60 degrees.

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If we make t as equal to two,

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this thing is going to be 60 degrees.

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Let me write that down.

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Let me write that over
here so we have some space.

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We have 60 is equal to...

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60 is equal to 60, is equal to 60

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e to the negative K T,

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all this color switching takes time,

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e to the negative K T, plus

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oh, and be careful, that's a times two.

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E to the negative K times two.

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That's a time equals two,

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I could write that E
to the negative two K.

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E to the negative two K,

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and then of course we have our plus 20.

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Then we have our plus 20.

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Now we just have to solve for K.

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Once again, at any point, if
you feel inspired to do so

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I encourage you to try
to solve it on your own.

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Alright, so let's do this.

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If we subtract 20 from both sides,

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we get 40 is equal to 60
e to the negative two K.

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Divide both sides by 60.

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You are left with two thirds.

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40 divided by 60 is two thirds.

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is equal to e to the negative two K.

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E to the negative two K.

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All this color changing takes work.

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Let me know if y'all
want me to keep changing.

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I enjoy changing colors.

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It just keeps it
interesting on the screen.

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Anyway, e to the negative two K.

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Actually, let me scroll down a little bit

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so I have some more real
estate to work with.

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Now I can take, let's see.

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I can take the natural log of both sides.

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So, I'll have the natural log.

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Natural log of two thirds is equal to

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the natural log of e
to the negative two K.

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That's the whole reason why

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I took the natural log of both sides.

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Then to solve for K, I divide
both sides by negative two.

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I get K is equal to negative one half.

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Negative one half the
natural log of two thirds.

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I just swapped sides.

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Natural log of two thirds.

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Let me do that since I kept
the colors going so long,

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let me keep it that way.

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Natural log of two thirds.

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So we have solved for
all of the constants.

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Now we can rewrite this
thing right over here.

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We can rewrite it as...

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We just need a mini drumroll here,

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we are not completely done yet.

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We get t of T is equal to 60 e...

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e to the negative K.

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Well, negative K, the
negative and negative

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is going to be positive.

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It will be one half
natural log of two thirds.

00:08:41.148 --> 00:08:45.023
So one half natural log of two thirds.

00:08:45.023 --> 00:08:47.080
Negative K, so negative of a negative.

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One half natural log of two thirds,

00:08:49.998 --> 00:08:51.541
which actually will be a negative value.

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Two thirds is less than e,
so you are going to have

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a natural log of it is
going to be negative

00:08:55.009 --> 00:08:56.826
so it makes you feel
good that the temperature

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is going to be going down over time.

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So, plus or times T, plus 20.

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Now, all we have to do is
figure out what T get us

00:09:10.836 --> 00:09:13.381
to a temperature of 40 degrees celsius.

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Let's solve for that.

00:09:14.078 --> 00:09:17.815
If we want this to be
40, 40 is equal to...

00:09:18.517 --> 00:09:20.555
Actually now I'm just
going to stick to one color

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as we march through this part.

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40 is going to be equal to 60
e to the one half natural log

00:09:29.813 --> 00:09:34.355
of two thirds T power plus 20.

00:09:34.355 --> 00:09:35.451
Now, let's see.

00:09:35.451 --> 00:09:37.556
We can subtract 20 from both sides.

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We get to 20 is equal to 60
e to all that crazy business,

00:09:42.054 --> 00:09:46.473
one half natural log
of two thirds times T.

00:09:46.959 --> 00:09:50.647
Now we can divide both sides
by 60 and we get one third.

00:09:50.647 --> 00:09:52.359
20 divided by 60 is one third,

00:09:52.359 --> 00:09:57.359
is equal to e to the one half natural log

00:09:58.077 --> 00:10:02.140
of two thirds times T.

00:10:02.014 --> 00:10:05.218
Now, let's see, we can take
the natural log of both sides.

00:10:05.344 --> 00:10:10.344
The natural log of one third is equal to

00:10:11.951 --> 00:10:16.951
one half natural log of two thirds times T

00:10:17.863 --> 00:10:19.252
and then home stretch to solve for T

00:10:19.252 --> 00:10:20.786
you just divide both sides by one half

00:10:20.786 --> 00:10:22.489
natural log of two thirds.

00:10:22.489 --> 00:10:25.447
We get T is equal to this, which is

00:10:25.447 --> 00:10:30.447
the natural log of one third divided by

00:10:30.054 --> 00:10:31.706
one half natural log of two thirds.

00:10:32.192 --> 00:10:33.698
Well, if you divide by one
half that's the same thing

00:10:33.698 --> 00:10:35.344
as multiplying by two.

00:10:35.344 --> 00:10:40.344
Then you are going to divide
by natural log of two thirds.

00:10:42.091 --> 00:10:46.067
Let's see if this actually
makes a sensical answer.

00:10:46.067 --> 00:10:47.600
Let me get a calculator out.

00:10:47.006 --> 00:10:52.006
Actually, I could just use Google here.

00:10:52.885 --> 00:10:56.573
I had ... I already forgot what it was.

00:10:56.573 --> 00:10:58.260
Natural log one--

00:10:58.026 --> 00:10:59.142
So I had natural log one third

00:10:59.376 --> 00:11:00.390
over natural log of two thirds

00:11:00.039 --> 00:11:01.403
and the whole thing times two.

00:11:01.754 --> 00:11:06.754
So I can type two times the
natural log of one third

00:11:17.904 --> 00:11:22.904
divided by the natural log of two thirds.

00:11:24.473 --> 00:11:28.086
Let's see what Google gets us.

00:11:28.663 --> 00:11:30.801
Alright, it didn't ...

00:11:30.801 --> 00:11:32.606
How did I mess up?

00:11:32.606 --> 00:11:33.684
This is equal to two
times the natural log--

00:11:34.386 --> 00:11:36.789
Oh, okay, it messed up the parenthesis.

00:11:36.789 --> 00:11:37.798
Let me make this clear.

00:11:37.879 --> 00:11:41.697
The natural log of one third

00:11:41.697 --> 00:11:45.385
divided by the natural log of two thirds.

00:11:47.308 --> 00:11:48.740
Then, there you go.

00:11:48.074 --> 00:11:49.733
If we were to round to
the nearest hundredth

00:11:50.399 --> 00:11:53.263
it would be five point four two.

00:11:53.263 --> 00:11:56.438
Five point four two minutes.

00:11:56.438 --> 00:11:58.537
Remember, everything we
were doing were in minutes.

00:11:58.537 --> 00:12:02.760
This right over here, this
is approximately equal to

00:12:02.076 --> 00:12:06.107
five point four two.

00:12:07.007 --> 00:12:09.964
Five point four two minutes.

00:12:10.027 --> 00:12:11.220
And we are done!

00:12:11.022 --> 00:12:16.022
That's how long it will take
us to cool to 40 degrees.


Newton's Law of Cooling

WEBVTT
Kind: captions
Language: en

00:00:00.213 --> 00:00:01.782
- [Voiceover] Let's think
about another scenario

00:00:01.782 --> 00:00:03.849
that we can model with the
differential equations.

00:00:04.452 --> 00:00:07.271
This is a scenario where we take an object

00:00:07.271 --> 00:00:10.996
that is hotter or cooler than
the ambient room temperature,

00:00:10.996 --> 00:00:15.871
and we want to model how
fast it cools or heats up.

00:00:15.871 --> 00:00:18.507
And the way that we'll think about it is

00:00:18.507 --> 00:00:20.206
the way that Newton thought about it.

00:00:20.206 --> 00:00:25.206
And it is described as
Newton's Law of Cooling.

00:00:28.023 --> 00:00:30.778
And in a lot of ways, it's common sense.

00:00:30.778 --> 00:00:34.902
It states that the rate
of change of temperature

00:00:34.902 --> 00:00:38.126
should be proportional
to the difference between

00:00:38.126 --> 00:00:42.139
the temperature of the object
and the ambient temperature.

00:00:42.139 --> 00:00:44.272
So let me write that
in mathematical terms.

00:00:44.272 --> 00:00:46.064
So Newton's Law of Cooling tells us,

00:00:46.064 --> 00:00:48.119
that the rate of change of temperature,

00:00:48.614 --> 00:00:50.157
I'll use that with a capital T,

00:00:50.157 --> 00:00:53.851
with respect to time, lower case t,

00:00:53.851 --> 00:00:56.809
should be proportional to the difference

00:00:56.809 --> 00:00:59.795
between the temperature of the object

00:00:59.795 --> 00:01:02.053
and the ambient temperature.

00:01:02.053 --> 00:01:04.197
So that is a mathematical
description of it.

00:01:04.197 --> 00:01:05.318
And once again, it's common sense.

00:01:05.318 --> 00:01:08.209
If something is much, much hotter

00:01:08.209 --> 00:01:10.216
than the ambient temperature,
the rate of change

00:01:10.279 --> 00:01:11.753
should be pretty steep, it should be

00:01:11.753 --> 00:01:13.499
declining in temperature quickly.

00:01:13.499 --> 00:01:15.786
If something is much, much cooler,

00:01:15.786 --> 00:01:19.427
it should be increasing
in temperature quickly.

00:01:19.427 --> 00:01:22.743
And if something is
close, if these two things

00:01:22.743 --> 00:01:24.747
are pretty close, well
maybe this rate of change

00:01:24.783 --> 00:01:26.329
shouldn't be so big.

00:01:26.329 --> 00:01:27.683
Now I know one thing that you're thinking.

00:01:27.683 --> 00:01:32.683
You're like, okay, if
the temperature is hotter

00:01:32.805 --> 00:01:36.184
than the ambient temperature,
then I should be cooling.

00:01:36.184 --> 00:01:38.969
My temperature should be decreasing.

00:01:38.969 --> 00:01:41.117
And a decreasing temperature would imply

00:01:41.117 --> 00:01:43.764
a negative instantaneous change.

00:01:43.764 --> 00:01:46.920
So how will this be a negative value

00:01:46.092 --> 00:01:48.693
in the case where our
temperature of our object is

00:01:49.521 --> 00:01:51.716
greater than our ambient temperature?

00:01:51.716 --> 00:01:53.167
And the way that that would happen is,

00:01:53.167 --> 00:01:55.225
you would have to have a negative k.

00:01:55.225 --> 00:01:57.830
If you don't like thinking
in terms of a negative k,

00:01:57.083 --> 00:01:59.089
you can just put a
negative right over here

00:01:59.089 --> 00:02:01.861
and now you would have a positive k.

00:02:02.662 --> 00:02:04.258
Now it makes sense.

00:02:04.258 --> 00:02:07.075
If our thing is hotter, if
it has a higher temperature

00:02:07.075 --> 00:02:10.139
than the ambient temperature,
so this is a positive,

00:02:10.139 --> 00:02:11.792
then our rate of change will be negative,

00:02:11.792 --> 00:02:12.740
will be getting cooler.

00:02:12.074 --> 00:02:14.307
If it was the other way around, if our

00:02:14.973 --> 00:02:16.881
temperature of our
object is cooler than our

00:02:16.881 --> 00:02:18.557
ambient temperature, then this thing

00:02:18.557 --> 00:02:21.132
is going to be a negative,
and then the negative of that

00:02:21.132 --> 00:02:23.915
is going to be a positive,
we're assuming a positive k,

00:02:23.915 --> 00:02:26.847
and our temperature will be increasing.

00:02:26.847 --> 00:02:30.204
So hopefully, this makes
some intuitive sense.

00:02:30.204 --> 00:02:32.470
And our constant k could depend on

00:02:32.047 --> 00:02:34.103
the specific heat of the
object, how much surface area

00:02:35.003 --> 00:02:37.276
is exposed to it, or whatever else.

00:02:37.303 --> 00:02:40.080
But now I'm given this,
let's see if we can solve

00:02:40.008 --> 00:02:42.344
this differential equation
for a general solution.

00:02:42.416 --> 00:02:44.487
And I encourage you to pause
this video and do that,

00:02:44.487 --> 00:02:46.196
and I will give you a clue.

00:02:46.196 --> 00:02:49.879
This is a separable differential equation.

00:02:50.903 --> 00:02:53.295
So I assume you've had a go at it,

00:02:53.295 --> 00:02:55.327
so let's now work through it together.

00:02:55.327 --> 00:02:59.602
So, we just have to
algebraically manipulate this

00:02:59.602 --> 00:03:03.387
so all my Ts and dTs are on one side.

00:03:03.387 --> 00:03:07.566
I should say, so all my capital
Ts and dTs are on one side,

00:03:07.566 --> 00:03:09.222
this is going to be a
little bit more confusing

00:03:09.222 --> 00:03:10.293
because I have a capital
T and a lower case t.

00:03:10.932 --> 00:03:13.316
Capital T for temperature,
lower case t for time.

00:03:13.316 --> 00:03:14.743
But hopefully we'll be
able to work through it.

00:03:14.743 --> 00:03:17.251
And then I'm going to have
all my time differentials

00:03:17.251 --> 00:03:19.423
and time variables on the other side.

00:03:19.423 --> 00:03:21.357
So one thing I could is
I could divide both sides

00:03:21.357 --> 00:03:25.513
by T minus ambient
temperature, minus T sub a.

00:03:25.513 --> 00:03:26.538
Remember this is just
going to be a constant

00:03:26.763 --> 00:03:28.098
based on what our ambient temperature is.

00:03:28.098 --> 00:03:29.165
We're going to assume
our ambient temperature

00:03:29.165 --> 00:03:30.838
doesn't change as a function of time,

00:03:31.056 --> 00:03:33.623
it's just such a big
room that our cup of tea

00:03:34.127 --> 00:03:36.635
is not going to actually warm up the room.

00:03:36.635 --> 00:03:38.030
So that's just one of these assumptions

00:03:38.003 --> 00:03:39.327
that we're going to make.

00:03:39.354 --> 00:03:42.083
So if we do that, if we
divide both sides by this,

00:03:42.083 --> 00:03:43.744
we are going to have...

00:03:43.744 --> 00:03:45.159
So I'm going to divide both sides,

00:03:45.159 --> 00:03:46.491
I'm going to do this in a new color.

00:03:46.491 --> 00:03:51.491
If I divide both sides
by that, I get one over T

00:03:51.934 --> 00:03:55.739
minus T sub a, and let
me multiply both sides

00:03:55.739 --> 00:03:57.750
times the time differential.

00:03:57.849 --> 00:04:02.849
So I'm going to have, that dT,
our temperature differential.

00:04:05.886 --> 00:04:10.299
Times our temperature differential,
is going to be equal to

00:04:10.073 --> 00:04:15.073
negative k times our time differential.

00:04:18.058 --> 00:04:19.147
So once again, to separate the variables,

00:04:20.047 --> 00:04:22.011
all I did was divide both sides by this,

00:04:22.434 --> 00:04:25.330
and multiply both sides by that.

00:04:25.929 --> 00:04:27.980
Now I can integrate both sides,

00:04:27.098 --> 00:04:29.406
we've seen this show before.

00:04:30.288 --> 00:04:33.785
So I can integrate both sides.

00:04:33.785 --> 00:04:37.300
And the integral of this is going to be

00:04:37.003 --> 00:04:39.824
the natural log of the absolute value

00:04:40.121 --> 00:04:41.606
of what we have in the denominator.

00:04:41.606 --> 00:04:43.515
And you can do u substitution if you want.

00:04:43.515 --> 00:04:46.531
If we said u is equal to T minus T sub a,

00:04:46.531 --> 00:04:49.687
then du is just going to be one dT,

00:04:49.687 --> 00:04:52.434
and so this is essentially,
you could say the integral

00:04:52.434 --> 00:04:54.899
of one over u du, and so
it would be the natural log

00:04:54.899 --> 00:04:56.135
of the absolute value of u,

00:04:56.135 --> 00:04:58.378
and this right over here is u.

00:04:58.378 --> 00:05:02.279
So this is the natural
log of the absolute value

00:05:02.279 --> 00:05:07.155
of T minus T sub a, is equal to,

00:05:07.155 --> 00:05:08.814
and once again I could
put a constant here,

00:05:08.814 --> 00:05:09.678
but I'm going to end up with a constant

00:05:09.678 --> 00:05:12.175
on the right hand side too so
I'm just going to merge them

00:05:12.175 --> 00:05:14.618
into the constant on the right hand side.

00:05:14.618 --> 00:05:18.782
So that is going to be equal to, now here,

00:05:18.782 --> 00:05:23.136
this is going to be negative kt,

00:05:23.136 --> 00:05:26.797
and once again we have plus C.

00:05:26.797 --> 00:05:31.370
And now we can raise e
to both of these powers,

00:05:31.037 --> 00:05:32.413
or another way of interpreting this

00:05:32.746 --> 00:05:36.119
is if e to this thing is
going to be the same as that.

00:05:36.119 --> 00:05:40.841
So we can write this
as, the absolute value,

00:05:40.841 --> 00:05:43.035
let me do that in that same blue color.

00:05:43.496 --> 00:05:47.345
We can write this as the
absolute value of T minus T sub a

00:05:47.345 --> 00:05:51.256
is equal to e, something about e I always

00:05:51.256 --> 00:05:53.104
think of the color green.

00:05:53.104 --> 00:05:58.104
e to the negative kt plus C.

00:06:00.361 --> 00:06:04.260
This of course is the same thing as,

00:06:04.026 --> 00:06:07.311
this is equal to e to the negative kt,

00:06:07.545 --> 00:06:09.213
we've done this multiple times before.

00:06:09.213 --> 00:06:14.213
Negative kt times e to the C power.

00:06:21.648 --> 00:06:24.709
And we could just call this
another arbitrary constant.

00:06:25.258 --> 00:06:28.993
If we called this C1,
then we could just call

00:06:28.993 --> 00:06:30.405
this whole thing C.

00:06:30.405 --> 00:06:35.405
So this we could say is
Ce to the negative kt.

00:06:38.882 --> 00:06:40.698
So at least it's starting
to resemble what we did

00:06:40.698 --> 00:06:41.726
when we were modelling population.

00:06:41.978 --> 00:06:43.074
We'll see it's a little bit different.

00:06:43.074 --> 00:06:44.937
Instead of just temperature
on this left hand side,

00:06:44.937 --> 00:06:48.190
we have temperature minus
our ambient temperature.

00:06:48.019 --> 00:06:51.164
And so, we can do a couple of things.

00:06:51.335 --> 00:06:55.048
If, in a world, say we were
dealing with a hot cup of tea,

00:06:55.048 --> 00:06:57.974
something that's hotter than
the ambient temperature.

00:06:57.974 --> 00:07:02.768
So we could imagine a world where T is

00:07:02.768 --> 00:07:05.623
greater than or equal to
our ambient temperature.

00:07:05.623 --> 00:07:09.570
So that means this is hot, or it's hotter,

00:07:09.057 --> 00:07:10.080
I guess we could say.

00:07:10.008 --> 00:07:10.095
So if we're dealing with something

00:07:11.067 --> 00:07:12.306
hotter than the ambient temperature,

00:07:12.909 --> 00:07:15.072
then this absolute value
is going to be positive

00:07:15.072 --> 00:07:17.081
or the thing inside the absolute value

00:07:17.162 --> 00:07:17.956
is going to be positive.

00:07:17.956 --> 00:07:19.861
So we don't need the absolute value.

00:07:19.861 --> 00:07:20.943
Or the absolute value of it is going to be

00:07:20.943 --> 00:07:22.482
the same thing as it.

00:07:22.482 --> 00:07:26.163
And then we can just add
T sub a to both sides,

00:07:26.163 --> 00:07:28.740
and then we would have our temperature,

00:07:28.074 --> 00:07:30.840
and I can even write this
as a function of time,

00:07:31.506 --> 00:07:33.549
is going to be equal to this business,

00:07:33.936 --> 00:07:38.936
is going to be equal to Ce, let me do that

00:07:39.014 --> 00:07:40.562
in that same color.

00:07:40.562 --> 00:07:45.562
Ce to the negative kt plus T sub a.

00:07:50.733 --> 00:07:52.431
All I did is I'm assuming that this

00:07:52.431 --> 00:07:54.670
inside the absolute value
is going to be positive,

00:07:54.067 --> 00:07:56.124
so the absolute value is not
going to change the value.

00:07:57.024 --> 00:08:00.591
And I added T sub a to
both sides to get this.

00:08:00.807 --> 00:08:04.508
So this right over here is going
to be our general solution,

00:08:04.508 --> 00:08:09.508
in the case where we start
with something that is hotter

00:08:09.744 --> 00:08:11.450
than the ambient room temperature.

00:08:11.045 --> 00:08:13.259
And if we want to look at the
case where something is cooler

00:08:13.664 --> 00:08:15.122
than the ambient room temperature,

00:08:15.122 --> 00:08:17.399
so that's the situation,
let's say T is less than

00:08:17.399 --> 00:08:19.050
our ambient room temperature.

00:08:19.005 --> 00:08:21.602
Then the absolute value of
T, then this thing over here

00:08:21.647 --> 00:08:24.221
is going to be negative, and
so the absolute value of it's

00:08:24.221 --> 00:08:25.843
going to be the negative of that.

00:08:25.843 --> 00:08:30.843
So then this up here
results in T sub a minus T,

00:08:31.046 --> 00:08:33.198
that's going to be the same
thing as the absolute value,

00:08:33.198 --> 00:08:35.481
it's going to be the
negative of the negative.

00:08:35.481 --> 00:08:40.481
So then that is going to be
equal to e to the negative k

00:08:50.116 --> 00:08:52.671
plus, actually let me just do it...

00:08:54.013 --> 00:08:59.013
T sub a minus T is going to be equal to Ce

00:09:00.573 --> 00:09:04.991
to the negative kt, so
this is equal to that.

00:09:04.991 --> 00:09:07.623
I'm just assuming that
T is less than T sub a.

00:09:07.623 --> 00:09:12.623
And so then, to solve for T,
you could add T to both sides

00:09:12.857 --> 00:09:15.195
and subtract this from both sides.

00:09:15.195 --> 00:09:19.951
You would have T as a function
of t is going to be equal to,

00:09:19.951 --> 00:09:21.598
let's see, if this went onto that side

00:09:21.598 --> 00:09:25.894
and this goes over here,
you would have T sub a

00:09:26.479 --> 00:09:31.479
minus Ce to the negative kt.

00:09:31.907 --> 00:09:33.265
Did I do that right?

00:09:33.265 --> 00:09:34.886
So yep, that looks right.

00:09:34.886 --> 00:09:38.912
So this is the situation where
you have something that is

00:09:38.912 --> 00:09:43.355
cooler than the ambient temperature.

00:09:43.355 --> 00:09:45.831
So this right over
here, based on the logic

00:09:45.831 --> 00:09:50.831
of Newton's Law of Cooling,
these are the general solutions

00:09:51.061 --> 00:09:52.935
to that differential equation.

00:09:53.484 --> 00:09:54.492
In the next video we can actually apply it

00:09:55.284 --> 00:09:59.962
to model how quickly something
might cool or heat up.


Particular solution given initial conditions for population

WEBVTT
Kind: captions
Language: en

00:00:00.354 --> 00:00:02.180
- [Voiceover] In the last
video, we established that

00:00:02.017 --> 00:00:04.519
if we say the rate of
change of a population

00:00:04.681 --> 00:00:07.319
with respect to time is
going to be proportional

00:00:07.319 --> 00:00:09.152
to the population, we were able to

00:00:09.152 --> 00:00:12.373
solve that differential equation,
find a general solution,

00:00:12.373 --> 00:00:13.910
which involves an exponential.

00:00:13.091 --> 00:00:16.827
That the population is going
to be equal to some constant

00:00:17.646 --> 00:00:21.132
times e to some other
constant, times time.

00:00:21.132 --> 00:00:24.111
And in the last video we
assumed time was days.

00:00:24.111 --> 00:00:27.300
So let's just apply
this, just to feel good

00:00:27.003 --> 00:00:29.675
that we can truly model
population this way.

00:00:29.972 --> 00:00:31.500
So let's use it with
some concrete numbers.

00:00:31.005 --> 00:00:33.254
Once again, you've
probably done that before.

00:00:33.749 --> 00:00:36.097
You've probably started
with the assumption

00:00:36.097 --> 00:00:38.421
that you can model with
an exponential function,

00:00:38.421 --> 00:00:41.532
and then you used some
information, some conditions,

00:00:41.532 --> 00:00:43.206
to figure out what the constants are.

00:00:43.206 --> 00:00:44.562
You probably did this earlier

00:00:44.562 --> 00:00:46.813
in pre-calculus or algebra class.

00:00:46.813 --> 00:00:49.181
Let's just do it again,
just so that we can feel

00:00:49.181 --> 00:00:51.907
that this thing right over here is useful.

00:00:51.907 --> 00:00:53.755
So let's give you some information.

00:00:53.755 --> 00:00:57.893
Let's say, that at time equals zero,

00:00:57.893 --> 00:01:02.893
the population is equal to 100 insects

00:01:03.807 --> 00:01:06.433
or whatever we're measuring
the population of.

00:01:06.433 --> 00:01:11.433
And let's say that at time
equals 50, the population,

00:01:15.553 --> 00:01:20.553
so after 50 days, the
population is at 200.

00:01:20.826 --> 00:01:22.861
So notice it doubled after 50 days.

00:01:23.176 --> 00:01:27.278
So given this information,
can we solve for C and k?

00:01:27.278 --> 00:01:28.691
I encourage you to pause the video

00:01:28.691 --> 00:01:31.759
and try to work through it on your own.

00:01:31.759 --> 00:01:33.764
So this first initial
condition's pretty straighforward

00:01:33.809 --> 00:01:36.905
to you, because when t is
equal to zero, P is 100.

00:01:36.905 --> 00:01:41.905
So we could say, based on this
first piece of information,

00:01:42.531 --> 00:01:46.134
we could say that 100 must be equal to

00:01:46.671 --> 00:01:51.671
C times e to the k times zero.

00:01:56.531 --> 00:01:59.617
Well that's just going
to be e to the zero.

00:01:59.617 --> 00:02:02.073
Well e to the zero is just one.

00:02:02.073 --> 00:02:05.165
So this is just the same
thing as C times one,

00:02:05.165 --> 00:02:07.242
and just like that we have
figured out what C is.

00:02:09.717 --> 00:02:11.743
We can now write that the population

00:02:11.968 --> 00:02:16.968
is going to be equal to 100e to the kt.

00:02:19.569 --> 00:02:21.105
So you can see, expressed this way,

00:02:21.105 --> 00:02:24.306
our C is always going to
be our initial population.

00:02:24.306 --> 00:02:29.306
So, e to the kt.

00:02:30.962 --> 00:02:33.343
And now, we can use this
second piece of information.

00:02:33.343 --> 00:02:36.370
So our population is 200,
let's write that down.

00:02:36.037 --> 00:02:40.780
So our population is 200,
when time is equal to 50.

00:02:41.113 --> 00:02:42.602
After 50 days.

00:02:42.602 --> 00:02:47.602
So 200 is equal to 100e
to the k times 50, right?

00:02:54.696 --> 00:02:56.347
t is now 50.

00:02:56.347 --> 00:02:57.265
Let me just write that.

00:02:58.603 --> 00:03:00.952
k times 50.

00:03:00.952 --> 00:03:04.434
Now we can divide both sides by 100,

00:03:04.434 --> 00:03:09.434
and we will get, two is
equal to e to the 50k.

00:03:15.033 --> 00:03:17.546
Then we can take the
natural log of both sides.

00:03:17.843 --> 00:03:19.742
Natural log on the left hand side,

00:03:19.742 --> 00:03:21.789
we get the natural log of two.

00:03:21.789 --> 00:03:25.185
And on the right hand side, the
natural log of e to the 50k,

00:03:25.185 --> 00:03:28.077
well that's just going
to be, that's the power

00:03:28.077 --> 00:03:31.211
that you need to raise e
to, to get e to the 50k.

00:03:31.211 --> 00:03:35.270
Well that's just going to be 50k.

00:03:35.801 --> 00:03:38.713
All you did was took the
natural log of both sides.

00:03:38.713 --> 00:03:40.687
Notice, that this equation
that I've just written

00:03:40.687 --> 00:03:41.620
expresses the same thing.

00:03:41.062 --> 00:03:42.329
Natural log of two is equal to 50k,

00:03:42.887 --> 00:03:45.540
that means e to the 50k is equal to two,

00:03:45.054 --> 00:03:47.262
which is exactly what
we had written there.

00:03:47.748 --> 00:03:49.491
And now we can solve for k.

00:03:49.491 --> 00:03:53.422
Divide both sides by 50,
and we are left with,

00:03:53.761 --> 00:03:58.761
k is equal to the natural
log of 2, over 50.

00:04:03.443 --> 00:04:04.151
And we're done.

00:04:04.151 --> 00:04:05.914
We can now write the particular solution

00:04:05.914 --> 00:04:09.066
that meets these conditions.

00:04:09.066 --> 00:04:11.562
So we can now write that our population,

00:04:11.562 --> 00:04:15.007
and I can even write our
population as a function of time,

00:04:15.007 --> 00:04:20.007
is going to be equal
to 100 times e to the,

00:04:25.642 --> 00:04:29.041
now k is natural log of two
over 50, so I'll write that.

00:04:29.041 --> 00:04:34.041
So, natural log of two over
50, and then that times t.

00:04:37.181 --> 00:04:40.237
If you assume rate of change of population

00:04:40.741 --> 00:04:42.874
is going to be proportional to population,

00:04:42.874 --> 00:04:45.140
you assume this information
right over here,

00:04:45.014 --> 00:04:47.099
this function is what is going to describe

00:04:47.099 --> 00:04:49.530
your growth of population.


Separable differential equations introduction

WEBVTT
Kind: captions
Language: en

00:00:00.002 --> 00:00:01.234
- [voiceover] So now that
we've spent some time

00:00:01.433 --> 00:00:03.436
thinking about what a
differential equation is

00:00:03.733 --> 00:00:05.666
and even visualizing solutions

00:00:05.666 --> 00:00:08.668
to a differential equations
using things like slope field,

00:00:08.686 --> 00:00:10.572
let's start seeing if we can actually

00:00:10.572 --> 00:00:12.005
solve differential equations.

00:00:12.005 --> 00:00:15.007
As we'll see, different types
of differential equations

00:00:15.007 --> 00:00:16.015
might require different techniques

00:00:17.005 --> 00:00:18.166
and some of them we might not be able

00:00:18.661 --> 00:00:20.633
to solve at all using analytic techniques.

00:00:20.633 --> 00:00:24.173
We'd have to resort to numeric techniques
to estimate the solutions.

00:00:24.767 --> 00:00:26.006
But let's go to what I would argue

00:00:26.006 --> 00:00:29.373
as the simplest form of
differential equation to solve

00:00:29.967 --> 00:00:32.466
and that's what's called a Separable.

00:00:32.466 --> 00:00:35.433
Separable differential equation.

00:00:35.433 --> 00:00:38.003
And we will see in a
second why it is called

00:00:38.003 --> 00:00:40.005
a separable differential equation.

00:00:40.005 --> 00:00:43.838
So let's say that we
have the derivative of Y

00:00:44.333 --> 00:00:53.836
with respect to X is equal to negative X
over Y E to the X squared.

00:00:54.133 --> 00:00:55.140
So we have this differential equation

00:00:55.833 --> 00:00:58.271
and we want to find
the particular solution

00:00:58.271 --> 00:01:03.271
that goes through the point 0,1.

00:01:04.004 --> 00:01:06.010
I encourage you to pause this
and I'll give you a hint.

00:01:07.000 --> 00:01:10.866
If you can on one side of
this equation through algebra

00:01:10.866 --> 00:01:13.005
separate out the Ys and the DYs

00:01:13.005 --> 00:01:17.222
and on the other side
have all the Xs and DXs,

00:01:17.267 --> 00:01:18.633
and then integrate.

00:01:18.633 --> 00:01:21.366
Perhaps you can find
the particular solution

00:01:21.366 --> 00:01:23.133
to this differential equation

00:01:23.133 --> 00:01:25.133
that contains this point.

00:01:25.133 --> 00:01:26.138
Now if you can't do it don't worry

00:01:26.633 --> 00:01:28.790
because we're about to work through it.

00:01:29.267 --> 00:01:31.006
So like I said, let's use
a little bit of algebra

00:01:31.006 --> 00:01:33.011
to get all the Ys and DYs on one side

00:01:34.001 --> 00:01:36.634
and all the Xs and DXs on the other side.

00:01:36.733 --> 00:01:38.739
So one way, let's say I
want to get all the Ys

00:01:39.333 --> 00:01:41.003
and DYs on the left hand side,

00:01:41.003 --> 00:01:43.969
and all the Xs and DXs
on the right hand side.

00:01:44.266 --> 00:01:46.338
Well, I can multiply both sides times Y.

00:01:46.986 --> 00:01:49.009
So I can multiply both sides times Y

00:01:49.009 --> 00:01:52.642
that has the effect of putting
the Ys on the left hand side

00:01:53.533 --> 00:01:56.008
and then I can multiple
both sides times DX.

00:01:56.008 --> 00:01:58.256
I can multiple both sides times DX

00:01:59.048 --> 00:02:00.366
and we kind of treat,

00:02:00.366 --> 00:02:01.372
you can treat these differentials

00:02:01.966 --> 00:02:03.923
as you would treat a variable

00:02:03.923 --> 00:02:07.973
when you're manipulating it
to essentially separate out the variables.

00:02:08.567 --> 00:02:12.333
And so, this will cancel with that.

00:02:12.333 --> 00:02:17.333
And so, we are left with Y, DY.

00:02:18.006 --> 00:02:23.006
Y, DY is equal to negative X.

00:02:23.666 --> 00:02:29.772
And actually let me write it this way.
Let me write it as negative X, E.

00:02:30.366 --> 00:02:31.096
Actually, I might need
a little more space.

00:02:31.096 --> 00:02:36.096
So negative X E to the
negative X squared DX.

00:02:40.994 --> 00:02:44.771
DX.
Now why is this interesting?

00:02:45.266 --> 00:02:47.733
Because we could integrate both sides.

00:02:47.733 --> 00:02:48.735
And now this also highlights

00:02:48.933 --> 00:02:50.505
why we call it the separable.
You won't be able to do this

00:02:51.003 --> 00:02:52.009
with every differential equation.

00:02:52.009 --> 00:02:53.588
You won't be able to algebraically

00:02:54.479 --> 00:02:57.533
separate the Ys and DYs on one side

00:02:57.533 --> 00:03:02.075
and the Xs and DXs on the other side.
But this one we were able to.

00:03:02.867 --> 00:03:04.333
And so that's why this is called

00:03:04.333 --> 00:03:07.584
a separable differential equation.

00:03:07.906 --> 00:03:12.906
Differential equation.

00:03:13.134 --> 00:03:15.376
And it's usually the first
technique that you should try.

00:03:15.376 --> 00:03:17.992
Hey, can I separate the Ys and the Xs

00:03:17.992 --> 00:03:19.779
and as I said, this is
not going to be true

00:03:19.779 --> 00:03:22.667
of many, if not most
differential equations.

00:03:22.667 --> 00:03:23.766
But now that we did this

00:03:23.766 --> 00:03:26.094
we can integrate both sides.

00:03:26.094 --> 00:03:27.178
So let's do that.

00:03:27.934 --> 00:03:30.733
So, I'll find a nice
color to integrate with.

00:03:30.733 --> 00:03:35.701
So, I'm going to integrate both sides.

00:03:35.701 --> 00:03:38.008
Now if you integrate the left hand side
what do you get?

00:03:38.008 --> 00:03:39.774
You get and remember, we're integrating

00:03:40.566 --> 00:03:41.899
with respect to Y here.

00:03:41.899 --> 00:03:47.807
So this is going to be Y squared over two
and we could put some constant there.

00:03:48.401 --> 00:03:50.867
I could call that plus C one.

00:03:50.867 --> 00:03:51.892
And if you're integrating that

00:03:51.892 --> 00:03:53.767
that's going to be equal to.

00:03:53.767 --> 00:03:54.841
Now the right hand side we're integrating

00:03:55.507 --> 00:03:59.003
with respect to X.
And let's see, you could do U substitution

00:03:59.003 --> 00:04:00.536
or you could recognize that look,

00:04:00.833 --> 00:04:05.274
the derivative of negative X squared
is going to be negative two X.

00:04:05.967 --> 00:04:07.333
So if that was a two there

00:04:07.333 --> 00:04:08.899
and if you don't want to change
the value of the integral

00:04:08.899 --> 00:04:11.594
you put the 1/2 right over there.
And so now you could either do

00:04:12.386 --> 00:04:13.921
U substitution explicitly

00:04:13.921 --> 00:04:15.005
or you could do it in your head

00:04:15.005 --> 00:04:19.110
where you said U is equal
to negative X squared

00:04:19.605 --> 00:04:22.869
and then DU will be negative to X, DX

00:04:22.869 --> 00:04:25.001
or you can kind of do this
in your head at this point.

00:04:25.001 --> 00:04:28.002
So I have something and it's derivative

00:04:28.002 --> 00:04:30.371
so I really could just
integrate with respect to

00:04:30.569 --> 00:04:33.601
that something too with respect to that U.

00:04:33.601 --> 00:04:37.672
So this is going to be 1/2.

00:04:37.672 --> 00:04:39.433
This 1/2 right over here.

00:04:39.433 --> 00:04:40.433
The anti-derivative.

00:04:40.433 --> 00:04:43.966
This is E to the negative X squared

00:04:43.966 --> 00:04:46.966
and then of course, I might
have some other constant.

00:04:46.966 --> 00:04:49.969
I'll just call that C two.

00:04:50.266 --> 00:04:51.271
And once again, if this part over here

00:04:51.766 --> 00:04:52.994
what I just did seemed strange,

00:04:52.994 --> 00:04:53.494
the U substitution,

00:04:53.904 --> 00:04:56.366
you might want to review that piece.

00:04:56.366 --> 00:04:58.267
Now, what can I do here?

00:04:58.267 --> 00:04:59.270
We'll have a constant
on the left hand side.

00:04:59.567 --> 00:05:00.533
It's an arbitrary constant.

00:05:00.533 --> 00:05:01.294
We don't know what it is.

00:05:01.294 --> 00:05:04.833
I haven't used this initial condition yet

00:05:04.833 --> 00:05:05.866
we could call it.

00:05:05.866 --> 00:05:08.871
So, let me just subtract
C one from both sides.

00:05:09.366 --> 00:05:12.368
So if I just subtract
C one from both sides

00:05:12.566 --> 00:05:15.000
I have an arbitrary so
this is gonna cancel,

00:05:15.000 --> 00:05:16.833
and I have C two, sorry.

00:05:16.833 --> 00:05:18.742
Let me.
So, this is C one.

00:05:19.633 --> 00:05:21.001
So these are going to cancel

00:05:21.001 --> 00:05:22.003
and C two minus C one.

00:05:22.003 --> 00:05:24.337
These are both constants,
arbitrary constants

00:05:24.634 --> 00:05:29.336
and we don't know what they are yet.
And so, we could just rewrite this as

00:05:29.633 --> 00:05:33.635
on the left hand side we
have Y squared over two

00:05:33.833 --> 00:05:35.921
is equal to on the right hand side.

00:05:35.921 --> 00:05:38.676
I'll write 1/2 E.

00:05:38.676 --> 00:05:39.176
Let me write that in blue

00:05:39.486 --> 00:05:41.509
just because I wrote it in blue before.

00:05:41.509 --> 00:05:48.628
1/2 E to the negative X squared
and I'll just say C two minus C one.

00:05:49.123 --> 00:05:50.715
Let's just call that C.

00:05:51.125 --> 00:05:56.006
So if you take the sum of those two things
let's just call that C.

00:05:56.006 --> 00:05:59.009
And so now, this is kind
of a general solution.

00:05:59.009 --> 00:06:01.085
We don't know what this constant is

00:06:01.976 --> 00:06:04.333
and we haven't explicitly solved for Y yet

00:06:04.333 --> 00:06:05.666
but even in this form

00:06:05.666 --> 00:06:07.533
we can now find a particular solution

00:06:07.533 --> 00:06:09.019
using this initial condition.

00:06:09.019 --> 00:06:09.519
Let me separate it out.

00:06:10.011 --> 00:06:12.008
This was a part of this

00:06:12.008 --> 00:06:13.017
original expression right over here

00:06:14.007 --> 00:06:15.568
but using this initial condition.

00:06:16.261 --> 00:06:20.440
So, it tells us when X is zero,
Y needs to be equal to one.

00:06:21.133 --> 00:06:24.566
So we would have one
squared which is just one

00:06:24.566 --> 00:06:31.975
over two is equal to 1/2.
E to the negative zero squared.

00:06:32.866 --> 00:06:34.332
Well, that's just going to be

00:06:34.332 --> 00:06:37.776
e to the zero is just one.
This is gonna be 1/2 plus C

00:06:38.667 --> 00:06:40.201
and just like that
we're able to figure out

00:06:40.201 --> 00:06:42.107
if you subtract 1/2 from both sides

00:06:42.107 --> 00:06:47.017
C is equal to zero.
So the relationship between Y and X

00:06:48.007 --> 00:06:49.016
that goes through this point,

00:06:50.006 --> 00:06:52.011
we could just set C is equal to zero.

00:06:53.053 --> 00:06:55.427
So that's equal to zero.

00:06:55.904 --> 00:07:02.006
That's zero right over there.
And so we are left with Y squared over two

00:07:02.633 --> 00:07:07.633
is equal to E to the
negative X squared over two.

00:07:08.006 --> 00:07:10.139
Now we can multiply both sides by two

00:07:10.733 --> 00:07:13.274
and we're going to get Y squared.

00:07:13.936 --> 00:07:14.934
Y squared.

00:07:14.934 --> 00:07:16.001
Let me do that.

00:07:16.001 --> 00:07:17.535
So we're gonna get Y squared

00:07:17.634 --> 00:07:22.634
is equal to E to the negative X squared.

00:07:22.967 --> 00:07:26.566
Now, we can take the
square root of both sides

00:07:26.566 --> 00:07:27.333
and you can say, well look,

00:07:27.333 --> 00:07:28.005
you know, Y squared is equal to this

00:07:28.005 --> 00:07:33.005
so Y could be equal to the
plus or minus square root

00:07:34.776 --> 00:07:37.792
of E to the negative X squared.

00:07:37.936 --> 00:07:40.651
Of E to the negative X squared.

00:07:40.651 --> 00:07:45.144
But they gave us an initial condition
where Y is actually positive.

00:07:45.936 --> 00:07:46.990
So we're finding the particular solution

00:07:47.476 --> 00:07:48.633
that goes through this point.

00:07:48.633 --> 00:07:51.366
That means Y is gonna be
the positive square root.

00:07:51.366 --> 00:07:53.370
If this was a point zero negative one

00:07:53.766 --> 00:07:55.008
then we would say Y is
the negative square root

00:07:55.008 --> 00:07:56.874
but we know that Y is
the positive square root,

00:07:57.666 --> 00:07:59.799
it's the principal root right over there.

00:07:59.799 --> 00:08:03.240
So let me do that a little bit neater
so we can get rid of, whoops.

00:08:03.933 --> 00:08:04.937
I thought I was writing in black.

00:08:05.333 --> 00:08:08.634
So we can get rid of this right over here.

00:08:08.634 --> 00:08:10.943
We're only going to be dealing with
the positive square root

00:08:11.834 --> 00:08:15.233
so we could write Y is equal to E

00:08:15.233 --> 00:08:19.952
to the negative X squared to the 1/2 power

00:08:19.952 --> 00:08:23.219
and that of course is equal to

00:08:23.219 --> 00:08:28.219
E to the negative X squared over two.

00:08:29.666 --> 00:08:31.071
So this right over here is

00:08:31.071 --> 00:08:35.130
or Y equals E to the
negative X squared over two

00:08:36.003 --> 00:08:37.020
is a particular solution

00:08:37.047 --> 00:08:39.342
that satisfies the initial conditions

00:08:39.765 --> 00:08:41.772
to this original differential equation.

00:08:42.366 --> 00:08:43.232
So just like that.

00:08:43.232 --> 00:08:45.166
Because we were able to just as a review,

00:08:45.166 --> 00:08:48.567
because this differential
equation was setup in a way

00:08:48.567 --> 00:08:49.574
or because we could algebraically

00:08:50.267 --> 00:08:53.267
separate the Y, DYs from the Xs, DXs,

00:08:53.267 --> 00:08:55.933
we're able to just separate
them out algebraically,

00:08:55.933 --> 00:08:57.266
integrate both sides

00:08:57.266 --> 00:08:59.271
and use the information given
in the initial condition

00:08:59.766 --> 00:09:02.933
to find the particular solution.


Particular solution to differential equation example

WEBVTT
Kind: captions
Language: en

00:00:00.251 --> 00:00:03.313
- Let's now get some practice
with separable differential

00:00:03.313 --> 00:00:06.346
equations, so let's say I have
the differential equation,

00:00:06.643 --> 00:00:10.680
the derivative of Y with
respect to X is equal to

00:00:10.068 --> 00:00:14.692
two Y-squared, and let's say
that the graph of a particular

00:00:15.304 --> 00:00:18.683
solution to this, the graph
of a particular solution,

00:00:18.683 --> 00:00:23.435
passes through the point one
comma negative one, so my

00:00:23.435 --> 00:00:28.435
question to you is, what is Y,
what is Y when X is equal to

00:00:34.993 --> 00:00:38.132
three for this particular
solution, so the particular

00:00:38.132 --> 00:00:40.164
solution to the differential
equation that passes through

00:00:40.452 --> 00:00:44.459
the point one comma negative
one, what is Y when X is

00:00:44.522 --> 00:00:47.258
equal to three, and I encourage
you to pause the video,

00:00:47.258 --> 00:00:49.484
and try to work through it on your own.

00:00:50.238 --> 00:00:52.722
So I'm assuming you had a
go at it, and the key with a

00:00:52.722 --> 00:00:55.437
separable differential
equation, and that's a big clue

00:00:55.437 --> 00:00:57.323
that I'm even calling it a
separable differential equation,

00:00:57.323 --> 00:00:59.997
is that you separate the Xs from the Ys.

00:00:59.997 --> 00:01:03.236
Or all the Xs and the
DXs from the Ys and DYs.

00:01:03.236 --> 00:01:04.712
So how do you do that here?

00:01:04.989 --> 00:01:06.490
Well, what I could do,
let me just rewrite it.

00:01:06.049 --> 00:01:11.049
So it's gonna be DY DX is
equal to two Y-squared,

00:01:12.068 --> 00:01:16.332
is equal to two Y, equal to two Y-squared.

00:01:16.332 --> 00:01:20.670
So let's see, we can
multiply both sides by DX,

00:01:21.917 --> 00:01:24.825
and let's see, so then we're
gonna have, that cancels with

00:01:24.825 --> 00:01:27.441
that if we treat it as just
a value, or as a variable.

00:01:27.441 --> 00:01:32.441
We're gonna have DY is
equal to two Y squared DX.

00:01:32.621 --> 00:01:33.695
Well, we're not quite done yet.

00:01:33.695 --> 00:01:36.126
We gotta get this two Y
squared on the left hand side.

00:01:36.126 --> 00:01:39.231
So we can divide both
sides by two Y-squared.

00:01:39.616 --> 00:01:44.616
So if we divide both sides by
two Y-squared, two Y-squared,

00:01:45.416 --> 00:01:50.416
the left hand side, we could
rewrite this as 1/2 Y to the

00:01:50.762 --> 00:01:55.762
negative two power, is going
to be equal to DY, DY is

00:01:59.992 --> 00:02:04.992
equal to DX, and now, we
can integrate both sides.

00:02:05.656 --> 00:02:08.651
So we can integrate both sides.

00:02:08.096 --> 00:02:10.935
Let me give myself a
little bit more space.

00:02:12.043 --> 00:02:17.043
And so, what is, what is this
left hand side going to be?

00:02:17.935 --> 00:02:20.183
Well, we increment the exponent,

00:02:20.183 --> 00:02:23.302
and then divide by that value,
so Y to the negative two,

00:02:23.302 --> 00:02:26.113
if your increment is
Y to the negative one,

00:02:26.113 --> 00:02:28.159
and then divide by negative one,

00:02:28.573 --> 00:02:33.573
so this is going to be -1/2
Y to the negative one power,

00:02:35.051 --> 00:02:37.728
and we could do a plus C like
we did in the previous video,

00:02:38.187 --> 00:02:39.808
but we're gonna have a
plus C on both sides,

00:02:39.808 --> 00:02:41.615
and you could subtract,
or you know, you have

00:02:41.615 --> 00:02:43.418
different arbitrary
constants on both sides and

00:02:43.418 --> 00:02:44.784
you could subtract them from each other,

00:02:44.784 --> 00:02:46.805
so I'm just gonna write the
constant only on one side.

00:02:47.041 --> 00:02:52.041
So you have that is equal to,
well if I integrate just DX,

00:02:53.067 --> 00:02:57.100
that's just going to give me
X, that's just gonna give me X.

00:02:57.397 --> 00:03:00.428
So this right over here is X,

00:03:00.428 --> 00:03:04.534
and of course I can have
a plus C over there,

00:03:04.811 --> 00:03:09.811
and If I want I can, I can
solve for Y if I multiply,

00:03:10.278 --> 00:03:13.695
let's see, I can multiply
both sides by negative two,

00:03:13.895 --> 00:03:16.240
and then I'm gonna have, the
left hand side you're just

00:03:16.024 --> 00:03:20.226
gonna have Y to the negative
one, or 1/Y is equal to,

00:03:20.442 --> 00:03:23.548
if I multiply the right hand
side times negative two,

00:03:23.548 --> 00:03:28.548
I'm gonna have negative
two times X plus, well

00:03:29.488 --> 00:03:31.537
it's some arbitrary constant,
it's still going to,

00:03:31.537 --> 00:03:33.814
it's gonna be negative two
times this arbitrary constant

00:03:33.814 --> 00:03:38.272
but I could still just call it
some arbitrary constant, and

00:03:38.272 --> 00:03:41.037
then if we want we can take
the reciprocal of both sides,

00:03:41.161 --> 00:03:46.161
and so we will get Y is equal
to, is equal to 1/-2X+C.

00:03:50.729 --> 00:03:55.215
And now we can use, we can use
the information they gave us

00:03:55.585 --> 00:03:58.449
right over here, the fact
that our particular solution

00:03:58.449 --> 00:04:00.928
needs to go through this
point to solve for C.

00:04:01.359 --> 00:04:06.359
So, when X is negative one,
so when X is negative one.

00:04:07.016 --> 00:04:12.016
Oh sorry, when X is one, when
X is one, Y is negative one,

00:04:13.099 --> 00:04:18.099
so we get negative one
is equal to 1/-2+C, or

00:04:20.866 --> 00:04:25.483
we could say C minus two, we
could multiply both sides times

00:04:25.483 --> 00:04:29.516
C minus two, if then we will
get, actually let me just

00:04:29.516 --> 00:04:32.507
scroll down a little bit, so
if you multiply both sides

00:04:32.507 --> 00:04:35.102
times C minus two, negative
one times C minus two is going

00:04:35.102 --> 00:04:40.102
to be negative C plus two or
two minus C is equal to one.

00:04:41.525 --> 00:04:45.476
All I did is I multiplied C
minus two times both sides,

00:04:46.938 --> 00:04:49.717
and then, let's see, I can
subtract two from both sides, so

00:04:49.717 --> 00:04:52.884
negative C is equal to negative
one, and then if I multiply

00:04:52.884 --> 00:04:57.884
both sides by negative one,
we get C is equal to one.

00:04:58.724 --> 00:05:03.724
So our particular solution
is Y is equal to 1/-2X+1.

00:05:09.736 --> 00:05:12.541
And we are almost done,
they didn't just ask for,

00:05:12.541 --> 00:05:14.409
we didn't just ask for
the particular solution,

00:05:14.409 --> 00:05:16.850
we asked, what is Y when
X is equal to three.

00:05:17.022 --> 00:05:20.901
So Y is going to be equal
to one over, three times

00:05:21.099 --> 00:05:26.099
negative two is negative six
plus one, which is equal to

00:05:26.305 --> 00:05:31.305
negative, is going to be
equal to 1/-5, or -1/5.

00:05:33.334 --> 00:05:36.686
And we are done.


Differential equation from slope field

WEBVTT
Kind: captions
Language: en

00:00:00.778 --> 00:00:01.786
- [Voiceover] We've already seen scenarios

00:00:02.578 --> 00:00:04.273
where we start with a
differential equation

00:00:04.273 --> 00:00:06.258
and then we generate a slope field

00:00:06.258 --> 00:00:08.475
that describes the solutions
to the differential equation

00:00:08.475 --> 00:00:11.029
and then we use that to
visualize those solutions.

00:00:11.029 --> 00:00:13.072
What I want to do in this
video is do an exercise

00:00:13.072 --> 00:00:17.553
that takes us the other way,
start with a slope field

00:00:17.553 --> 00:00:19.829
and figure out which differential equation

00:00:19.829 --> 00:00:22.694
is the slope field
describing the solutions for.

00:00:23.137 --> 00:00:26.295
And so I encourage you to
look at each of these options

00:00:26.295 --> 00:00:29.418
and think about which of
these differential equations

00:00:29.418 --> 00:00:32.296
is being described by this slope field.

00:00:32.296 --> 00:00:33.304
I encourage you to pause
the video right now

00:00:34.096 --> 00:00:35.170
and try it on your own.

00:00:36.975 --> 00:00:40.132
So I'm assuming you have had a go at it.

00:00:40.132 --> 00:00:41.206
So let's work through each of them.

00:00:41.874 --> 00:00:44.289
And the way I'm going to do
it is I'm just going to find

00:00:44.289 --> 00:00:47.806
some points that seem to be
easy to do arithmetic with,

00:00:47.806 --> 00:00:51.533
and we'll see if the slope
described by the differential

00:00:51.533 --> 00:00:54.505
equation at that point is
consistent with the slope

00:00:54.505 --> 00:00:56.548
depicted in the slope field.

00:00:56.548 --> 00:00:57.551
And, I don't know, just for simplicity,

00:00:57.848 --> 00:00:58.851
maybe I'll do x equals one and

00:00:59.148 --> 00:01:01.029
y equals one for all of these.

00:01:01.029 --> 00:01:05.928
So, when x equals one
and y is equal to one.

00:01:05.928 --> 00:01:08.134
So, this first differential
equation right over here,

00:01:08.134 --> 00:01:11.188
if x is one and y is
one, then dy/dx would be

00:01:11.674 --> 00:01:14.426
negative one over one or negative one.

00:01:14.426 --> 00:01:17.495
Dy/dx would be negative one.

00:01:18.116 --> 00:01:19.623
Now, is that depicted here?

00:01:19.623 --> 00:01:22.308
When x is equal to one
and y is equal to one,

00:01:22.308 --> 00:01:24.419
our slope isn't negative one.

00:01:24.419 --> 00:01:26.798
Our slope here looks positive.

00:01:26.798 --> 00:01:29.502
So we can rule this one out.

00:01:29.782 --> 00:01:31.419
Now, let's try the next one.

00:01:31.419 --> 00:01:35.018
So, if x is equal to one
and y is equal to one,

00:01:35.018 --> 00:01:40.018
well then dy/dx would be equal
to one minus one or zero.

00:01:40.985 --> 00:01:44.084
And, once again, I just
picked x equals one

00:01:44.084 --> 00:01:45.826
and y equals one for convenience.

00:01:45.826 --> 00:01:46.754
I could have picked any other.

00:01:46.754 --> 00:01:49.065
I could have picked negative
five and negative seven.

00:01:49.065 --> 00:01:51.082
This just makes the
arithmetic a little easier.

00:01:51.235 --> 00:01:53.000
Once again, when you look at that point

00:01:53.000 --> 00:01:54.161
that we've already looked at,

00:01:54.161 --> 00:01:55.682
our slope is clearly not zero.

00:01:55.682 --> 00:01:59.035
We have a positive slope
here, so we can rule that out.

00:01:59.234 --> 00:02:02.879
Once again, for this magenta
differential equation,

00:02:02.879 --> 00:02:05.514
if x and y are both equal to one,

00:02:05.514 --> 00:02:10.514
then one minus one is once
again going to be equal to zero.

00:02:11.423 --> 00:02:14.941
And we've already seen this
slope is not zero here,

00:02:14.941 --> 00:02:17.598
so rule that one out.

00:02:17.855 --> 00:02:19.794
And now here we have x plus y,

00:02:19.794 --> 00:02:22.278
so when x is one and y is one,

00:02:22.278 --> 00:02:25.296
our derivate of y with
respect to x is going to be

00:02:25.296 --> 00:02:29.371
one plus one, which is equal to two.

00:02:29.371 --> 00:02:31.054
Now, this looks interesting.

00:02:31.054 --> 00:02:33.933
It looks like this slope
right over here could be two.

00:02:33.933 --> 00:02:34.699
This looks like one.

00:02:34.699 --> 00:02:36.220
This looks like two.

00:02:36.022 --> 00:02:37.439
I would want to validate
some other points,

00:02:37.637 --> 00:02:40.980
but this looks like a really,
really good candidate.

00:02:40.098 --> 00:02:42.873
And you can also see
what is happening here.

00:02:43.755 --> 00:02:46.018
When dy/dx is equal to x plus y,

00:02:46.018 --> 00:02:48.564
you would expect that as
x increases for a given y

00:02:48.564 --> 00:02:50.605
your slope would increase

00:02:50.605 --> 00:02:53.169
and as y increases for a
given x your slope increases.

00:02:53.169 --> 00:02:54.242
And we see that.

00:02:54.899 --> 00:02:57.973
If we were to just hold y constant at one

00:02:57.973 --> 00:03:00.907
but increase x along this line,

00:03:00.907 --> 00:03:02.974
we see that the slope is increasing.

00:03:02.974 --> 00:03:04.587
It is getting steeper.

00:03:04.587 --> 00:03:06.329
And if we were to keep x constant

00:03:06.329 --> 00:03:08.465
and increase y across this line,

00:03:08.465 --> 00:03:10.531
we see that the slope increases.

00:03:10.531 --> 00:03:12.533
And, in general, we see
that the slope increases

00:03:12.551 --> 00:03:15.175
as we go to the top right.

00:03:15.175 --> 00:03:20.175
And we see that it decreases
as we go to the bottom left

00:03:20.271 --> 00:03:23.684
and both x and y become
much, much more negative.

00:03:23.684 --> 00:03:25.646
So, I'm feeling pretty good about this,

00:03:25.646 --> 00:03:28.188
especially if we can
knock this one out here,

00:03:28.188 --> 00:03:30.011
if we can knock that one out.

00:03:30.011 --> 00:03:32.553
So, dy/dx is equal to x over y.

00:03:32.553 --> 00:03:35.746
Well, then when x equals
one and y equals one,

00:03:35.746 --> 00:03:40.482
dy/dx would be equal to one,

00:03:40.482 --> 00:03:42.769
and this slope looks larger than one.

00:03:42.769 --> 00:03:45.045
It looks like two, but since

00:03:45.045 --> 00:03:46.368
we are really just eyeballing it,

00:03:46.368 --> 00:03:47.703
let's see if we can find something

00:03:47.703 --> 00:03:51.604
where this more clearly falls apart.

00:03:51.604 --> 00:03:52.869
So, let's look at the situation

00:03:52.869 --> 00:03:56.503
when they both equal negative one.

00:03:56.503 --> 00:04:00.682
So, x equals negative one and
y is equal to negative one.

00:04:00.682 --> 00:04:04.466
Well, in that case, dy/dx
should still be equal to one

00:04:04.466 --> 00:04:06.544
because you have negative one over one.

00:04:06.544 --> 00:04:08.205
Do we see that over here?

00:04:08.205 --> 00:04:10.666
So, when x is equal to negative one,

00:04:10.666 --> 00:04:13.045
y is equal to negative one.

00:04:13.045 --> 00:04:16.608
Our derivative here looks negative.

00:04:16.608 --> 00:04:17.910
It looks like negative two,

00:04:17.091 --> 00:04:20.422
which is consistent with this
yellow differential equation.

00:04:21.241 --> 00:04:24.747
The slope here is definitely
not a positive one,

00:04:24.747 --> 00:04:28.323
so we could rule this one out as well.

00:04:28.323 --> 00:04:30.726
And so we should feel pretty confident

00:04:30.726 --> 00:04:34.267
that this is the differential
equation being described.

00:04:34.267 --> 00:04:36.520
And now that we've done it,
we can actually think about

00:04:36.052 --> 00:04:37.479
well, okay, what are the solutions

00:04:37.947 --> 00:04:39.688
for this differential
equation going to look like.

00:04:39.688 --> 00:04:42.080
Well, it depends where they start

00:04:42.008 --> 00:04:43.170
or what points they contain.

00:04:43.242 --> 00:04:45.841
If you have a solution
that contains that point,

00:04:45.841 --> 00:04:50.841
it looks like it might
do something like this.

00:04:57.241 --> 00:05:02.241
If you had a solution
that contained this point,

00:05:03.963 --> 00:05:08.963
it might do something like that.

00:05:11.684 --> 00:05:12.960
And, of course, it keeps going.

00:05:12.096 --> 00:05:13.733
It looks like it would asymptote towards

00:05:14.597 --> 00:05:19.136
y is equal to negative
x, this downward sloping.

00:05:19.136 --> 00:05:24.136
This essentially is the line
y is equal to negative x.

00:05:24.964 --> 00:05:27.518
Actually, no that is not the
line y equals negative x.

00:05:27.518 --> 00:05:31.674
This is the line y is equal
to negative x minus one,

00:05:31.674 --> 00:05:33.601
so that's this line right over here.

00:05:33.601 --> 00:05:38.601
And it looks like if
the solution contained,

00:05:41.715 --> 00:05:43.503
say, this point right over here,

00:05:43.503 --> 00:05:46.138
that would actually be a solution to

00:05:46.138 --> 00:05:51.138
the differential equation y is
equal to negative x minus one

00:05:55.298 --> 00:05:56.215
and you can verify that.

00:05:56.215 --> 00:06:00.499
If y is equal to negative x minus one,

00:06:00.499 --> 00:06:01.741
then the x and negative x cancel out

00:06:01.741 --> 00:06:04.156
and you are just left with
dy/dx is equal to negative one,

00:06:04.156 --> 00:06:05.351
which is exactly what is being

00:06:05.351 --> 00:06:07.058
described by this slope field.

00:06:07.058 --> 00:06:09.297
Anyway, hopefully you
found that interesting.


Slope field to visualize solutions

WEBVTT
Kind: captions
Language: en

00:00:00.132 --> 00:00:02.485
- [Voiceover] So we have
the differential equation,

00:00:02.485 --> 00:00:04.404
the derivative of y with respect to x

00:00:04.404 --> 00:00:08.442
is equal to y over six times four minus y.

00:00:08.442 --> 00:00:09.794
And what we have plotted right over here

00:00:09.794 --> 00:00:11.412
is the slope field or a slope field

00:00:11.412 --> 00:00:14.433
for this differential
equation and we can verify

00:00:14.433 --> 00:00:18.654
that this indeed is a slope field

00:00:18.654 --> 00:00:20.013
for this differential equation,

00:00:20.013 --> 00:00:22.157
let's draw a little table here,

00:00:22.157 --> 00:00:25.181
so let's just verify a few points,

00:00:25.181 --> 00:00:30.181
so let's say x, y, and dy/dx.

00:00:32.502 --> 00:00:34.896
So let's say we start with, I don't know,

00:00:34.896 --> 00:00:39.896
let's start with this point
right over here, one comma one,

00:00:40.002 --> 00:00:43.769
when x is one, and y is
one, well when I look at

00:00:43.967 --> 00:00:47.405
the differential equation,
1/6 times four minus one,

00:00:47.405 --> 00:00:50.959
so it's 1/6 times three, which is 3/6,

00:00:50.959 --> 00:00:54.603
which is 1/2, and we see
indeed on the slope field,

00:00:54.603 --> 00:00:58.454
they depicted the slope
there if a solution goes

00:00:58.454 --> 00:01:02.259
to that point, right at that
point, its slope would be 1/2.

00:01:02.259 --> 00:01:05.275
And as you see, it's actually
only dependent on the y-value,

00:01:05.429 --> 00:01:08.050
it doesn't matter what x
is as long as y is one,

00:01:08.005 --> 00:01:10.511
dy/dx is going to be 1/2,
and you see that's why

00:01:10.556 --> 00:01:15.556
when x is 1 1/2, and y is one,
you still have a slope of 1/2

00:01:16.395 --> 00:01:20.454
and as long as y is one,
all of these sampled points

00:01:20.985 --> 00:01:25.985
right over here, all have a slope of 1/2,

00:01:25.995 --> 00:01:28.411
so that, just looking at
that, that makes us feel

00:01:28.411 --> 00:01:30.877
that this slope field is consistent

00:01:30.877 --> 00:01:32.583
with this differential
equation, but let's try a few

00:01:32.583 --> 00:01:34.530
other points just to feel a
little bit better about it,

00:01:34.053 --> 00:01:36.364
and then we will use a slope field

00:01:36.841 --> 00:01:39.605
to actually visualize some solutions.

00:01:39.605 --> 00:01:41.559
So let's say, let's do
an interesting point,

00:01:41.559 --> 00:01:44.587
let's say we have this point, actually no,

00:01:44.587 --> 00:01:47.010
that's at a half point, let's say we have

00:01:47.001 --> 00:01:51.480
this, let's see, let's say we
do this point right over here,

00:01:51.489 --> 00:01:54.658
so that's x is equal to
one, and y is equal to six,

00:01:54.658 --> 00:01:56.090
and we see the way the
differential equation

00:01:56.009 --> 00:01:57.058
is defined, it doesn't
matter what our x is,

00:01:57.058 --> 00:01:58.753
it's really dependent
on the y that's going

00:01:59.275 --> 00:02:02.720
to drive the slope, but
we have 6/6, which is one,

00:02:02.072 --> 00:02:05.043
times four minus six,
which is negative two.

00:02:05.691 --> 00:02:07.193
So it's negative two, so we should

00:02:07.193 --> 00:02:08.822
have a slope of negative two and it

00:02:08.822 --> 00:02:11.370
looks like that's what they
depicted, so as long as

00:02:11.037 --> 00:02:14.201
y is six, we should have
a slope of negative two.

00:02:14.534 --> 00:02:17.127
Have a slope of negative two,

00:02:17.127 --> 00:02:19.500
and you see that in the slope field.

00:02:19.005 --> 00:02:20.112
So hopefully you feel
pretty good that this

00:02:20.607 --> 00:02:22.783
is the slope field for
this differential equation,

00:02:22.783 --> 00:02:26.182
if you don't, I encourage
you to keep verifying

00:02:26.182 --> 00:02:28.741
these points here, but
now let's actually use

00:02:28.741 --> 00:02:30.882
the slope field, let's actually use this,

00:02:30.882 --> 00:02:34.193
to visualize solutions to
this differential equation

00:02:34.193 --> 00:02:37.154
based on points that the
solution might go through.

00:02:37.154 --> 00:02:40.490
So let's say that we have a solution

00:02:40.049 --> 00:02:42.135
that goes through this
point right over here.

00:02:43.035 --> 00:02:45.498
So what is that solution
likely to look like?

00:02:45.813 --> 00:02:48.134
And once again this is going
to be a rough approximation,

00:02:48.134 --> 00:02:48.223
well right at that point it's going

00:02:49.024 --> 00:02:51.852
to have a slope just as
the slope field shows,

00:02:51.852 --> 00:02:55.370
and as our y increases,
it looks like our slope,

00:02:55.037 --> 00:02:56.500
it looks like our slope...

00:02:56.833 --> 00:02:59.147
so at this point I should be, actually

00:02:59.147 --> 00:03:04.147
let me undo that, so if I
keep going up at this point

00:03:05.177 --> 00:03:07.819
when y is equal to two
I should be parallel

00:03:07.819 --> 00:03:10.862
to all of these segments
on the slope field

00:03:10.862 --> 00:03:14.411
that y is equal to two,
and then it looks like

00:03:14.411 --> 00:03:18.474
the slope starts to
decrease as we approach

00:03:19.041 --> 00:03:23.430
y is equal to four, and
so if I had a solution

00:03:23.043 --> 00:03:24.615
that went through this point,

00:03:25.002 --> 00:03:28.157
my guess is that would look something,

00:03:28.157 --> 00:03:30.201
and then now the slope decreases again

00:03:30.201 --> 00:03:35.201
as we approach y is equal to zero.

00:03:35.394 --> 00:03:37.662
And of course we see that,
because if when y equals zero

00:03:37.662 --> 00:03:38.947
this whole thing is zero so our

00:03:38.947 --> 00:03:41.052
derivative is going to be zero.

00:03:41.052 --> 00:03:44.256
So, a reasonable solution might

00:03:44.256 --> 00:03:46.705
look something like this,
so this gives us a clue

00:03:46.736 --> 00:03:49.540
of, look, if a solution
goes through this point,

00:03:49.054 --> 00:03:53.545
this right over here might
be what it looks like.

00:03:54.031 --> 00:03:56.172
But what if it goes through, I don't know,

00:03:56.172 --> 00:03:58.237
what if it goes through
this point right over here?

00:03:58.822 --> 00:04:01.723
Well then, it might look like...

00:04:01.723 --> 00:04:06.723
It might look like this
by the same exact logic.

00:04:07.921 --> 00:04:09.953
So it might look like
this, so just like that,

00:04:10.241 --> 00:04:11.908
we're trying to get a
sense, we don't know the

00:04:11.908 --> 00:04:13.949
actual solution for this
differential equation,

00:04:14.318 --> 00:04:15.832
but we're starting to get a sense of what

00:04:15.832 --> 00:04:19.418
type of functions or
the class of functions,

00:04:19.418 --> 00:04:22.363
that might satisfy the
differential equation.

00:04:22.363 --> 00:04:24.136
But, what's interesting
about this slope field,

00:04:24.136 --> 00:04:27.209
is it looks like there's
some interesting stuff

00:04:27.209 --> 00:04:31.888
if our solution includes points between

00:04:31.888 --> 00:04:34.439
where the y value is
between zero and four,

00:04:34.439 --> 00:04:35.987
it looks like we're going
to have solutions like this,

00:04:35.987 --> 00:04:39.582
but what if we had y-values
that were larger than that

00:04:39.582 --> 00:04:44.356
or that were less than that,
or exactly zero or four?

00:04:44.356 --> 00:04:46.672
So for example, what if
we had a solution that

00:04:46.672 --> 00:04:48.828
went through this point right over here?

00:04:48.828 --> 00:04:50.546
Well, that point right
over here, the slope field

00:04:50.546 --> 00:04:53.065
tells us that our slope is zero.

00:04:53.065 --> 00:04:54.476
So our y-value is not going to change,

00:04:54.476 --> 00:04:57.018
and as long as our y-value doesn't change,

00:04:57.018 --> 00:04:58.581
our y-value is going to stay at four,

00:04:58.581 --> 00:05:00.643
so our slope is going to stay at zero,

00:05:00.643 --> 00:05:02.326
so we actually already
found this is actually

00:05:02.326 --> 00:05:04.848
a solution to the differential
equation, y is equal

00:05:04.848 --> 00:05:09.112
to four is a solution to
this differential equation.

00:05:09.112 --> 00:05:14.112
So, y is equal to four,
and you can verify that

00:05:14.745 --> 00:05:17.991
that is a solution,
when y is equal to four,

00:05:17.991 --> 00:05:21.354
this right-hand side is going to be zero,

00:05:21.354 --> 00:05:24.923
and the derivative is zero
for y is equal to four.

00:05:24.923 --> 00:05:28.308
So that is a solution to
the differential equation.

00:05:28.308 --> 00:05:31.372
And the same thing for y is equal to zero.

00:05:31.948 --> 00:05:35.292
That is also a solution to
the differential equation.

00:05:35.292 --> 00:05:37.674
Now what if we included points,

00:05:37.674 --> 00:05:39.312
what if we included this point up here,

00:05:39.312 --> 00:05:41.336
and actually, let me do
it in a different color,

00:05:41.336 --> 00:05:44.273
so that you could see it, let's say our

00:05:44.273 --> 00:05:46.188
solution included that point.

00:05:46.188 --> 00:05:51.188
Well then it might look
something like this,

00:05:51.696 --> 00:05:53.724
and once again, I'm just
using the slope field

00:05:53.976 --> 00:05:57.837
as a guide to give me an
idea of what the slope

00:05:57.837 --> 00:06:02.837
might be as my curve progresses,
as my solution progresses.

00:06:03.247 --> 00:06:06.754
So solution that includes
the point zero five,

00:06:06.754 --> 00:06:08.663
might look something like this,

00:06:08.663 --> 00:06:11.704
and once again, it's just another clue.

00:06:11.704 --> 00:06:16.704
A solution that includes the
point zero negative 1 1/2,

00:06:17.403 --> 00:06:19.746
might look something like...

00:06:19.746 --> 00:06:23.697
might look something like this.

00:06:23.697 --> 00:06:27.903
So anyway, hopefully this
gives you a better appreciation

00:06:27.903 --> 00:06:30.121
for why slope fields are interesting.

00:06:30.121 --> 00:06:31.919
If you have a differential equation

00:06:31.919 --> 00:06:34.160
that just involves the first derivative,

00:06:34.016 --> 00:06:35.750
and some x's and y's,
this one only involves

00:06:35.894 --> 00:06:37.857
the first derivative and y's.

00:06:37.857 --> 00:06:39.826
We can plot a slope field like this,

00:06:39.826 --> 00:06:42.961
not too much trouble if we
essentially just keep solving

00:06:42.961 --> 00:06:45.825
for the slopes, and then
we can use that slope field

00:06:45.825 --> 00:06:47.799
to get a conceptual or
visual understanding

00:06:47.799 --> 00:06:49.861
of what the solutions might look like

00:06:49.861 --> 00:06:54.861
given points that the solutions
might actually contain.


Creating a slope field

WEBVTT
Kind: captions
Language: en

00:00:00.428 --> 00:00:01.795
- [Voiceover] Let's say that we have

00:00:01.795 --> 00:00:05.536
the differential equation
dy dx or the derivative

00:00:05.536 --> 00:00:10.536
of y with respect to x is
equal to negative x over y.

00:00:11.519 --> 00:00:14.844
Let's say we don't know how
to find the solutions to this,

00:00:14.844 --> 00:00:16.845
but we at least want to get a sense

00:00:16.944 --> 00:00:19.012
of what the solutions might look like.

00:00:19.441 --> 00:00:22.337
And to do that what we
could do is we could look

00:00:22.337 --> 00:00:26.811
at a coordinate plane, so
let me draw some axes here.

00:00:27.305 --> 00:00:31.171
So let me draw relatively
straight line alright.

00:00:31.459 --> 00:00:36.288
So that's my y-axis and this is my x-axis.

00:00:38.826 --> 00:00:43.820
Let me mark this as one, that's
two, that's negative one,

00:00:43.082 --> 00:00:48.082
negative two, one, two,
negative one, and negative two.

00:00:51.304 --> 00:00:54.210
And what I could do is since
this differential equation

00:00:54.021 --> 00:00:57.962
is just in terms of xs and
ys and first derivatives

00:00:58.151 --> 00:01:02.476
of y with respect to x, I
could sample points on the

00:01:02.476 --> 00:01:05.382
coordinate plane, I could look
at the x and y coordinates

00:01:05.382 --> 00:01:07.079
substitute them in here, figure out

00:01:07.079 --> 00:01:08.686
what the slope is going to be.

00:01:08.686 --> 00:01:11.323
And then I could visualize
the slope if a solution

00:01:11.323 --> 00:01:14.161
goes through that point what
the slope needs to be there

00:01:14.161 --> 00:01:16.176
and I can visualize that
with a line segment,

00:01:16.176 --> 00:01:17.642
a little small line segment that has

00:01:17.642 --> 00:01:20.364
the same slope as the slope in question.

00:01:20.784 --> 00:01:22.745
So let's actually do that.

00:01:23.175 --> 00:01:24.926
So let me setup a little table here.

00:01:25.978 --> 00:01:27.083
I'm going to do a little table here

00:01:27.083 --> 00:01:29.110
to do a bunch of x and y values.

00:01:29.011 --> 00:01:30.455
Once again I'm just sampling some points

00:01:30.554 --> 00:01:32.647
on the coordinate plane
to be able to visualize.

00:01:33.133 --> 00:01:38.133
So x, y and this is dy dx.

00:01:38.983 --> 00:01:43.983
So let's say when x is zero and y is one,

00:01:44.979 --> 00:01:47.137
what is the derivative
of y with respect to x?

00:01:47.453 --> 00:01:49.017
It's going to be negative zero over one

00:01:49.017 --> 00:01:50.814
so it's just going to be zero.

00:01:51.387 --> 00:01:54.543
And so at the point zero one
if a solution goes through

00:01:54.543 --> 00:01:57.740
this point its slope is going to be zero.

00:01:57.977 --> 00:02:00.480
And so we can visualize
that by doing a little

00:02:00.048 --> 00:02:02.317
horizontal line segment right there.

00:02:03.159 --> 00:02:04.412
So let's keep going.

00:02:04.697 --> 00:02:08.860
What about when x is one and y is one?

00:02:09.634 --> 00:02:12.138
Well then dy dx, the derivative
of y with respect to x

00:02:12.138 --> 00:02:15.390
is negative one over one, so
it's going to be negative one.

00:02:15.781 --> 00:02:19.377
So at the point one comma
one if a solution goes

00:02:19.377 --> 00:02:24.377
through that point, it would
have a slope of negative one.

00:02:26.236 --> 00:02:27.558
And so I draw a little line segment

00:02:27.558 --> 00:02:29.436
that has a slope of negative one.

00:02:31.358 --> 00:02:33.071
Let me do this in a new color.

00:02:33.071 --> 00:02:36.693
What about when x is one and y is zero?

00:02:37.041 --> 00:02:40.656
Well then it's negative one over zero,

00:02:41.025 --> 00:02:44.726
so this is actually
undefined, but it's a clue

00:02:44.726 --> 00:02:49.501
that maybe the slope there
if you had a tangent line

00:02:49.501 --> 00:02:51.472
at that point, maybe it's vertical.

00:02:51.472 --> 00:02:56.472
So I'll put that as a
question mark, vertical there.

00:02:56.887 --> 00:02:59.936
So maybe it's something
like that if you actually

00:03:00.377 --> 00:03:02.378
did have, I guess it
wouldn't be a function,

00:03:02.477 --> 00:03:05.208
if you had some kind of
relation that went through it,

00:03:05.208 --> 00:03:08.419
but let's not draw that just yet.

00:03:08.419 --> 00:03:09.815
But let's try some other points.

00:03:12.728 --> 00:03:16.434
Let's try the point
negative one negative one.

00:03:17.303 --> 00:03:20.325
So now we have negative negative one,

00:03:20.325 --> 00:03:21.927
which is one, over negative one.

00:03:21.927 --> 00:03:24.245
Well you'd have a slope
of negative one here.

00:03:24.245 --> 00:03:25.970
So negative one negative one

00:03:25.097 --> 00:03:30.097
you'd have a slope of negative one.

00:03:31.896 --> 00:03:35.620
What about if you had one negative one?

00:03:36.279 --> 00:03:37.670
Well now it's negative one over

00:03:37.067 --> 00:03:39.278
negative one, your slope is now one.

00:03:39.881 --> 00:03:43.418
So one negative one, if a
solution goes through this

00:03:43.418 --> 00:03:46.097
its slope would look like that.

00:03:46.049 --> 00:03:49.024
And we could keep going, we
could even do two negative two.

00:03:49.465 --> 00:03:52.586
That's going to have a
slope of one as well.

00:03:52.978 --> 00:03:56.937
If you did positive two positive two,

00:03:56.937 --> 00:03:58.262
that would be negative two over two.

00:03:58.262 --> 00:04:00.790
You'd have a slope of
negative one right over here.

00:04:01.181 --> 00:04:05.230
And so we could do a bunch
of points, just keep going.

00:04:05.023 --> 00:04:06.855
Now I'm just doing them in my head,

00:04:07.062 --> 00:04:08.284
I'm not going on the table.

00:04:08.284 --> 00:04:10.272
But you get a sense of
what's going on here.

00:04:10.272 --> 00:04:13.177
Here you're slope, what if
it was negative one one?

00:04:13.177 --> 00:04:16.161
It's going to have a slope of one.

00:04:16.022 --> 00:04:19.079
So at this point your
slope negative one one

00:04:19.079 --> 00:04:21.236
so negative negative one is one over one,

00:04:21.947 --> 00:04:23.464
so you're going to have a slope like that.

00:04:23.464 --> 00:04:26.332
At negative two two same exact idea,

00:04:26.332 --> 00:04:27.505
it would look like that.

00:04:27.505 --> 00:04:30.268
And so when you keep
drawing these line segments

00:04:30.268 --> 00:04:33.351
over these kind of sample
points in the cartesian

00:04:33.351 --> 00:04:37.023
or in the x-y plane, you
start to get a sense of

00:04:37.023 --> 00:04:39.770
well what would a solution have to do.

00:04:40.127 --> 00:04:43.443
And you can start to visualize
that hey maybe a solution

00:04:44.464 --> 00:04:49.464
would have to do something like this.

00:04:49.501 --> 00:04:52.820
This would be a solution,
so maybe it would

00:04:52.082 --> 00:04:53.731
have to do something like this.

00:04:54.469 --> 00:04:58.861
Or if we're looking only at
functions and not relations

00:04:58.861 --> 00:05:02.276
I'll make it so it's very clear.

00:05:02.691 --> 00:05:04.562
So maybe it would have to
do something like this.

00:05:04.562 --> 00:05:09.006
Or if the function started
here based on what we've

00:05:09.006 --> 00:05:14.006
seen so far maybe it would
have to do something like this.

00:05:15.424 --> 00:05:20.212
Or if this were a point
on the function over here,

00:05:20.212 --> 00:05:23.524
it would have to do something like this.

00:05:23.524 --> 00:05:25.020
And once again I'm doing this based on

00:05:25.002 --> 00:05:27.336
what the slope field is telling me.

00:05:28.246 --> 00:05:31.430
So this field that I'm
creating where I'm sampling

00:05:31.043 --> 00:05:32.531
a bunch of points and I'm visualizing

00:05:32.918 --> 00:05:34.635
the slope with a line segement.

00:05:34.635 --> 00:05:39.635
Once again this is called a slope field.

00:05:42.805 --> 00:05:43.996
So hopefully that gives you kind of

00:05:43.996 --> 00:05:45.894
the basic idea of what a slope field is.

00:05:45.894 --> 00:05:50.127
And the next two videos we'll
explore this idea even deeper.


Differential equation introduction

WEBVTT
Kind: captions
Language: en

00:00:00.473 --> 00:00:02.424
- [Voiceover] Let's
now introduce ourselves

00:00:02.424 --> 00:00:05.656
to the idea of a differential equation.

00:00:05.656 --> 00:00:06.721
And as we'll see, differential equations

00:00:07.306 --> 00:00:10.688
are super useful for
modeling and simulating

00:00:10.688 --> 00:00:13.573
phenomena and understanding
how they operate.

00:00:13.589 --> 00:00:15.038
But we'll get into that later.

00:00:15.038 --> 00:00:17.772
For now let's just think
about or at least look at

00:00:17.772 --> 00:00:20.523
what a differential equation actually is.

00:00:21.572 --> 00:00:24.938
So if I were to write, so
let's see here is an example

00:00:24.938 --> 00:00:27.139
of differential equation,
if I were to write that

00:00:27.139 --> 00:00:31.224
the second derivative of y plus two times

00:00:31.224 --> 00:00:36.007
the first derivative of y
is equal to three times y,

00:00:36.007 --> 00:00:38.189
this right over here is
a differential equation.

00:00:38.189 --> 00:00:39.722
Another way we could write it if we said

00:00:39.722 --> 00:00:42.305
that y is a function of
x, we could write this

00:00:42.305 --> 00:00:43.907
in function notation.

00:00:43.907 --> 00:00:45.855
We could write the second
derivative of our function

00:00:45.855 --> 00:00:49.920
with respect to x plus two
times the first derivative

00:00:50.505 --> 00:00:54.108
of our function is equal to
three times our function.

00:00:54.788 --> 00:00:58.240
Or if we wanted to use
the Leibniz notation,

00:00:58.024 --> 00:01:03.024
we could also write, the
second derivative of y

00:01:03.987 --> 00:01:08.938
with respect to x plus two
times the first derivative

00:01:08.938 --> 00:01:13.422
of y with respect to x is
equal to three times y.

00:01:14.037 --> 00:01:17.755
All three of these equations
are really representing

00:01:17.755 --> 00:01:20.856
the same thing, they're saying
OK, can I find functions

00:01:20.856 --> 00:01:23.023
where the second
derivative of the function

00:01:23.023 --> 00:01:25.025
plus two times the first
derivative of the function

00:01:25.223 --> 00:01:27.937
is equal to three times
the function itself.

00:01:28.355 --> 00:01:31.339
So just to be clear,
these are all essentially

00:01:31.339 --> 00:01:33.655
saying the same thing.

00:01:33.806 --> 00:01:36.089
And you might have just
caught from how I described it

00:01:36.089 --> 00:01:38.092
that the solution to a
differential equation

00:01:38.389 --> 00:01:41.238
is a function, or a class of functions.

00:01:41.323 --> 00:01:44.322
It's not just a value or a set of values.

00:01:44.354 --> 00:01:47.673
So the solution here, so the solution

00:01:47.673 --> 00:01:52.673
to a differential equation is a function,

00:01:53.738 --> 00:01:56.672
or a set of functions,
or a class of functions.

00:01:56.756 --> 00:01:58.805
It's important to contrast this relative

00:01:58.805 --> 00:02:00.027
to a traditional equation.

00:02:00.027 --> 00:02:01.421
So let me write that down.

00:02:01.474 --> 00:02:04.549
So a traditional equation,
maybe I shouldn't say

00:02:05.224 --> 00:02:06.687
traditional equation,
differential equations

00:02:06.687 --> 00:02:08.306
have been around for a while.

00:02:08.589 --> 00:02:11.787
So let me write this as
maybe an algebraic equation

00:02:11.787 --> 00:02:13.355
that you're familiar with.

00:02:13.622 --> 00:02:18.622
An algebraic equation
might look something like,

00:02:19.389 --> 00:02:21.623
and I'll just write up a simple quadratic.

00:02:21.856 --> 00:02:26.856
Say x squared plus three x
plus two is equal to zero.

00:02:30.139 --> 00:02:32.673
The solutions to this algebraic equation

00:02:32.673 --> 00:02:35.255
are going to be numbers,
or a set of numbers.

00:02:35.255 --> 00:02:38.271
We can solve this, it's
going to be x plus two

00:02:38.271 --> 00:02:41.639
times x plus one is equal to zero.

00:02:41.924 --> 00:02:45.506
So x could be equal to
negative two or x could

00:02:45.506 --> 00:02:47.105
be equal to negative one.

00:02:47.171 --> 00:02:52.171
The solutions here are
numbers, or a set of values

00:02:53.003 --> 00:02:55.225
that satisfy the equation.

00:02:55.522 --> 00:02:58.390
Here it's a relationship
between a function

00:02:58.039 --> 00:02:59.286
and its derivatives.

00:02:59.637 --> 00:03:01.655
And so the solutions, or the solution,

00:03:01.655 --> 00:03:03.637
is going to be a function
or a set of functions.

00:03:03.637 --> 00:03:05.456
Now let's make that a
little more tangible.

00:03:05.456 --> 00:03:08.439
What would a solution to
something like any of these three,

00:03:08.439 --> 00:03:10.072
which really represent the same thing,

00:03:10.072 --> 00:03:12.355
what would a solution actually look like?

00:03:12.355 --> 00:03:16.623
Actually let me move
this over a little bit.

00:03:17.121 --> 00:03:19.170
Move this over a little bit.

00:03:19.307 --> 00:03:21.508
So we can take a look at what some

00:03:21.508 --> 00:03:22.601
of these solutions could look like.

00:03:23.438 --> 00:03:24.722
Let me erase this a little.

00:03:25.038 --> 00:03:27.022
This little stuff that
I have right over here.

00:03:27.637 --> 00:03:31.071
So I'm just gonna give you
examples of solutions here.

00:03:31.071 --> 00:03:34.539
We'll verify that these
indeed are solutions

00:03:34.539 --> 00:03:36.888
for I guess this is really
just one differential equation

00:03:36.888 --> 00:03:38.190
represented in different ways.

00:03:38.019 --> 00:03:39.835
But you'll hopefully
appreciate what a solution

00:03:40.006 --> 00:03:42.304
to a differential equation looks like.

00:03:42.389 --> 00:03:45.272
And that there is often
more than one solution.

00:03:45.272 --> 00:03:49.087
There's a whole class of functions
that could be a solution.

00:03:49.689 --> 00:03:52.556
So one solution to this
differential equation,

00:03:52.556 --> 00:03:54.224
and I'll just write it as our first one.

00:03:54.605 --> 00:03:57.339
So one solution, I'll call it y one.

00:03:57.339 --> 00:03:59.523
And I could even write it as y one of x

00:03:59.523 --> 00:04:01.548
to make it explicit that
it is a function of x.

00:04:02.005 --> 00:04:05.364
One solutions is y one of x is equal to e

00:04:05.364 --> 00:04:08.022
to the negative three x.

00:04:08.022 --> 00:04:10.423
And I encourage you to
pause this video right now

00:04:10.423 --> 00:04:13.355
and find the first derivative of y one,

00:04:13.355 --> 00:04:16.437
and the second derivative of
y one, and verify that it does

00:04:16.437 --> 00:04:19.288
indeed satisfy this differential equation.

00:04:20.137 --> 00:04:22.455
So I'm assuming you've had a go at it.

00:04:22.588 --> 00:04:23.671
So let's work through this together.

00:04:23.671 --> 00:04:25.473
So that's y one.

00:04:25.622 --> 00:04:29.555
So the first derivative of y one,

00:04:29.555 --> 00:04:31.289
so we just have to do the chain rule here,

00:04:31.289 --> 00:04:33.706
the derivative of negative
three x with respect to x

00:04:33.706 --> 00:04:35.024
is just negative three.

00:04:35.256 --> 00:04:36.972
And the derivative of e
to the negative three x

00:04:36.972 --> 00:04:39.975
with respect to negative three x is just e

00:04:39.975 --> 00:04:41.922
to the negative three x.

00:04:42.256 --> 00:04:46.273
And if we take the second
derivative of y one,

00:04:46.273 --> 00:04:48.368
this is equal to the same exact idea,

00:04:49.223 --> 00:04:50.688
the derivative of this is
three times negative three

00:04:50.688 --> 00:04:55.273
is going to be nine e
to the negative three x.

00:04:55.756 --> 00:04:56.762
And now we could just
substitute these values

00:04:57.356 --> 00:04:59.223
into the differential
equation, or these expressions

00:04:59.223 --> 00:05:01.007
into the differential
equation to verify that this

00:05:01.007 --> 00:05:05.089
is indeed going to be
true for this function.

00:05:05.089 --> 00:05:06.372
So let's verify that.

00:05:06.605 --> 00:05:11.605
So we first have the
second derivative of y.

00:05:12.856 --> 00:05:14.723
So that's that term right over there.

00:05:14.837 --> 00:05:19.789
So we have nine e to the negative three x

00:05:19.789 --> 00:05:24.789
plus two times the first derivative.

00:05:26.187 --> 00:05:29.888
So that's going to be two
times this right over here.

00:05:30.005 --> 00:05:33.507
So it's going to be minus
six, I'll just write

00:05:33.507 --> 00:05:37.654
plus negative six e to
the negative three x.

00:05:37.654 --> 00:05:40.256
Notice I just took this two
times the first derivative.

00:05:40.256 --> 00:05:44.805
Two times the first derivative
is going to be equal to,

00:05:44.805 --> 00:05:48.203
or needs to be equal to, if
this indeed does satisfy,

00:05:48.203 --> 00:05:52.340
if y one does indeed satisfy
the differential equation,

00:05:52.034 --> 00:05:54.682
this needs to be equal to three times y.

00:05:54.988 --> 00:05:59.023
Well three times y is three
times e to the negative three x.

00:05:59.188 --> 00:06:02.189
Three e to the negative three x.

00:06:02.606 --> 00:06:04.305
Let's see if that indeed is true.

00:06:04.523 --> 00:06:07.937
So these two terms right over here,

00:06:07.937 --> 00:06:09.755
nine e to the negative three x,

00:06:09.755 --> 00:06:12.788
essentially minus six e
to the negative three x,

00:06:12.788 --> 00:06:15.724
that's gonna be three e
to the negative three x.

00:06:15.938 --> 00:06:20.938
Which is indeed equal to three
e to the negative three x.

00:06:22.923 --> 00:06:27.923
So y one is indeed a solution
to this differential equation.

00:06:28.288 --> 00:06:31.223
But as we'll see, it is
not the only solution

00:06:31.223 --> 00:06:32.588
to this differential equation.

00:06:32.839 --> 00:06:37.839
For example, let's say y
two is equal to e to the x

00:06:44.056 --> 00:06:46.488
is also a solution to this
differential equation.

00:06:46.488 --> 00:06:47.973
And I encourage you to
pause the video again

00:06:47.973 --> 00:06:49.756
and verify that it's a solution.

00:06:50.572 --> 00:06:51.889
So assuming you've had a go at it.

00:06:52.055 --> 00:06:54.272
So the first derivative of this
is pretty straight-forward,

00:06:54.272 --> 00:06:55.521
is e to the x.

00:06:55.521 --> 00:06:58.173
Second derivative, one
of the profound things

00:06:58.173 --> 00:07:00.306
of the exponential function,
the second derivative here

00:07:00.306 --> 00:07:01.773
is also e to the x.

00:07:02.224 --> 00:07:05.623
So the second derivative,
let me just do it

00:07:05.623 --> 00:07:06.974
in those same colors.

00:07:07.022 --> 00:07:11.423
So the second derivative
is going to be e to the x

00:07:11.423 --> 00:07:13.439
plus two times e to the x

00:07:17.621 --> 00:07:22.621
is indeed going to be equal
to three times e to the x.

00:07:23.255 --> 00:07:24.522
This is absolutely going to be true.

00:07:24.522 --> 00:07:27.523
E to the x plus two e to the
x is three e to the x.

00:07:27.589 --> 00:07:32.170
So y two is also a solution
to this differential equation.

00:07:32.622 --> 00:07:33.855
So that's a start.

00:07:33.855 --> 00:07:35.839
In the next few videos,
we'll explore this more.

00:07:35.839 --> 00:07:37.706
We'll start to see what
the solutions look like,

00:07:37.706 --> 00:07:38.761
what classes of solutions are,

00:07:39.256 --> 00:07:40.474
techniques for solving them,

00:07:40.474 --> 00:07:43.140
visualizing solutions to
differential equations,

00:07:43.014 --> 00:07:47.511
and a whole toolkit for
kind of digging in deeper.