Main content

## Calculus, all content (2017 edition)

### Unit 8: Lesson 1

AP Calculus AB questions- Sal interviews the AP Calculus Lead at College Board
- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c
- 2011 Calculus AB free response #1a
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2011 Calculus AB free response #5c

Solving a differential equation using separation of variables. Created by Sal Khan.

## Want to join the conversation?

- Whats the diference between a diferential equation and a implicit derivative. (?)(3 votes)
- Great question!

I'll give you my informal understanding.

They are basically opposites or going in opposite directions.

That is with implicit differentiation you take an ordinary equation like:

x^2 + y^2 = 1

and differentiate it to

2x + 2y dy/dx = 0

and solve for dy/dx.

So we are solving for a derivative.

But the next to last step:

2x + 2y dy/dx = 0

is a differential equation.

And we can take that and solve it and get

x^2 + y^2 = C

(Which is almost what we started with)

With a differential equation we are solving for y or in this case just an ordinary equation with no derivatives.(4 votes)

- i dont understand the theory behind the SEPERATION OF VARIABLES method for integration using dy/dx = g(y).f(x)

1/g(y).dy/dx = f(x)

then integrating both sides ie S 1/g(y).dy/dx .dx =S f(x) .dx w.r.t x

reduces to S 1/g(y).ds = S f(x) .dx but you cant reduce dy/dx .dx to dy by cancelling out algebriacaly dx dy/dx is not a fraction!

YET THIS WHAT SEEMS TO HAVE BEEN DONE HERE!

can anyone explain the theory behind how this was done?(1 vote)- It turns out that the dx/dy notation (which was invented by Leibniz) very closely mirrors actual operations. To cancel out the dx's this way is not as simple as cancelling out the fraction. There are actually a number of steps you need to do, but it turns out that S 1/g(y) · dy/dx · dx is equal to S 1/g(y) · dy. So, you are
**not**canceling out fractions, you are applying a rule and Leibniz's notation is so brilliant, that 'canceling out the fraction' actually gives you an expression that is equal.

Disclaimer: I have had this same question for a long time now. I don't actually understand the steps you need to do to show that S 1/g(y) · dy/dx · dx is equal to S 1/g(y) · dy, but I was able to answer your question because I have read this page:

http://math.stackexchange.com/questions/47092/physicists-not-mathematicians-can-multiply-both-sides-with-dx-why

Yeah, I know, it really bugs me to just accept stuff, but I'm gonna have to do the same until I have learned more.

Cheers(3 votes)

- The transition from ln(W-300) to e^(ln(W-300)) was very clever. What suggested that transition, and would it have been acceptable to leave the equation in terms of ln W?(2 votes)
- The presence of a natural logarithm itself invites us to raise e to the power of the two sides of the equation. And in general we want to solve for a variable to create a graph-able equation. If we had solved for ln W our y-axis would have to represent ln W rather than W. In real life we don't care about the natural log of how much trash we have, just how much we have.(1 vote)

- Differential equations in Calc AB is relatively easy, but how hard are they in Calc BC/Calc 3 and beyond?(1 vote)
- if you can get a hang of this than the others shouldent be that hard. but everyone is different so honestly every answer put here is right and wrong its one of those few questions that can go eather way.

i wish the best of luck to you ;)(1 vote)

## Video transcript

Part C, find the
particular solution-- W as a function of t-- to
the differential equations, derivative of W
with respect to t is equal to 1 over
25 times w minus 300, with initial condition, at
time 0, W is equal to 1,400. And the units in this problem,
it was 1,400 tons they told us. So it might seem strange to
see a differential equations problem on an AP exam. You say, wait, this isn't fair. This isn't a differential
equations class. And the thing to keep in
mind is if, on an AP exam, you see differential
equations, they're really asking you to solve
a differential equation that can be solved using
the skills that you would learn in an AP class. And the main class of
differential equations that you could
solve are the ones that are separable
differential equations. The ones where you can
separate the independent and the dependent variables. And if you don't have any
idea what I'm talking about, I'm about to show you. So let me rewrite this
differential equation. We have dw, dt is equal to
1 over 25 times W minus 300. Now this differential
equation, we have W here. And we have this dt. But we don't have
t anywhere else. So what I want to do is I want
to get the parts that involve W onto the left-hand
side, and I want to get anything that deals
with dt on the right-hand side. So let's divide both sides of
this equation by w minus 300. So we would have 1
over W minus 300 dw dt-- the derivative of
W with respect to t-- is equal to-- I divide
this side by W minus 300, so it's equal to 1 over 25. Now I want to get the
dt's on this side. So let's multiply
both sides by dt. [INAUDIBLE] the
differential is just a very, very small change in t. And so we get-- and I'm a little
bit imprecise with my notation here. But this is one way that
you can think about it when you are dealing with the
differential equations like this, separable
differential equations. You have 1 over W minus
300 dw-- let me make sure that the dw doesn't look like
it's in the denominator-- dw is equal to 1 over 25 dt. And now we can integrate both
sides, which is essentially just saying that we're taking--
since this is equal to this-- that each little
increment of W times this is equal to each little
increment of dt times 1 over 25. We can then say, well, if we sum
up all of the increments of W times this quantity--
if we sum up the infinite number of
infinitesimally small increments-- that would be
the same thing as summing up all of these characters. Because each of these
characters is really the same as each of
these characters. So let's do that sum. We're essentially just
integrating both sides of this. And here, you might be able
to do this in your head. Or we could do the
formal u substitution, just to make it clear
what we're doing. If you said u is equal to W
minus 300, then what is du? If you wanted the differential. So if you want to take
the derivative of u with respect to W, you would
say du with respect to W is equal to what? It is equal to 1. The derivative of this
with respect to W is 1. Or if you multiply
both sides times dw, you have du is equal to dw. And so this left-hand side right
over here-- this integral-- could be rewritten as the
integral of 1 over u du. We define this as u. And we just saw that du
is the same thing as dw, so this could be
rewritten as du. And this is one of
the basic integrals, so hopefully we've learned. And this is equal to the natural
log of the absolute value of u. Or if you unwind
the substitution, this is equal to the natural
log of the absolute value of W minus 300. So let me write this down. So the left-hand
side right over here. And you might be able
do that in your head. You say, hey, look. You have W minus 300. Its derivative is 1, so
it's essentially there. So I can just take this
as if this was 1 over-- I could just take the
antiderivative here and treat this whole thing as a variable. So it would just be the natural
log of this entire thing, or the absolute value
of that entire thing. So the integral of
the left-hand side is the natural log of the
absolute value of W minus 300. And the derivative of the
right-hand side-- and this is a little bit more
straightforward-- or sorry, the antiderivative with respect
to t on the right-hand side is 1 over 25 t, or we
could say t over 25. And then these are both
indefinite integrals. You could have a constant
on both sides or one side. We could just put it on
one side right over here. You could have put
plus some constant. You could have added
some C1 over here. And then you could've
had C2 over here. But then you could've
subtracted C1 from both sides, and then just have a
C3 constant over here. But to simplify
it, you could just say, look, this is going
to be equal to this plus some constant. You could've put the constant
on that side as well. It doesn't matter. All the constants
would merge into one. Now we've gotten pretty far. We want to solve the
particular solution. Right now, we only
have this constant. And we also haven't written W as
a function of t just right yet. They're implicitly expressed. It is more of a relationship
than a function right now. So the first thing
we can do is try to solve for C. We're told
in the problem that W of 0 is equal to 1400 tons. They told us that. That's one of our
initial conditions. So we know that the natural
log, that when t is equal to 0, W is equal to 1400. So W is equal to 1,400 when
t is equal to 0 plus C. Or another way to
think about it is, the natural log of the absolute
value of 1,100 is equal to C. And obviously, this
is a positive value. So we could just say that C
is equal to the natural log of 1,100. We can drop the
absolute value sign because the absolute
value of 1,100 is 1,100. So now we can rewrite
this part right over here. Let me do it in yellow. Let me start--
that's not yellow. Let me start right over here. So we could write
the natural log of the absolute
value of W minus 300. And actually, we know that
W is an increasing function. W starts at 1,400 and
only goes up from there as t gets larger and larger. So if we're starting at
1,400 and only getting larger and larger values
for W, this expression right here is never going to get
negative for positive t values. And that's all we care about. And so we can actually drop
the absolute value sign. This will always be
a positive value. So we could say the absolute
value of W minus 300 is equal to t over
25-- and we've just solved for our constant--
plus the natural log of 1,100. And now we can just solve
this explicitly for W. We can raise e to both sides. This value is equal
to this value. The left hand side is equal
to the right hand side. So e raised to this power
is the same thing as e raised to that power. So we have e to the
natural log of W minus 300 is going to be equal
to e to the t over 25 plus the natural log of 1,100. Let me take this-- actually,
just to get the real estate, let me delete that. All right. Now I can keep working
right below that. And so e to the natural
log of something is just going to be
equal to that something. The natural log is what
power do I have to raise e to to get this thing. Well I'm raising
e to that power, so I'm going to get this thing. So the left-hand side
just becomes W minus 300, and the right-hand side,
we could rewrite this as e to the t over 25 times e to
the natural log of 1,100. We have the same base,
different exponents. If you wanted to
simplify this, you would just add the
exponents, which is exactly what we did up here. So these two things
are equivalent. e to the natural log of
1,100, that is 1,100. The natural log of
1,100 is a power that you raise e
to to get to 1,100. So if you raise e to that
power, you get 1,100. So that is 1,100. So we get W minus 300 is equal
to 1,100 e to the t over 25. And then you just add
300 to both sides. W is equal to 1,100 e to
the t over 25 plus 300. And we're done. We have solved the
differential equation. We found the
particular solution. We had to solve for C to find
that particular solution. We're done. And we did it using
just basic integration. We didn't have to use any
of the fancy differential equation solving
tools that you might learn in a more advanced class.