General Differential Equation Advice:
"Do you have any tips on how to write differential equations?"
Hmmm, there are a couple things you can do to make it a little easier. Can you look at the wording in Problem #5 on pg. 323? There are a couple key words to look for when you’re writing a differential equation:
- Whenever it says “rate”, it’s telling you how to set up the dy/dt variable in the equation.
- This problem mostly talks about volume. So, when it says, “The rate at which [the water] flows…”, keep in mind that flowing water is measured by its volume. Here, if you’re talking about the rate of volume, then it’s the same thing as saying the derivative of volume. Hence, dV/dt.
- Other examples you could possibly see:
- “Rate at which temperature is changing” would be dT/dt
- “Rate at which a plant is growing” would be dh/dt
- “Spending rate of money” would be dM/dt
- Problems will usually follow the word “rate” with the word “is”. “Is” can usually indicate the place your equal sign should go in the equation.
- Anytime you see the words “proportional to”, you need to include a “times k” in your equation. Something being directly “proportional” to something else means that you can multiply one thing by a constant to get the other thing.
- Lastly, the sentence will list whatever object or item that is proportional to the given rate. This will usually just be a plain variable, but if you see any other mathematical words (like “difference”, “square root”, etc.), include them in the equation here.
Here are a couple examples. I’ll try to break them down:
Example #1
“The rate at which [the water] flows…”
dV/dt
“…is…”
=
“…proportional to…”
k*
“…the square root of the depth, y feet.”
sqrt(y)
Full equation: dV/dt = k*sqrt(y)
Example #2 (see pg. 345 #T1)
“The rate of change of pressure…”
dP/dt
“…is…”
=
“…directly proportional to…”
k*
“…the pressure of the remaining oxygen.”
P
Full equation: dP/dt = k*P
Example #3 (see #3 on our in-class Sec 7.3 worksheet)
“The rate of change in the temperature of an object…”
dT/dt
“…is directly proportional to…”
= k*
“…the difference between its temperature and the temperature of the surrounding medium.”
T - 72 (because the surrounding medium was the room temperature of 72 degrees)
Full equation: dT/dt = k*(T - 72)
Chapter 7 Review Worksheet:
Answers are posted in my Google Drive folder! Please let me know if you can't access them for some reason, and I can email them to you instead.
#R3a) When you separate the variables, you don't have to divide by 6. You can, but you don't have to. Start like this:
dy/dx = 6y^(1/2)
Then divide by y^(1/2) and multiply by dx:
dy/y^(1/2) = 6dx
Now roll the y^(1/2) up to the numerator and make the exponent negative.
y^(-1/2) dy = 6 dx
Now integrate:
(integral sign) y^(-1/2) dy = (integral sign) 6 dx
2y^(1/2) = 6x + c
Now solve for y. Start by dividing by 2 on both sides.
y^(1/2) = 3x + c
Now square both sides to get rid of the exponent.
y = (3x + c)^2 This is the general solution.
#R3b) Plug in x = 3 and y = 25, then solve for c.
25 = (3*3 + c)^2
25 = (9 + c)^2
5 = 9 + c
-4 = c
y = (3x - 4)^2 This is the particular solution.
y = (3x - 4)^2 This is the particular solution.
3a) dy/dt = k(y - 42)
3b) Separate the variables:
dy / (y - 42) = k dt
Integrate each side:
ln | y - 42 | = kt + c
Use "e" as a base on each side:
| y - 42 | = e^(kt + c)
Split the right side into two multiplied terms using exponent properties:
| y - 42 | = e^kt * e^c
Rewrite e^c as C, which allows you to drop the absolute value sign from the left side, because C can be either positive or negative:
y - 42 = Ce^kt
Add 42 to the other side to solve for y:
y = Ce^kt + 42
3c) Let's start from the general solution you found in 1b.
y = Ce^kt + 42
This problem tells you two important pieces of information. First, the soda is 70 degrees at t = 0. Second, the soda is 61 degrees at t = 1/2. Let’s start by plugging in the initial condition to solve for one of our unknown variables:
70 = Ce^k(0) + 42
70 = Ce^0 + 42
70 = C + 42
28 = C
Now plug C = 28 back into your general equation:
y = 28e^kt + 42
Next, plug in the other given data point in order to find k:
61 = 28e^k(1/2) + 42
19 = 28e^0.5k
19/28 = e^0.5k
ln (19/28) = 0.5k
2 ln (19/28) = k
-0.776 = k
So, your final particular solution equation is:
y = 28e^-0.776t + 42
5) There are a couple ways to do this, one of which is to plug in points to each equation, and see which one seems to match the trend. A way to use this strategy in a more effective way is to look for slopes that are all the same. For example, in this slope field, all the x = 1 points have a slope of 0. All the y = 0 points also have a slope of 0. So, look for the equation that would always give you 0 when plugging in x = 1 or when plugging in y = 0.
The other way would be to try to separate the variables and integrate every single answer choice, then see which of those graphs seems to fit the slope field. This way takes a long time, and you can’t always integrate these equations by hand!
6) In this problem, they’re giving you an equation with just x’s and y’s in it, and they’re asking you for dy/dx. In other words, they’re asking you to find the derivative! But, since this isn’t a regular “y =“ equation, we’ll have to use an implicit derivative.
sin (xy) = x
When you find the derivative of the left side, remember to use chain rule. The most outside function is sin, so we’ll find the derivative of that (while leaving the inside the same)…
cos (xy) = x
…then we need to multiply by the derivative of the inside function. To find the derivative of (xy), we’ll need to use product rule, because it’s two functions multiplied together. BIG THING TO REMEMBER: when you’re doing an implicit derivative, and you get to the derivative of a term with “y” in it, you MUST multiply by dy/dx.
Product rule for (xy)…derivative of x is 1, and derivative of y is dy/dx. Then use u’v + uv’.
So, here’s what it looks like when we multiply our problem by the derivative of (xy):
cos (xy) * (1*y + x * dy/dx) = x
We also need to find the derivative of the right side, which will be 1. Here’s our final derivative equation:
cos (xy) * (y + x * dy/dx) = 1
Unfortunately, this isn’t one of the answer choices, and it’s because we haven’t yet solved for dy/dx. Let’s start by dividing both sides by cos (xy):
cos (xy) * (y + x * dy/dx) = 1
y + x * dy/dx = 1 / cos (xy)
Next, we can subtract y from each side:
x * dy/dx = 1 / cos (xy) - y
Before we go too far though, it looks like all the answer choices are single fractions, not two things subtracted. So, let’s combine 1 / cos (xy) and - y by getting a common denominator:
x * dy/dx = 1 / cos (xy) - y * cos (xy) / cos (xy)
Then combine the fractions on the right side:
x * dy/dx = (1 - y * cos (xy) ) / cos (xy)
Lastly, we need to get rid of the “x” on the left side. Multiply both sides by 1 / x:
dy / dx = (1 - y * cos (xy) ) / cos (xy) * 1/x
dy / dx = (1 - y * cos (xy) ) / x cos (xy)
This equation is answer choice D.
10) dy / dt = ky
Separate the variables:
dy / y = k dt
Integrate each side:
ln | y | = kt + c
Use “e” as a base on each side to solve for “y”:
e^ln | y | = e^(kt + c)
| y | = e^(kt + c)
| y | = e^kt * e^c
y = Ce^kt
The only correct answer choice is “B”: 2e^kt. They just picked C = 2.
Sec 7.4 Slope Fields packet
17) Here's an explanation I found elsewhere on the internet:
The slope of the tangent line is simply given by dy/dx, which you are told is equal to:
dy/dx = x*y/2
At the point (1,1),
dy/dx = 1*1/2 = 1/2
The tangent line must also pass through the point (1,1), so using the point-slope form of the equation of a straight line, we obtain the following equation for the tangent at (1,1):
y - 1 = (1/2)*(x - 1)
y = x/2 + 1/2
Using this equation to estimate the value of y at x = 1.2 gives:
y(1.2) ~ = 1.2/2 + 0.5 = 1.1
---------------
The original differential equation is separable:
dy/dx = x*y/2
dy/y = x/2 dx
ln(y/c) = (x^2)/4
y = c*exp((x^2)/4))
where c is a constant of integration
Using the initial condition to solve for the constant c, we have:
1 = c*exp(1/4)
c = exp(-1/4)
So the particular solution is:
y(x) = exp(-1/4)*exp((x^2)/4)
y(x) = exp((x^2 - 1)/4)
The actual value of y at x = 1.2 is:
y(1.2) = exp((1.2^2 - 1)/4) = 1.116
You can do the comparison and determine if the linear extrapolation overestimated or underestimated the value at x = 1.2 (is 1.1 larger or smaller than 1.116?)
18a) Draw a tiny line that illustrates the slope at each point. For example, at the point (2, 2), when you plug in x = 2 and y = 2 to the given dy/dx equation, you get dy/dx = 2/2 = 1. So, draw a tiny line at (2, 2) that has a slope of 1. Do this at each point.
18b) Start by drawing a point at (0, 1). Then, slowly, SLOWLY draw a curve starting from that point. Every time you pass one of the “tiny slopes” you drew in the slope field, your new curve should be parallel to that tiny slope. It is NOT like connecting the dots as you draw the curve. You don’t need to connect the tiny slope lines. Just be sure to draw your curve parallel to them.
18c) To solve for y, we need to separate the variables, then integrate.
dy / dx = x / y
y * dy = x * dx
Then integrate each side. Make the new exponents “one bigger” and divide by the new exponent:
1/2y^2 = 1/2x^2 + c
y^2 = x^2 + c
y = +-sqrt(x^2 + c)
To find the “c” value, you could plug in the given initial value of (0, 1) at any point in your work. Here’s what it looks like to plug it in to the last step:
1 = +-sqrt(0^2 + c)
1 = +-sqrt(c)
1 = c
y = sqrt(x^2 + 1)
We only need the “positive” square root in the answer because we can see the solution curve is in the positive y-values on the graph.
18d) Follow the same steps as Part B in this problem
18e) Follow the same steps as Part C in this problem. The difference is when you’re solving for “c” and you plug in (0, -1):
-1 = +-sqrt(0^2 + c)
-1 = +-sqrt(c)
1 = c
y = -sqrt(x^2 + 1)
We only need the “negative square root in the answer because we can see the solution curve is in the negative y-values on the graph.
Sec 7.3 (pg. 320):
#1c) Let's look at each step as we integrate:
(integral sign) dm / (100-km) = (integral sign) dt
Here, you may want to use u-substitution…
u = 100 - km
du = -k dm
-1/k du = dm
So, here it is with u-substitution:
-1/k * (integral sign) 1 / u * du = (integral sign) dt
Then integrate:
-1/k * ln | u | = t + c
Then substitute back in for u:
-1/k * ln | 100 - km | = t + c
Now solve for m. Start by multiplying both sides by -k. The "+c" can still be "+c", because it's just a constant multiplied by another constant k.
ln | 100 - km | = -kt + c
| 100 - km | = e^(-kt + c)
| 100 - km | = e^-kt * e^c
100 - km = Ce^-kt
-km = Ce^-kt - 100
Now divide both sides by -k. The first term on the right (with C) will be a constant divided by a constant k, so we can just leave it as C.
m = Ce^-kt + 100/k
Lastly, it gave you the initial value M = 0 when t = 0. Plug this in to the equation:
0 = Ce^-k(0) + 100/k
0 = Ce^0 + 100/k
0 = C + 100/k
-100/k = C
Now substitute this value for C back into the equation you had found.
m = -100/k * e^-kt + 100/k
You could leave your answer like this, or factor out 100/k like the textbook did:
m = 100/k (-e^-kt + 1)
#7a) Your starting equation is dy/dx = ky^n
#7a) Your starting equation is dy/dx = ky^n
It wants you to plug in n = 1 first:
dy/dx = ky
Then, to solve for y, you'll need to separate the variables and integrate:
Multiply by dx, and divide by y:
dy/y = k*dx
Now integrate:
(integral sign) dy/y = (integral sign) k*dx
ln | y | = kx + c
It also says to make k = 1 and let the constant of integration C = -3. This is referring to the +c you just got from integrating (hence, "constant of integration).
ln | y | = k - 3
| y | = e^(x - 3)
| y | = e^(x - 3)
| y | = e^x * e^-3
| y | = 0.04978e^x
y = +-0.04978e^x [This is supposed to be a + or - in front of the coefficient.]
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