Differential equations are equations that include both a function and its derivative (or higher-order derivatives). For example, y=y' is a differential equation. Learn how to find and represent solutions of basic differential equations. Differential Equations 2019 AB4/BC4 Rain barrel: A cylindrical barrel collects rainwater, with questions relating the rates of the water height and volume, and a separable differential equation to solve explicitly for the height as a function of time t.
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B. $y = frac{7}{2}x^2 + C$
D. $frac{1}{2}y^2$ $= frac{7}{2}x^2 + C$
SOLUTION
This is a separable differential equation. Multiplying both sides by $dx$:
$dy = 7x dx$
To solve for $y$, we must take the integral of both sides (don’t forget the constant of integration +C):
$displaystyleint dy = displaystyleint (7x)dx$
$y = dfrac{7}{2} x^2 + C$
$dfrac{dy}{dx}$ $= dfrac{2x}{4 + x^2}$
B. $y = frac{1}{2}arctan(2x) + C$
D. $y = 2x ln |x^2 + 4| + C$
SOLUTION
This is a separable differential equation. Moving the dx to the right side gives:
$dy = dfrac{2x}{(4 + x^2)} dx$
Taking the integral of both sides:
$displaystyleint dy = displaystyleint dfrac{2x}{(x^2 + 4)} dx$
Note that the right-hand integral is in the form $frac{u'}{u'}$, so it can be expressed as $ln |u| + C$.
$y = ln |x^2 + 4| + C$
$dfrac{dy}{dx} cdot dfrac{1}{x}$ $= 7x^2 + 5$
B. $y = frac{7}{4}x^4 + C$
D. $y = frac{7}{4}x^4 + frac{5}{2}x^2 + C$
SOLUTION
Start by multiplying both sides of the equation by $x$. This gives:
$dfrac{dy}{dx} = 7x^3 + 5x$
Separating the differential equation gives:
$dy = (7x^3 + 5x)dx$
Integrating both sides to solve for $y$:
$displaystyleint dy = displaystyleint (7x^3 + 5x)dx$
$y = dfrac{7}{4} x^4 + dfrac{5}{2} x^2 + C$
B. $y^3 = Cx^7$
D. $y^7 = Cx^{frac{1}{3}}$
SOLUTION
Start by moving all the $y$ terms to the left side and all the $x$ terms to the right side:
$dfrac{dy}{3y} = dfrac{dx}{7x}$
Integrating both sides to solve for $y$:
$displaystyleint dfrac{1}{3y} dy =displaystyleint dfrac{1}{7x} dx$
$dfrac{1}{3} ln|y| = dfrac{1}{7} ln|x| + C$
Using the exponent log rule, we can rewrite the equation above as:
$ln |y^{frac{1}{3}}|$ $= ln|x^{frac{1}{7}}| + C$
Exponentiating both sides by $e$ to get rid of the logarithms:
$y^{frac{1}{3}}$ $= e^Ccdot x^{frac{1}{7}}$
Since $e^C$ is still considered an unknown constant, the equation above is equivalent to:
$y^{frac{1}{3}} = Cx^{frac{1}{7}}$
B. $x^2 + y^2 = C$
D. $x^2 + ln|y| = C$
SOLUTION
Start by moving the $y$ terms to the left side and the $x$ terms to the right side:
$ydy = −xdx$
Integrating both sides:
$displaystyleint ydy = displaystyleint −xdx$
$dfrac{1}{2} y^2 = −dfrac{1}{2} x^2 + C$
$dfrac{1}{2} y^2 + dfrac{1}{2} x^2 = C$
We can multiply through by $2$, since $C$ is an arbitrary constant.
$y^2 + x^2 = C$
Fun Fact: This is the equation of a circle!
B. $y = −dfrac{1}{frac{1}{2}x + C}$
D. $y = −dfrac{1}{frac{1}{2}x^2} + C$
SOLUTION
Begin by separating the $y$ and $x$ variables to different sides of the equation:
$dfrac{1}{y^2}dy = xdx$
Integrating both sides:
$displaystyleint dfrac{1}{y^2}dy = displaystyleint xdx$
$ −dfrac{1}{y} = dfrac{1}{2}x^2 + C$
Rearranging and solving for $y$:
$y = −dfrac{1}{frac{1}{2} x^2 + C}$
The $C$ must be divided in the fraction as well while rearranging.
B. $ln|y| = frac{1}{3}x^3 + C$
D. $ln|y| = frac{1}{5}x^5 + C$
SOLUTION
Start by separating the variables:
$dfrac{1}{y} dy = x^2 dx$
Integrate both sides to solve for $y$:
$ln|y| = dfrac{1}{3} x^3 + C$
We could exponentiate both sides, but the given answer choices are not simplified.
$dfrac{dy}{dx} = dfrac{y}{1 + x^2},$ $text{ and }$ $y(0) = 1$
B. $y = 2ln |1 + x^2|$
D. $y = e^{arctan(x)}$
SOLUTION
Begin by separating the variables.
$dfrac{1}{y}dy = dfrac{1}{1 + x^2}dx$
Integrating both sides gives the equation below.
$ln(y) = arctan(x) + C$
Exponentiating both sides gives the equation below.
$y = Ce^{arctan(x)}$
Using the initial condition given allows us to solve for the constant of integration.
$1 = C cdot e^{arctan(0)}$
$C = 1$
Substitute this value back into the function.
$y = e^{arctan(x)}$
The process of solving for the constant of integration using an initial condition is very simple but should only be done after the differential equation is solved.
$dfrac{dy}{dx} = 7ytan(x)$, $y(0) = 128$
B. $y = 64 cdot sec^7(x)$
D. $y = 128 cdot cos^7(x)$
SOLUTION
Begin by separating the variables:
$dfrac{1}{7y} dy = tan(x)dx$
Integrate both sides to solve for $y$.
$dfrac{1}{7} ln|y| = −ln |cos(x)| + C$
Exponentiate both sides by power $e$.
$y^{frac{1}{7}} = C cdot sec(x)$
Plugging in the initial value for $y(0)$:
$128^{frac{1}{7}} = C cdot sec(0)$
$C = 128^{frac{1}{7}}$
Using the new value for $C$:
$y^{frac{1}{7}}128^{frac{1}{7}} cdot sec(x)$
$y = 128 cdot sec^7(x)$
$dfrac{dy}{dx} = e^{2x}$, $y(0) = 0$
B. $y = frac{1}{2}e^{2x} − frac{1}{2}$
D. $y = frac{1}{2}e^x − frac{1}{2}$
SOLUTION
Begin by separating variables:
$dy = e^{2x} dx$
Integrating both sides gives:
$y = dfrac{1}{2} e^{2x} + C$
Using the initial value given:
$0 = dfrac{1}{2} e^0 + C$
$C = −dfrac{1}{2}$
Substituting this value of $C$ back into the solution:
$y = dfrac{1}{2} e^{2x} −dfrac{1}{2}$
$y' − y' − 30y = 0$ $text{Identify}$ $text{if}$ $y = e^{−5x}$ $text{is}$ $text{a}$ $text{solution}$ $text{to}$ $text{the}$ $text{differential}$ $text{equation.}$
B. $text{No}$
SOLUTION
The first and second derivatives of $y$ are $−5e^{−5x}$ and $25e^{−5x}$ respectively. Plugging in the values for $y$, $y'$ and $y'$ into the given differential equation gives:
$(25e^{−5x}− ( −5e^{−5x} ) − 30(e^{−5x} )$ $= 0$
$0 = 0$
Since the left side is equal to $0$, the equation is true. This means $y=e^{−5x}$ is a valid solution to the differential equation.
$y'' + 12y' + 47y' + 60y = 0$ $text{Identify}$ $text{if}$ $y = e^{−6x}$ $text{is}$ $text{a}$ $text{solution}$ $text{to}$ $text{the}$ $text{differential}$ $text{equation.}$
B. $text{No}$
SOLUTION
Taking the first three derivatives of $y$:
$y = e^{−6x}$, $, y' = −6e^{−6x}$, $, y' = 36e^{−6x}$, $, y'' = −216e^{−6x}$
Plugging in the values above into the differential equation:
$(−216e^{−6x} ) + 12(36e^{−6x} )$ $+ ~47(−6e^{−6x} ) + 60(e^{−6x})$
$= e^{−6x} (−216 + 12 cdot 36 + 47 cdot −6 + 60)$ $= −6e^{−6x} ≠ 0$
Since the result is not zero, $e^{−6x}$ is not a valid solution for the differential equation.
Question 13: $y' − 4y' + 4y = 0$ $text{Identify}$ $text{if}$ $y = xe^{2x}$ $text{is}$ $text{a}$ $text{solution}$ $text{to}$ $text{the}$ $text{differential}$ $text{equation.}$
B. $text{No}$
SOLUTION
Taking the first two derivatives of $y$ using the product rule:
$y = xe^{2x}$, $, y' = e^{2x} + 2xe^{2x}$, $, y' = 4e^{2x} + 4xe^{2x}$
Plugging these values back into the differential equation:
$4e^{2x} + 4xe^{2x} − 4(e^{2x} + 2xe^{2x} ) + 4(xe^{2x} )$
$= 4e^{2x} + 4xe^{2x} − 4e^{2x} − 8xe^{2x} + 4xe^{2x}$ $= 0$
Since the result is $0$, $y = xe^{2x}$ is a valid solution for the differential equation.
$x^2y' + xy' = 0$ $text{Confirm}$ $text{that}$ $y = ln(x)$ $text{is}$ $text{a}$ $text{solution}$ $text{to}$ $text{the}$ $text{differential}$ $text{equation.}$
B. $text{No}$
SOLUTION
The derivatives of the solution are as follows.
$y = ln(x)$
$y' = dfrac{1}{x}$
$y'= −dfrac{1}{x^2}$
Substituting these values back into the differential equation gives the following.
$x^2 cdot −dfrac{1}{x^2} + x cdot dfrac{1}{x} = 0$
$−1 + 1 = 0$
$0 = 0$
This confirms that the given solution is correct.
$y^{(4)} − y = 0$ $text{Identify}$ $text{if}$ $y = cos(x)$ $text{is}$ $text{a}$ $text{solution}$ $text{to}$ $text{the}$ $text{differential}$ $text{equation.}$
B. $text{No}$
SOLUTION
Taking the first $4$ derivatives of $y$:
$y = cos(x)$
$y' = −sin(x)$
$y' = −cos(x)$
$y''= sin(x)$
$y^{(4)} = cos(x)$
Substituting into the differential equation:
$cos(x) − cos(x) = 0$Since the result is $0$, we see that $y = cos(x)$ is a valid solution for the differential equation.
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
10.5 Differential Equations Ap Calculus Multiple Choice
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
10.5 Differential Equations Ap Calculus Practice
Plug this value into the other equation, and you can find k.
Finally, we can find the value of N by plugging k into
10.5 Differential Equationsap Calculus Transcendentals
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.
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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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