Instead of putting the equation in exponential form, I differentiated each side of the equation: (1/y) dy = 3 dx ln y = 3x + C Therefore C = ln y - 3x So, plugging in the given values of x = 1 and y = 2, I get that C = ln(2) - 3. If you put this in a calculator, it's a very different value (about -2.307) than what Sal got by raising both sides to the power of e: 2 * e^-3 evaluates to about .0996. Why did my method not work?

It's important to note that the C you work with is not the same as the C Sal works with. At 2:06, Sal gets e^C, which he says is just going to be another arbitrary constant. Thus Sal replaces e^c with C. If you perform e^(-2.307) you should get 0.0996.

Sal changed e^C (which any arbitrary constant) to C (which is any another constant) and at least according to my knowledge C can't be any another constant as it will always be positive coz e^C can never be negative. We actually limited the value of our arbitrary constant to positive numbers, then shouldn't we mention that C is any constant number greater than zero i.e(C>0) ?

"Any arbitrary constant" includes complex numbers. Recall that e^(x*j) = cos(x) + j*sin(x). C could be, for example, 5 + π*j, in which case e^C ≈ -7.4. However, even with complex exponents, e^C can _never_ be zero! In this case that does not matter, because we have already limited ourselves to y ≠ 0 by using 1/y as part of the solution. But it is certainly something to keep in mind for other situations.

When dealing with separable equations why in some videos do you put y and the constant on one side and in others (like this one) you only move y over?

It doesn't matter, the result will be the same. It just changes how you get to that result.

Wouldn't it be possible to solve for C right after you integrated both sides of the separated equation? ln |y| = 3x + C , when x = 1, y = 2. Plug in these values and you'd get ln (2) = 3 + C, so C = ln (2) - 3 Then you'd get ln |y| = 3x +ln (2) - 3 So that y = e^(3x+ln (2) -3), because e^ln(2) = 2 you'd get 2e^(3x-3), and this is correct, but the method is different from the one Sal uses, can someone explain to me, why this works, or if there are any steps that aren't correct, so that you'd have to use Sal's method?

What exactly does Sal do (and yes what you did is correct)

If for whatever reason I had chosen to multiply both sides of the original integral in orange by -1, I would get very different solutions. I have always had a hard time understanding this. can someone help?

−∫(1 ∕ 𝑦)𝑑𝑦 = −∫3𝑑𝑥 ⇔ ⇔ −ln|𝑦| = −3𝑥 + 𝐶, where we can just as well write the right-hand side as −3𝑥 − 𝐶 Multiplying both sides by (−1), we get; ln|𝑦| = 3𝑥 + 𝐶, which is the exact same equation that Sal solved for in the video.

when I use y=-ce^3x then also I got the same ans y=e^(3x-3) as Sal sir got using y=ce^3x that means no effect of absolute value here?

I think you may have miscalculated somewhere. 2 = -ce^3 gets a different result than 2 = ce^3

I solved the integral by using U-substitution (int(1/3y)dy=int(1)dx), but after evaluating, I got ln|3y|=3x+C instead of ln|y|=3x+C. What did I do wrong? Here are my steps: int(1/3y)dy=int(1)dx int(1/3y)dy=x+C (d/dy)(3y)=3 (1/3)int(3*1/3y)dy=x+C u=3y (1/3)int(1/u)du=x+C (1/3)ln|u|+C=x+C (1/3)ln|u|=x+C ln|u|=3x+C ln|3y|=3x+C

There's nothing wrong. ln|3𝑦| = 3𝑥 + 𝐶 ⇒ 𝑦 = (1∕3)𝑒^(3𝑥 + 𝐶) ⇒ 𝑑𝑦∕𝑑𝑥 = 𝑒^(3𝑥 + 𝐶) = 3𝑦, which is the equation we started with. Similarly, ln|𝑦| = 3𝑥 + 𝐶 ⇒ 𝑦 = 𝑒^(3𝑥 + 𝐶) ⇒ 𝑑𝑦∕𝑑𝑥 = 3𝑒^(3𝑥 + 𝐶) = 3𝑦

at 2:38, does it mean that y=+/- Ce^3x could lead to the same solution of this differential equation? by substituting y=2 and x=1 into y=-Ce^3x, you get C=-2e^-3. sustituting this C back into y=-Ce^3x, you'll get the same y=2e^(3x-3)

So you can just write Ce^3x. It is understood C can take any value.

Main content

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 7

Lesson 7: Exponential models with differential equations

Worked example: exponential solution to differential equation

Name: Worked example: exponential solution to differential equation
Uploaded: 2017-01-12T12:16:36Z
Description: The solution of the general differential equation dy/dx=ky (for some k) is C⋅eᵏˣ (for some C). See how this is derived and used for finding a particular solution to a differential equation.

Google Classroom

The solution of the general differential equation dy/dx=ky (for some k) is C⋅eᵏˣ (for some C). See how this is derived and used for finding a particular solution to a differential equation.

Want to join the conversation?

Sort by:

G Y
Posted 6 years ago. Direct link to G Y's post “Instead of putting the eq...”
Instead of putting the equation in exponential form, I differentiated each side of the equation:

(1/y) dy = 3 dx
ln y = 3x + C

Therefore

C = ln y - 3x

So, plugging in the given values of x = 1 and y = 2, I get that C = ln(2) - 3. If you put this in a calculator, it's a very different value (about -2.307) than what Sal got by raising both sides to the power of e: 2 * e^-3 evaluates to about .0996.

Why did my method not work?
Button navigates to signup pageButton navigates to signup page
(15 votes)
Answer
- Tejas
  Posted 6 years ago. Direct link to Tejas's post “It's important to note th...”
  It's important to note that the C you work with is not the same as the C Sal works with. At
  2:06
  , Sal gets e^C, which he says is just going to be another arbitrary constant. Thus Sal replaces e^c with C. If you perform e^(-2.307) you should get 0.0996.
  Comment on Tejas's post “It's important to note th...”
  (25 votes)
Sidharth Gat
Posted 6 years ago. Direct link to Sidharth Gat's post “Sal changed e^C (which an...”
Sal changed e^C (which any arbitrary constant) to C (which is any another constant) and at least according to my knowledge C can't be any another constant as it will always be positive coz e^C can never be negative. We actually limited the value of our arbitrary constant to positive numbers, then shouldn't we mention that C is any constant number greater than zero i.e(C>0) ?
Button navigates to signup pageComment on Sidharth Gat's post “Sal changed e^C (which an...”
(11 votes)
Answer
- Alexander Bessman
  Posted 4 years ago. Direct link to Alexander Bessman's post “"Any arbitrary constant" ...”
  "Any arbitrary constant" includes complex numbers. Recall that e^(x*j) = cos(x) + j*sin(x). C could be, for example, 5 + π*j, in which case e^C ≈ -7.4.
  
  However, even with complex exponents, e^C can never be zero! In this case that does not matter, because we have already limited ourselves to y ≠ 0 by using 1/y as part of the solution. But it is certainly something to keep in mind for other situations.
  Button navigates to signup page
  (2 votes)
Insatiable
Posted 4 years ago. Direct link to Insatiable's post “At around 2:50, Sal decid...”
At around
2:50
, Sal decides to consider only the first scenario that he broke the absolute value function into, saying that "it will take the [other, negative part of it] into consideration". How did he know this would happen?
Button navigates to signup pageComment on Insatiable's post “At around 2:50, Sal decid...”
(9 votes)
Answer
brian harris
Posted 6 years ago. Direct link to brian harris's post “When dealing with separab...”
When dealing with separable equations why in some videos do you put y and the constant on one side and in others (like this one) you only move y over?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Michele Biondi
  Posted 6 years ago. Direct link to Michele Biondi's post “It doesn't matter, the re...”
  It doesn't matter, the result will be the same. It just changes how you get to that result.
  Button navigates to signup page
  (3 votes)
Dirk Kuiper
Posted 6 years ago. Direct link to Dirk Kuiper's post “Wouldn't it be possible t...”
Wouldn't it be possible to solve for C right after you integrated both sides of the separated equation? ln |y| = 3x + C , when x = 1, y = 2.
Plug in these values and you'd get ln (2) = 3 + C, so C = ln (2) - 3
Then you'd get ln |y| = 3x +ln (2) - 3
So that y = e^(3x+ln (2) -3), because e^ln(2) = 2 you'd get 2e^(3x-3), and this is correct, but the method is different from the one Sal uses, can someone explain to me, why this works, or if there are any steps that aren't correct, so that you'd have to use Sal's method?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Logan Joseph
  Posted 6 years ago. Direct link to Logan Joseph's post “What exactly does Sal do ...”
  What exactly does Sal do (and yes what you did is correct)
  Button navigates to signup page
  (2 votes)
Bryan
Posted 4 years ago. Direct link to Bryan's post “Why do we ultimately get ...”
Why do we ultimately get the same specific solution when we solve for C with
y = Ce^(3x)
and substitute C back in and when we use
y = -Ce^(3x)
and substitute C back in?

I might be missing something fairly obvious, but since both functions satisfy our original differential equation and are the reflections of each other over the x-axis, there should be two functions that both pass through the point (1, 2) and satisfy dy/dx = 3y right?

Edit: I think (??) I answered my own question, but I would appreciate it if someone could verify it. In this case, since our two functions have C as a coefficient, and not as something that you add on, the two functions lead to the same specific solution. If you wanted to have the reflection over the x-axis of the solution also pass through the point, you would need to add a constant to it; but then, it wouldn't satisfy our differential equation.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
Jose Molina
Posted 6 years ago. Direct link to Jose Molina's post “If for whatever reason I ...”
If for whatever reason I had chosen to multiply both sides of the original integral in orange by -1, I would get very different solutions. I have always had a hard time understanding this. can someone help?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Jerry Nilsson
  Posted 6 years ago. Direct link to Jerry Nilsson's post “−∫(1 ∕ 𝑦)𝑑𝑦 = −∫3𝑑𝑥 ...”
  −∫(1 ∕ 𝑦)𝑑𝑦 = −∫3𝑑𝑥 ⇔
  ⇔ −ln|𝑦| = −3𝑥 + 𝐶, where we can just as well write the right-hand side as −3𝑥 − 𝐶
  Multiplying both sides by (−1), we get; ln|𝑦| = 3𝑥 + 𝐶, which is the exact same equation that Sal solved for in the video.
  Button navigates to signup page
  (2 votes)
Abhinay Singh
Posted 4 years ago. Direct link to Abhinay Singh's post “when I use y=-ce^3x then ...”
when I use y=-ce^3x then also I got the same ans y=e^(3x-3) as Sal sir got using y=ce^3x
that means no effect of absolute value here?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- loumast17
  Posted 4 years ago. Direct link to loumast17's post “I think you may have misc...”
  I think you may have miscalculated somewhere.
  
  2 = -ce^3 gets a different result than 2 = ce^3
  Comment on loumast17's post “I think you may have misc...”
  (2 votes)
Louis Ullman
Posted 5 years ago. Direct link to Louis Ullman's post “I solved the integral by ...”
I solved the integral by using U-substitution (int(1/3y)dy=int(1)dx), but after evaluating, I got ln|3y|=3x+C instead of ln|y|=3x+C. What did I do wrong?

Here are my steps:

int(1/3y)dy=int(1)dx
int(1/3y)dy=x+C
(d/dy)(3y)=3
(1/3)int(3*1/3y)dy=x+C
u=3y
(1/3)int(1/u)du=x+C
(1/3)ln|u|+C=x+C
(1/3)ln|u|=x+C
ln|u|=3x+C
ln|3y|=3x+C
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Jerry Nilsson
  Posted 5 years ago. Direct link to Jerry Nilsson's post “There's nothing wrong. l...”
  There's nothing wrong.
  
  ln|3𝑦| = 3𝑥 + 𝐶 ⇒ 𝑦 = (1∕3)𝑒^(3𝑥 + 𝐶) ⇒ 𝑑𝑦∕𝑑𝑥 = 𝑒^(3𝑥 + 𝐶) = 3𝑦, which is the equation we started with.
  
  Similarly,
  ln|𝑦| = 3𝑥 + 𝐶 ⇒ 𝑦 = 𝑒^(3𝑥 + 𝐶) ⇒ 𝑑𝑦∕𝑑𝑥 = 3𝑒^(3𝑥 + 𝐶) = 3𝑦
  Button navigates to signup page
  (3 votes)
Liang
Posted 4 months ago. Direct link to Liang's post “at 2:38, does it mean tha...”
at
2:38
, does it mean that y=+/- Ce^3x could lead to the same solution of this differential equation?

by substituting y=2 and x=1 into y=-Ce^3x, you get C=-2e^-3. sustituting this C back into y=-Ce^3x, you'll get the same y=2e^(3x-3)
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- cossine
  Posted 4 months ago. Direct link to cossine's post “So you can just write Ce^...”
  So you can just write Ce^3x. It is understood C can take any value.
  Comment on cossine's post “So you can just write Ce^...”
  (3 votes)

Video transcript

- [Teacher] So, we've got the differential equation, the derivative of y with respect to x is equal to three times y. And we want to find the particular solution that gives us y being equal to two when x is equal to one. So, I encourage you to pause this video and see if you can figure this out on your own. All right, now, let's work through it together. So, some of you might have immediately said, "Hey, this is the form of a differential equation "where the solution is going to be an exponential," and you just got right to it. But, I'm not going to go straight to that, I'm just gonna recognize that this is a separable differential equation and that I'm gonna solve it that way. So, when I say that it's separable, that means that we can separate the y's, dy's, on one side, and all the x's, dx's on the other side. And so, what I can do is if I divide both sides of this equation by y and multiply both sides by dx, I get one over y, dy, is equal to three-dx. Now, on the left and right hand sides, I have these clean things that I can now integrate. That's what people talk about when they say separable differential equation. Now, here on the left, if I wanted to write it in a fairly general form, I could write, well, the anti-derivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the anti-derivative with respect to y, here. Now, I could add a constant, but I'm gonna add in a constant on the right-hand side, so there's no reason to add two arbitrary constants on both sides. I could just add one on one side. So, that is going to be equal to the anti-derivative here is going to be three-x and I'll add the promised constant, plus c, right over there. And now, let's think about it a little bit. Well, we can rewrite this in exponential form. We could say, we could write, that e to the three-x plus c is equal to the natural log of y. I could write the natural of y is equal to e to three-x plus c. Now, I could rewrite this is equal to e to three-x times e to the c. Now, e to the c is just gonna be some other arbitrary constant, which I could still denote by c. They're are going to be different values, but we're just trying to just get a sense of what the structure of this thing looks like. So, we could say this is going to be some constant times e to the three-x. So, another way of thinking about it. Saying the absolute value of y is equal to this. This isn't a function yet. We're trying to find this function solution to this differential equation. So, this would tell us either y is equal to c, e to the three-x, or y is equal to negative c, e to the three-x. Well, we've kept it in general terms. I haven't put any... We don't know c is. So, what we could do, instead, is just pick this one, and then we can solve for c assuming this one right over here. And so, we will see if we can meet these constraints using this and it'll essentially take the other one into consideration, whether we're going positive or negative. So, let's do that. So, when y is equal to two, I'm now going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two. So, I could write it like that, and we get two is equal to c times e to the third power, three times one. And so, to solve for c, I could just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it back in and our particular solutions is gonna be y is equal to c. C is two-e to the negative third power times e to the three-x. Now, I'm taking the product of two things with the same base. I can add the exponents. So, I could say y is equal to two times e to the three-x and I'll add the exponents to three-x minus three, and there you go. This is one way that you could write the particular solution that meets these constraints for this separable differential equation.