2015 AP Calculus AB/BC 4cd | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

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 2? Justify your answer.

Well, to think about whether we have a relative minimum or a relative maximum, we could see what the derivative at that point is. If it's 0, then it's a good candidate that we're dealing with a relative minimum or maximum. If it's not zero, then it's neither. Then, if it is zero, if we want to figure out relative minimum or relative maximum, we can evaluate the sign of the second derivative.

So let’s just think about this. We want to evaluate f prime. We want to figure out what f prime of 2 is equal to. We know that f prime of x, which is the same thing as dy/dx, is equal to 2 times x minus y. We saw that in the last problem.

So, f prime of 2 can be written this way: f prime of 2 is going to be equal to 2 times 2, which is two times two minus whatever y is when x is equal to two. Well, do we know what y is when x equals two? Sure, they tell us right over here y is equal to f of x. So, when x is equal to 2, y is equal to 3.

Thus, 2 times 2 minus 3 gives us 4 minus 3, which is equal to 1. Since the derivative at 2 is not 0, this is not going to be a minimum, a relative minimum, or a relative maximum.

So, we could say, since f prime of 2 does not equal 0, we have that f has neither a relative minimum nor relative maximum at x equals 2.

Alright, let’s do the next one: Find the values of the constants m and b for which y equals mx plus b is a solution to the differential equation.

Alright, this one is interesting. So let’s actually write down everything we know before we even think that y equals mx plus b could be a solution to the differential equation. We know that dy/dx is equal to 2x minus y. They told us that. We also know that 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. Then we could also express this; we saw that we could also write that as 2 minus 2x plus y if you just substitute this in for that.

So, this is also equal to 2 minus 2x plus y. That’s everything we know before we even thought that maybe there’s a solution of y equals mx plus b.

So now let’s start with y equals mx plus b. If y is equal to mx plus b, this is the equation of a line. Then, dy/dx is going to be equal to the derivative of this with respect to x, which is just m. The derivative of this with respect to x is constant; it’s not going to change with respect to x, it’s just a 0.

And that makes sense; the rate of change of y with respect to x is the slope of our line.

So can we use this? And this is really all that we know. We could go even further; we could take the second derivative here. The second derivative of y with respect to x is going to be zero. The second derivative of a linear function is going to be zero, and we see that here.

So, this is all of the information that we have; we get this from the previous parts of the problem and we get this just by taking the first and second derivatives.

Given this, can we figure out what m and b are?

Alright, so we could say m is equal to 2x minus y. That doesn’t seem right. This one is a tricky one. 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.

So there you have it; we have enough information to solve for m. We know that 0 is equal to 2 minus m. Therefore, 0 is equal to 2 minus m. We can add m to both sides and we get m is equal to 2.

So that by itself was quite useful. Then we could say, let’s see; can we solve this further? Well, we know that this right over here dy/dx, is m and it’s equal to 2.

So we could say that 2 is equal to 2x minus y. Then, let’s see if we solve for y: add y to both sides and subtract 2 from both sides, we get y is equal to 2x minus 2.

And there we have our whole solution. 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 have to do something like this and it doesn’t just jump out at you—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. They wrote this before and then we say, okay, this is going to be a solution. Then let me see if I can somehow solve.

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, 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 b. This is pretty neat, this is a solution, 2x minus 2.

If we go to our slope field above, it wouldn't have jumped out at me, but if you think about it, 2x minus 2, its y-intercept would be negative 2.

So, the line would look something like this. The line would look something like this, and you can verify that at any one of these points, the slope is equal to two.

If we're at the point (2, 2), well it’s going to be 2 times 2 minus 2 is 2. For (1, 0), 2 times 1 minus 0 is 2. For (0, -2), 0 minus (-2) equals 2.

So you see, this is pretty neat. The slope is changing all around it, but this is a linear solution to that original differential equation. That was pretty cool.