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### Course: 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

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# 2015 AP Calculus AB/BC 4cd

Solutions for the differential equation.

## Want to join the conversation?

- Sal, I'm not sure if you did the problem wrong or if this is an oversight by the college board (or, more likely, I'm not understanding this.)

if y = 2x-2, the initial condition of f(2)=3 is incorrect.(4 votes)- Part c, the f(x) is not necessary a linear function. Part d function must be a linear function, as it is in the form y=mx+b. So they could be different functions, and so the conditions in part c cannot apply to part d.(5 votes)

- To get the linear equation, I tried plugging in the initial conditions f(2) = 3 into the linear equation using m=2 to find the y intercept. 3 = 2(2) + b, so b = -1. Why doesn't that approach work? Thanks, ben(4 votes)
- Well, the general solution to this differential equation is

y = f(x) = 2x - 2 + Ce^(-x), where C ∈ ℝ.

(Don't worry if you don't know how to get this solution.)

With the initial condition f(2) = 3 the constant C is not 0. (It is actually e².)

Because C ≠ 0, we can't just ignore that exponential part.

However, if the initial condition was f(2) = 2, your method would work, because in this case C = 0.

This is a good question, but I don't know how to explain this without talking about the general solution.(1 vote)

- Why can't I understand most of this.....oh wait, I'm still an 8th grader(0 votes)
- lf you learnt about slopes, as most 7th graders do, you should be able to understand the majority of this(0 votes)

## Video transcript

- [Voiceover] Part c, let y
equals f of x be the particular solution to the differential
equation with the initial condition f of
two is equal to three. Does f have a relative
minimum, a relative maximum, or neither at x equals two? Justify your answer. Well, to think about
whether we have a realtive minimum or relative maximum, we can say, well, what's the derivative at that point? If it's zero, then it's a good candidate that we're dealing with a, that it could be a relative
minimum or maximum, if it's not zero, then it's neither. And then if it is zero,
if we wanna figure out relative minimum or relative maximum, we can evaluate the sign
of the second derivative. So let's just think about this. So, we want to evaluate f prime, we wanna figure out what f prime of, f prime of two is equal to. So, we know that f prime of x, f prime of x, which is
the same thing as dy dx, is equal to two times x minus y. We saw that in the last problem. And so f prime of two,
I'll write it this way, f prime of two is going to be equal to two times two, two times two minus whatever y is when x is equal to two. Well do we know what y
is when x equals to two? Sure, they tell us right over here. Y is equal to f of x, so when x is equal to two, when x is equal to two, y is equal to three, so two times two minus three. And so this is going to be
equal to four minus three is equal to one. And so since the derivative
at two is not zero, this is not going to be a
minimum, a relative minimum or a relative maximum, so you could say since, since f prime of two, f prime of two does not equal zero, this, we have a, f has,
let me write it this way f has neither, neither minimum, or relative minimum I guess I could say, relative min or relative max at x equals two. Alright, let's do the next one. Find the values of the constants m and b, for which y equals m
x plus b is a solution to the differential equation. Alright, this one is interesting. So let's, actually let's
just write down everything we know before we even think
that y equals m x plus b could be a solution to
the differential equation. So, we know that, we know that dy over dx is equal to two x minus y, they told us that. We also know that the second derivative, the second derivative of y,
with respect to x, is equal to two minus dy dx. We
figured that out in part b of this problem. And then, we could also express this we saw that we could also write that as two minus two x plus
y, if you just substitute, if you substitute this in for that. So it's two minus two x plus y. So let me write it that way. This is also equal to
two minus two x plus y. So that's everything that we know before we even thought that
maybe there's a solution of y equals m x plus b. So now let's start with
y equals m x plus b. So if y is equal to m x
plus b, y equals m x plus b, so this is the equation of a line, then dy dx is going to be equal to, well the derivative of this
with respect to x is just m, the derivative of this with
respect to x, this is constant this is not going to
change with respect to x, it's just zero. And that makes sense, the rate of change of y with
respect to x is the slope, is the slope of our line. So, can we use, and this
is really all that we know, we could keep, actually
we could go even further, we could take the second derivative here, the second derivative
of y with respect to x. Well, that's going to be zero. The second derivative of
a, of a linear function, well it's going to be
zero, you see that here. So this is, this is all of
the information that we have. We get this from the previous
parts of the problem, and we get this just taking the
first and second derivatives of, of y equals m x plus b. So given this, can we figure out, can we figure out what m and b are? Alright, so we could,
if we said m is equal to two x minus y, that doesn't seem right. This one is a tricky one. Well, let's see, we know
that the second derivative is going to be equal to zero. We know that this is
going to be equal to zero for this particular solution. And we know dy dx is equal to m. We know this is m. And so there you have it,
we have enough information to solve for m. We know that zero is equal to two minus m. So zero is equal to two minus m. And so we can add m to
both sides and we get m is equal to, m is equal to two. So, that by itself, was quite useful. And then, what we could say, let's see, can we solve this further? Well we know that this,
right over here, dy dx, this is m, this is m. And it's equal to two. So, we could say that two
is equal to 2 x minus y. Two is equal to 2 x minus y. And then let's see, if we solve for y, add y to both sides,
subtract two from both sides, we get y is equal to two x minus two. And there we have our whole solution. And so you have your m, right over there. That is m. And then we also have our b. This one was a tricky one. Anytime that you, you have to, ya know, do something like this and it
doesn't just jump out at you, and if it wasn't obvious, it
didn't jump out at me at first, when I looked at this problem,
I said, well let me just write down everything that they told us, so they wrote this before, and then we say okay this
is going to be a solution. And so let me see if I
can somehow solve, so, let's see what I didn't use. I didn't use, I didn't use that. I did use this. I absolutely used that. I did use that, I did use
that, and I did use that. So this was a little bit
of a fun little puzzle where I just wrote down all
the information they gave us and I tried to figure out, based on that, whether I could figure out m and m and b. And this is pretty neat,
that this a solution two x minus two. If we go to our slope
field above, it's not, it wouldn't have jumped out at me. But if you think about, if you think about so two x minus two, it's
y intercept would be negative two like that, let me
do this in a different color, and so the line would look something, would like something like this. The line would look something like this. And you can verify that
any one of these points, at any one of these points, the slope, the slope is equal, the
slope is equal to two. If we're at the point two comma two, well it's gonna be two
times two minus two is two. One comma zero, two times
one minus zero is two. Negative two comma, or sorry
zero comma negative two well, zero minus negative two, that's two. So you see this is
pretty neat, the slope is changing all around it, but this is this is a linear solution to that original differential
equation, that was, that was pretty cool.