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# Simple differential equation example

Video transcript

Let f be a real valued
differentiable function on R-- the set of all real
numbers-- such that f of 1 is equal to 1. If the y-intercept of the
tangent at any point P-- which has the coordinates x
and y-- on the curve y equals f of x is equal to
the cube of the abscissa-- I haven't seen the word
abscissa in a while-- of P, then the value of f of
negative 3 is equal to. So let's just decode
what they're saying. So let me draw a
potential f right here. So that's our y-axis. And then, let's call
that our x-axis. And let me just draw a
simple, potential f of x. So let's say it looks
something like that. I don't know what it looks like. We don't know what
it looks like. We just know it's some real
valued differentiable function on R. So that is f of x. So what is the tangent
going to look like? So let's say we have some point
here, which is x, 0 and y, 0. So it's a particular x and y. What would the tangent
there look like? Well, it would be
a line that has the same slope as the
function does at that point. So it would be a line. I can draw a better
version than that. It would be a line
that looks like that. This line right here
would have the equation y is equal to mx plus
b, where m is the slope and b is the y-intercept. Now, we know that
the slope is going to be the same as the slope
at this point in the function. And what is the slope at
that point in the function? Well, the slope at that
point in that function is the derivative
of the function. So this is going to
be f prime of x, 0. So we take the derivative
of valuated x, 0. That will give us the
slope right at that point. So this is going to
be y is equal to f prime of x, 0 times x plus b. Now, if we wanted to solve
for the y-intercept-- for b right there-- we
just have to find out some point that's on this line. And we know a point that's
on the line. x, 0 and y, 0 is on the line. So let's substitute x, 0
and y, 0 into this equation and then solve for b. So we have y, 0 is
equal to f prime of x, 0-- that's just the
slope-- times x, 0-- we're just substituting y, 0 and x, 0
into this equation-- plus b. So I just put x, 0 and y, 0 into
this equation to solve for b. If you solve for b, you
could just subtract this from both sides of the equation. We get b. b is equal to y, 0 minus f
prime of x, 0 times x, 0. Now, that's the
y-intercept of the tangent. They tell us the y-intercept of
the tangent at any point P x, y on the curve y
equals f of x is equal to the cube of
the abscissa of P. An abscissa is just a fancy
way of saying the x-coordinate. Clearly, they're
just using the word to filter out people who don't
know the word "abscissa." Abscissa just means
the x-coordinate. If I tell you I have the point
2 comma 3, the abscissa of this is just 2. It's just the x-coordinate. So all this sentence is telling
us is that the y-intercept is equal to the cube
of the x-coordinate. So the y-intercept--
it's this-- is going to be equal to the
cube of the x-coordinate. The x-coordinate here is x, 0. So it's going to be
equal to x, 0 cubed. So for any x, 0 and y,
0 for any x, 0 and y, 0 that's on the function itself,
we now get this expression. I'll do it in a new color. We get y, 0 minus f
prime of x, 0 times x, 0 is equal to x, 0 to the third. Now, these are
particular x's and y's. It's just for a case of example. It applies to all of them. So we could say it applies to
all of the points on the curve. So we could say, for
any point on the curve, we get y minus f
prime of x times x is equal to x to the third. And we know that y
is equal to f of x. Or another way of
thinking about it is that y prime is
equal to f prime of x. So we could write this. And what we're getting
here is a little bit of a differential equation. We get y minus y prime times
x is equal to x to the third. Let's see. We can could divide both
sides of this equation by x. We get y over x minus y
prime is equal to x squared-- just divided everything by x. And then we can subtract
this from both sides and then add this
to both sides-- kind of skipping a
step right there. So we get y over x-- actually,
let me write it this way. I don't want to skip steps. So let's subtract y
over x from both sides. We get negative y prime is
equal to negative y over x plus x squared. And then, multiply both
sides times negative 1. We get y prime is equal to
y over x minus x squared. And so a little bit of a
differential equation here. And solving any
differential equation is really a bit of an art. And so we have to
think about it-- if we could separate the y's
from the x's-- if we could put the y's all on this side--
then it would be pretty easy to solve, but it's
not so obvious. And just to give
an example-- just to inform our intuition--
if this was a simpler version, if this was
just dy/dx-- which is the same thing as y
prime-- if it was just equal to y over x,
if it was just that, so if this negative x
squared term wasn't there, then this would be a separable
differential equation. You could divide both sides by
y and multiply both sides times dx. So if you multiplied
this times dx over y and you multiplied this times dx
over y, these would cancel out. Those would cancel out. And you would get ydy
is equal to x-- sorry. Not ydy. You'd get 1 over y. 1 over y dy is equal
to 1 over x dx. You integrate both
sides of that. You would get the
natural log of y-- that's the antiderivative
of 1 over y-- is equal to the natural log
of x plus some constant. And we could write any constant
as plus the natural log of some constant. It could just be any
constant over there. And then this is the same
thing as the natural log of y is equal to the natural log
of-- when you add exponents like this, that's essentially
what a logarithm is. It's essentially taking a
product of these two terms. This is equal to cx or
that y is equal to cx. Now, this is all just to
inform our intuition here. If this negative x
squared term wasn't here, then we would be able
solve it pretty easily. y would be equal to cx. If this term wasn't here, taking
the antiderivative of this is pretty straightforward. It would be negative 1/3
x to the third power. Our intuition here is--
maybe some combination of the antiderivatives
of this and this-- and you can't just
take them separately because this y is
going to have to be equal to the antiderivative
of the whole thing. It's not just the antiderivative
of this part over here. But if this guy
wasn't here, then it could be y is equal to cx. If this guy wasn't here,
you would have something to the third power. So just to try things out-- and
solving differential equations is always a bit
of an art-- let's just say that y is
equal to-- we know we're going to have
to have something to the third power in there. So let's call it Ax
to the third power. And we know we're going to
have a x power in there. Let's call it plus Bx. And then, let's call it plus C. And that's our intuition--
that somehow that this will help you get to this
term right over here. And then, this term-- when
I take the derivative-- will help to get
that term over there. So what would be the
derivative of this? The derivative of this--
y prime, in this case-- would be 3Ax squared plus B. And now, what would y over x
be? y over x would be equal to-- notice, I'm just
saying y prime is this. y over x would be Ax squared
plus B plus C over x. This differential equation
here would reduce to-- and we just suspect
that this might be a solution-- this y prime
is the same thing as this. So we have 3Ax squared plus
B is equal to y over x. It's equal to this thing
over here-- Ax squared plus B plus C over x minus
this x squared. So what does this simplify to? If we wanted to
solve this equation, we could combine
the x squared term. So let me write it this way. We have 3Ax squared
plus B is equal to-- we have Ax squared
minus x squared. So that's equal to A minus 1. This is equal to A
minus 1 x squared. And then, I have
plus B plus C over x. This is what the right-hand
side simplifies to, and this is what the
left-hand side is. In this, we have no constrain
on B. B so far can be anything. B could be anything. There's no C term over here. And actually, more
importantly, there's no 1 over x term over here. So C would have
to be equal to 0. There's no C. C is equal to 0. This term does not exist. There is no C. And then, when you
look at this, this coefficient-- the 3A-- the
coefficient of the x squared on the left-hand side has to
be equal to the coefficient of the x squared on the
right-hand side in order for these 2 to be equal. So we'd get 3A is
equal to A minus 1. Subtract A from both sides. 2A is equal to negative 1. A is equal to negative 1/2. So our hunch paid off. We were able to figure
out A's, B's, and C's so that this function does
satisfy our little differential equation. And our value we got
is y is equal to-- I keep using that green color. I don't want to. We get y is equal to
negative 1/2 x to the third. That's our value for A.
Plus, B could be anything. Plus Bx. So this is our value for f of x. And they gave us
some information. They told us in the problem
that f of 1 is equal to 1. We can use that information
to solve for B. f of 1 is equal to 1. So when x is equal to
1, y is equal to 1. So we get 1 is equal
to negative 1/2 times 1 plus B times 1 is just B. So
we can add 1/2 to both sides. And we get 3/2 is equal to B. So we're now able to fully
describe our function. Our function is y is
equal to negative 1/2 x to the third plus 3/2x. So that's our function,
but that's not what they're asking us. They're not asking
us for the function. They're asking us for the
value of f of negative 3. f of negative 3 is equal
to negative 1/2 times negative 3 to the third is
negative 27 plus 3/2 times negative 3. So minus 9/2. This will simplify to--
well, everything's over 2. Negative 27 is positive 27. And then you have a minus 9. This is 27 over 2, 9 over 2. negative 9 over 2. So this is equal to 18 over
2, which is equal to 9. And we are done. This is equal to 9.