10.5 Differential Equationsap Calculus

Unit 7 - Differential Equations 7.1 Modeling Situations with Differential Equations 7.2 Verifying Solutions for Differential Equations 7.3 Sketching Slope Fields 7.4 Reasoning Using Slope Fields 7.5 Euler's Method (BC topic) 7.6 General Solutions Using Separation of Variables 7.7 Particular Solutions using Initial Conditions. Maths Book back answers and solution for Exercise questions - Mathematics: Differential Calculus- Differentiability and Methods of Differentiation. Now put these into the differential equation to see if the equality holds.1116. Y” that is going to be 4e ⁺x + 6, that is the y” part, and then, + 6x.1129. The question is, does it equal the first derivative which is 4e ⁺x + 6x + 6.1145. Yes, this is a solution of that differential equation.1157. Tomorrow's answer's today! Find correct step-by-step solutions for ALL your homework for FREE!

Solve the separable, first-order differential equation for: First collect all the terms with the derivative to one side of the equation. Important Conceptual Note: often in texts on differential equations differentials often appear to have been rearranged algebraically as if is a 'fraction,' making it appear as if we 'multiplied both sides'.

A differential equation is a relation that involves an unknown function and its derivative. There are many kinds of differential equations and tons of specialized techniques we may use to solve them. Fortunately, on the AP Calculus exams you will only encounter a handful of the most basic kinds.

Basic Differential Equations: Integration

You have probably worked out hundreds of differential equations without even realizing it!

It’s true! Let me give you an example.

What makes this a differential equation?

Well think about what the notation means. You know that integration is the opposite of differentiation. So what we’re looking for here is a function f(x) whose derivative is equal to 2x. In other words, we have to solve:

f ‘(x) = 2x

That’s a relation involving an unknown function f and its derivative.

But this differential equation is trivial to solve! Just use the power rule for integrals (or guess-and-check).

f(x) = x2 + C

(For a full review of integration, check out: AP Calculus Exam Review: Antidifferentiation or AP Calculus Exam Review: Integrals.)

Initial Value Problems

Now suppose you have more information about an unknown function, such as its value at a certain point. Then you may be able to solve for the function explicitly, rather than getting stuck with an unknown constant of integration at the end.

An initial value problem typically gives a derivative expression along with a function value. The goal is to produce the original function.

Example 1: Initial Value Problem

If f ‘(x) = 1/x2 for x > 0, and f(1) = 5, find the expression for f(x) for x > 0.

Solution

Because the derivative expression is given, we integrate to find the original function. Don’t forget about your constant of integration!

Next, plug in the known info and solve for the unknown constant of integration.

This implies that the original function must be: f(x) = -1/x + 6.

Separation of Variables

Of course not every differential equation (or DE) problem can be handled by simple integration. Often more advanced techniques must be used, especially if both x and y show up in the equation.

For example, the following DE cannot be cracked by integration (…not in its current form, that is).

The key is to somehow break up the dy and dx.

The derivative notation dy/dx looks like a fraction. However, it’s not really a fraction, because the two quantities dy and dx (called differentials) are supposed to represent the idea of taking Δx to zero in the limit. In other words, if you had to identify the numerical “value” of a differential, it would make sense to say dx = dy = 0. And we all know that 0/0 is undefined.

But those differential gadgets are more subtle than that. They really don’t carry a numerical value. Instead, they stand for a limiting process. And if we’re very careful, we can work with dx and dy individually.

The method of Separation of Variables works by manipulating dy/dx as if it were a fraction, and then using integration to get rid of the differentials.

Now let’s take another look at our example above.

Example 2: Separation of Variables

If , and y(0) = -3, find an equation for y in terms of x.

Solution

First multiply dx to both sides. Then multiply or divide as necessary so that only expressions of y are on the left, and only those with x are on the right.

Next, we use the magic of calculus! All you have to do is to apply the integral symbol to each side of the equation. Now you have two separate integrals to work out.

Now even though both sides generated a constant of integration, those can be combined into a single constant on the right hand side. The next few steps are just for isolating y.

Notice that the quantity eC is replaced by another constant k. This is because C is still arbitrary (and unknown). So just think of k as a related, but still arbitrary constant.

Finally we solve for k using the known information, y(0) = -3.

(By the way, we will drop the “±” notation at this point because k itself can take care of that choice.)

Therefore, replacing k by its computed value -3, we obtain the final form of the function.

Setting Up Differential Equations

Fortunately you won’t encounter any DE problems on the AP Calculus exam that can’t be handled by either integration or separation of variables.

However, you may be asked to set up and/or analyze a DE (without solving it).

Example 3: Setting up a DE

A certain bacteria culture has P cells at time t and grows in proportion to the square root of the amount present. Set up a differential equation that models A(t).

Solution

Because the culture grows at a certain rate with respect to time, we know we’ll be working with the derivative dA/dt.

Let k be the constant of proportionality. Then translate the given information into mathematics:

dA/dt = k × √(A)

Remember, we don’t have to solve for A(t); just set up the DE that could be used to solve for it. So we’re done at this step!

Example 4: Analyzing a DE

The number y of people infected with the flu in a certain town at time t can be modeled by the differential equation, dy/dt = 0.00002y(21500-y), where y(0) = 1. Determine the limit of y as t → ∞.

Solution

It would be difficult to solve this DE for y explicitly. Instead, look closely at the equation. It has the form of a logistics equation.

That means we can answer this question with no work at all! You just have to remember that logistics equations always have solutions that tend toward the carrying capacityM as t → ∞. Here, M = 21500. That’s all you need to answer the question!

The limit will be 21500.

(For more details about the logistics equation, I recommend: AP Calculus BC Review: Logistics Growth Model.)

Conclusion

Differential equations show up only sparingly on an AP Calculus exam. But it’s important to be aware of the techniques for solving them, setting them up, and analyzing them.

  • Remember the basic methods of integration and separation of variables.
  • Know how to set up a differential equation.
  • Look for situations in which you may avoid solving the DE.

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About Shaun Ault

Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!

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Master AP Calculus AB & BC

Part II. AP CALCULUS AB & BC REVIEW

CHAPTER 10. Differential Equations

EXPONENTIAL GROWTH AND DECAY

You have probably alluded to exponential growth in everyday conversation without even realizing it. Perhaps you’ve said things like, “Ever since I started carrying raw meat in my pockets, the number of times I’ve been attacked by wild dogs has increased exponentially.' Exponential growth is sudden, quick, and relentless. Mathematically, exponential growth or decay has one defining characteristic (and this is key)', the rate of y’s growth is directly proportional toy itself. In other words, the bigger y is, the faster it grows; the smaller y is, the slower it decays.

Mathematically, something exhibiting exponential growth or decay satisfies the differential equation

where k is called the constant of proportionality. A model ship might be built to a 1:35 scale, which means that any real ship part is 35 times as large as the model. The constant of proportionality in that case is 35. However, k in exponential growth and decay is never so neat and tidy, and it is rarely (if ever) evident from reading a problem. Luckily, it is quite easy to find.

In the first problem set of this chapter (problem 3), you proved that the general solution to is I find the formula easier to remember, however, if you call the constant N instead of C (although that doesn’t amount to a hill of beans mathematically). Why is it easier to remember? It sounds like Roseanne pronouncing “naked”—“nekkit.”

In this formula, N stands for the original amount of material, k is the proportionality constant, t is time, and y is the amount of N that remains after time t has passed. When approaching exponential growth and decay problems, your first goals should be to find N and k; then, answer whatever question is being posed. Don’t be intimidated by these problems—they are very easy.

Example 3: The new theme restaurant in town (Rowdy Rita’s Eat and Hurl) is being tested by the health department for cleanliness. Health inspectors find the men’s room floor to be a fertile ground for growing bacteria. They have determined that the rate of bacterial growth is proportional to the number of colonies. So, they plant 10 colonies and come back in 15 minutes; when they return, the number of colonies has risen to 35. How many colonies will there be one full hour after they planted the original 10?

Solution: The key phrase in the problem is “the rate of bacterial growth is proportional to the number of colonies,” because that means that you can apply exponential growth and decay. They started with 10 colonies, so N = 10 (starting amount). Do not try to figure out what k is in your head—it defies simple calculation. Instead, we know that there will be 35 colonies after t = 15 minutes, so you can set up the equation

Solve this equation for k. Divide by 10 to begin the process.

Now you have a formula to determine the amount of bacteria for any time t minutes after the original planting:

We want the amount of bacteria growth after 1 hour; since we calculated k using minutes, we’ll have to express 1 hour as t = 60 minutes. Now, find the number of colonies.

So, almost 1,501 colonies are partying along the surface of the bathroom floor. In one day, the number will grow to 1.7 X 1053 colonies. You may be safer going to the bathroom in the alley behind the restaurant.

NOTE. All half-life problems automatically satisfy the property by their very nature.

Example 4: The Easter Bunny has begun to express his more malevolent side. This year, instead of hiding real eggs, he’s hiding eggs made of a radioactive substance Nb-95, which has a half-life of 35 days. If the dangerous eggs have a mass of 2 kilograms, and you don’t find the one hiding under your bed, how long will it take that egg to decay to a “harmless” 50 grams?

Solution: The egg starts at a mass of 2,000 g. A half-life of 35 days means that in 35 days, exactly half of the mass will remain. After 70 days, one fourth of the mass will remain, etc. Therefore, after 35 days, the mass will be 1,000. This information will allow us to find k.

TIP. In an exponential decay problem such as this, the k will be negative.

Now that we know N and k, we want to find t when only 50 grams are left. In this case, t will be in days (since days was the unit of time we used when determining k).

You should be safe by Thanksgiving. (Nothing wrong with a little premature hair loss and a healthy greenish glow, is there?)

EXERCISE 4

Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.

YOU MAY USE A GRAPHING CALCULATOR FOR ALL OF THESE PROBLEMS.

1. If Pu-230 (a particularly stinky radioactive substance) has a half-life of 24,360 years, find an equation that represents the amount of Pu-230 left after time t, if you began with N grams.

2. Most men in the world (except, of course, for me, if my wife is reading this) think that Julia Roberts is pretty attractive. If left unchecked (and the practice were legal), we can assume the number of her husbands would increase exponentially. As of right now, she has one husband, but if legal restrictions were lifted she might have 4 husbands 2 years from now. How many years would it take her to marry 100 men if the number of husbands is proportional to the rate of increase?

3. Assume that the world population’s interest in the new boy band, “Hunks o’ Love,” is growing at a rate proportional to the number of its fans. If the Hunks had 2,000 fans one year after they released their first album and 50,000 fans five years after their first album, how many fans did they have the moment the first album was released?

4. Vinny the Talking Dog was an impressive animal for many reasons during his short-lived career. First of all, he was a talking dog, for goodness sakes! However, one of the unfortunate side-effects of this gift was that he increased his size by 1/3 every two weeks. If he weighed 5 pounds at birth, how many days did it take him to reach an enormous 600 pounds (at which point his poor, pitiable, poochie heart puttered out)?

ANSWERS AND EXPLANATIONS

1. Because the rate of decrease is proportional to the amount of substance (the amount decreases by half), we can use exponential growth and decay. In other words, let’s get Nekt. In 24,360 years, N will decrease by half, so we can write

Divide both sides by N, and you get

Therefore, the equation will give you the amount of Pu-230 left after time t if you began with N grams.

2. This problem clearly states the proportionality relationship required to use exponential growth and decay. Here, N = 1, and y = 4 when t = 2 years, so you can set up the equation:

Now that we have k, we need to find t when y = 100.

3. Our job in this problem will be to find N, the original number of fans. We have the following equations based on the given information:

Solve the first equation for N, and you get

Plug this value into the other equation, and you can find k.

10.5

Finally, we can find the value of N by plugging k into

10.5 Differential Equationsap Calculus Equation

NOTE. It should be no surprise that the left-hand side of the equation is 4/3 in the second step, as Vinny’s weight is 4/3 of his original weight every 14 days. In the half-life problems, you may have noticed that this number always turns out to be ½.

4. Oh, cruel fate. If Vinny weighed 5 pounds at birth, he weighed or 6.667 pounds 14 days later. Notice that we will use days rather than weeks as our unit of time, since the final question in the problem asks for days.

We want to find t when y = 600.

The poor guy lived almost 8 months. The real tragedy is that even though he could talk, all he wanted to talk about were his misgivings concerning contemporary U.S. foreign policy. His handlers were relieved at his passing. “It was like having to talk to a furry John Kerry all the time,” they explained.


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