Integrating dy/dt = ky: Understand dx, dy & e^{kt} + C

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In summary: But for now, I think it's best to just think of it as notation.Hope this helps, and good luck!In summary, the conversation discusses the concept of integration and its application in solving equations involving differentials. The conversation also touches on the use of integration measure such as "dx" and "dy". It concludes with a discussion on the notation used in integration and the importance of understanding the concept behind it.
  • #1
UrbanXrisis
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[tex]\frac{dy}{dt}=ky[/tex]

I need to solve for y

[tex]\int\frac{dy}{y}=k \int dt[/tex]
[tex]ln(y)=kt[/tex]
[tex]y=e^{kt} +C[/tex]

is this correct? also, I've always wondered why the integral of [itex]dy[/itex] goes away? the integration of [itex]dx[/itex] becomes x, but what about dy? Also, if I were to integrate x^2 dx... the dx goes away after that integratin as well, why is that?
 
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  • #2
No, you need to add the constant of integration as soon as you have integrated, you will see why.
 
  • #3
Could u reformulate your question...?I don't understand "the 'dy' goes away"...:confused:

It's justa plain simple integration,

Daniel.
 
  • #4
[tex] \int dy [/tex]

dy represents an infinitesimally small size of y's. the integral sums an infinite amount of these infinitesimally small y's. This evaluates to y.

In incorrect algebra this translates to

[tex] \infty * \left(\frac{y}{\infty}\right) = y [/tex]
 
  • #5
okay.. here's the question that's bothering me...

k=constant

[tex]\int k dx=kx[/tex]
[tex]\int x^2 dx = \frac{1}{3} x^3 [/tex]

does this mean that... [tex]\int x^2 dx =\frac{1}{3} x^2 * x = \frac{1}{3} x^3 [/tex]

so that the dx turned into the x so that it could multiply the x^2?

I was never explained why...

[tex]\int x^n dx = \frac{x^{n+1}}{n+1} [/tex]
as in... when you add the n+1... I was never told where the dx went.. did the dx become the 1 added to the n? also, why do you have to divide by n+1?
 
  • #6
Zurtex said:
No, you need to add the constant of integration as soon as you have integrated, you will see why.

then you mean...
[tex]\int\frac{dy}{y}=k \int dt[/tex]
[tex]ln(y)=kt+C[/tex]
[tex]y=e^{kt}*e^C[/tex]
[tex]y=Ce^{kt}[/tex]
correct?


also.. for [tex]\int\frac{dy}{y}=k \int dt[/tex]

why doesn't it become
[tex]yln(y)=kt+C[/tex]

since [tex]\int dy = y[/tex] just as [tex]\int dt = t[/tex]
 
  • #7
UrbanXrisis said:
does this mean that... [tex]\int x^2 dx =\frac{1}{3} x^2 * x = \frac{1}{3} x^3 [/tex]

so that the dx turned into the x so that it could multiply the x^2?

This is correct pretty much. Remember the Riemann sum? It's something close to

[tex] \sum_{i=1}^{\infty} f(x_i)\Delta x_i [/tex]

One element draws a rectangle of height f(x) and width delta x. I'm sure you remember the approximation techniques drawing several rectangles gives a better approximation, and the sum of an infinite amount of rectangles is exact, well the infinite amount of rectangles becomes the integral. The small width's of [itex] \Delta x [/itex] turn the dx into x.

I was never explained why...

[tex]\int x^n dx = \frac{x^{n+1}}{n+1} [/tex]
as in... when you add the n+1... I was never told where the dx went.. did the dx become the 1 added to the n? also, why do you have to divide by n+1?

[tex] x^{n+1} = x*x^n [/tex]

Do you see why?

Take [itex] f(x) = x^n [/tex]

[tex] \int f(x) dx = \sum_{i=1}^{\infty}(x_i^n)\Delta x_i [/tex]

The [itex] \Delta x_i [/itex] turn into x, and when multiplied by [itex] x^{n} \mbox{ gives } x^{n+1} [/itex] I don't remember the exact reasoning for the n+1 in the denominator, but it follows from some kind of triange ratio I think. Someone else would have to take that one.
 
  • #8
That "dx","dy","dt" is what we call "integration measure".Maybe you should check again your notes on Riemann integration,i think you're unclear what this means...

Daniel.
 
  • #9
like I said, I was given the formulas but never explained why they do what they do... so I'm clueless and just accepted it for a while now, but I am really going to have a hard time understanding advanced math if I don't understand the basic concept. Could someone please explain the 'dy' and 'dx' concepts? btw, I have not learned summation and will not for another half a year (sry whozum)
 
  • #10
UrbanXrisis said:
then you mean...
[tex]\int\frac{dy}{y}=k \int dt[/tex]
[tex]ln(y)=kt+C[/tex]
[tex]y=e^{kt}*e^C[/tex]
[tex]y=Ce^{kt}[/tex]
correct?
This is correct. Anytime you see something in the form of [tex]\frac{dy}{dt} = ky[/tex] you should spot that this will integrate to the form of [tex] y = Ce^{kt}[/tex]. This is the general model for exponential growth / decay.



also.. for [tex]\int\frac{dy}{y}=k \int dt[/tex]

why doesn't it become
[tex]yln(y)=kt+C[/tex]

since [tex]\int dy = y[/tex] just as [tex]\int dt = t[/tex]

When you integrate, the "dy" will kind of go away. You would call this integrating with respect to y.

Do a simple integral like [tex]\int2xdx[/tex] You get x^2 + C, not x*x^2 + C.
Try your integral again. [tex]\int\frac{dy}{y} = \int{y}^{-1}dy = \ln |y|+ C[/tex]

Jameson
 
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  • #11
You can delete your own posts. It's the first box on the 'edit' screen.

[tex]y=e^{kt}*e^C[/tex]
[tex]y=Ce^{kt}[/tex]

Almost. The only thing that you need to be careful about is to make sure that you understand that the C in the first line is not the same as the C in the second line, and you should typically label them differently (call one [itex]C_1[/itex], for example).

As well, keep in mind that

[tex]\int \frac{1}{x} \ dx = \ln x+C[/tex]

is not quite true. It's really

[tex]\int \frac{1}{x} \ dx = \ln |x|+C,[/tex]

and this changes things sometimes.

Now, to your other questions. For now, when you see

[tex]\int f(x) \ dx,[/tex]

you should interpret it just to mean the most general form of the antiderivative of [itex]f(x)[/itex], and the [itex]\int[/itex] and [itex]dx[/itex] as nothing more than notational conveniences (the [itex]dx[/itex] tells you precisely what variable you want an antiderivative with respect to). In particular,

[tex]\int x^n \ dx = \frac{x^{n+1}}{n+1} + C, \ n \geq 0[/tex]

is true precisely because the right side is the most general form of an antiderivative of [itex]x^n[/itex] (read: every antiderivative of [itex]x^n[/itex] has this form for some [itex]C[/itex], and every function with this form is an antiderivative).

Now, there certainly is reasoning behind using this particular notation and you will learn about it when you study differentials in more detail.
 
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  • #12
Jameson said:
When you integrate, the "dy" will kind of go away. You would call this integrating with respect to y.

Do a simple integral like [tex]\int2xdx[/tex] You get x^2 + C, not x*x^2 + C.
Try your integral again. [tex]\int\frac{dy}{y} = \int{y}^{-1}dy = \ln |y|+ C[/tex]

Jameson

you say this... but what about my example...

UrbanXrisis said:
okay.. here's the question that's bothering me...

k=constant

[tex]\int k dx=kx[/tex]
[tex]\int x^2 dx = \frac{1}{3} x^3 [/tex]

does this mean that... [tex]\int x^2 dx =\frac{1}{3} x^2 * x = \frac{1}{3} x^3 [/tex]

so that the dx turned into the x so that it could multiply the x^2?

the dx turned into an x, it did not just dissapear. Also, Whozum agreed:

whozum said:
This is correct pretty much. Remember the Riemann sum? It's something close to

[tex] \sum_{i=1}^{\infty} f(x_i)\Delta x_i [/tex]

One element draws a rectangle of height f(x) and width delta x. I'm sure you remember the approximation techniques drawing several rectangles gives a better approximation, and the sum of an infinite amount of rectangles is exact, well the infinite amount of rectangles becomes the integral. The small width's of [itex] \Delta x [/itex] turn the dx into x.

why does the dx turn into a x and the dy goes away?
 
  • #13
edit: Come to think of it, there's a lot more to this than I can properly explain so I'll abstain.
 
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  • #14
UrbanXrisis said:
also, I've always wondered why the integral of [itex]dy[/itex] goes away? the integration of [itex]dx[/itex] becomes x, but what about dy? Also, if I were to integrate x^2 dx... the dx goes away after that integratin as well, why is that?

It's my understanding that dy, dx, etc. have no meaning in and of themselves they are merely parts of the symbol [tex]\int\dx[/tex]. But it turns out they can be manipulated algebraicly anyways--that's why the Leibnitz notation is better than the f', f'' stuff. Anyhow, you aren't integrating dx you are integrating 1. [tex]\int1dx[/tex]
 
  • #15
okay, that makes more sense now...
 
  • #16
so...

[tex]\int dy = y?[/tex]
 
  • #17
Yeah,

[tex]\int d(insert anything here) = (insert anything here)[/tex]
 
  • #18
Wrong.

[tex]\int \ d y = y + K[/tex]
 
  • #19
The d(some variable) does not turn into that variable and then multiply with the other arguments.

The way you seem to be doing some integrals is in the form of:

[tex]\int ax^ndx = \frac{ax^n}{n+1}*x + C[/tex]

This will correct evaluate any integral in that form where x is not -1, but you may be using the wrong reasoning. The constant turning dy into y or dx into x and then multiplying may work out for integrals using the general power rule, but it does not always work.

Look at my earlier example of [tex]\int \frac{1}{y} dy[/tex]

Using your logic, I would do this [tex] \frac{y^{(-1+1)}}{(-1+1)}*y + C[/tex]
which in this case is incorrect.

Does that make sense? I hope this helps.

Jameson
 
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  • #20
well, a simpler example is just

[tex]\int e^x \ dx = e^x + C.[/tex]

Explain that one using this reasoning!

The reason that the integral I put above is true is precisely because the right side is the most general form of the antiderivative of the integrand, [itex]e^x[/itex] - which is precisely what an indefinite integral is defined to be!
 

Related to Integrating dy/dt = ky: Understand dx, dy & e^{kt} + C

What is the meaning of dx, dy, and e^{kt} + C in the equation dy/dt = ky?

In the differential equation dy/dt = ky, dx represents a small change in the independent variable x, while dy represents the corresponding change in the dependent variable y. e^{kt} is the exponential function with a growth rate of k, and C is the constant of integration.

What is the purpose of integrating dy/dt = ky?

The process of integration allows us to find the function y in terms of x, given the differential equation dy/dt = ky. This helps us understand the relationship between the variables and how they change over time.

What does the constant k represent in the equation dy/dt = ky?

The constant k represents the rate of change of the dependent variable y with respect to the independent variable x. It determines the slope of the graph of y in terms of x.

How do you solve for y when integrating dy/dt = ky?

To solve for y, we first rearrange the equation to isolate dy/dt on one side. Then, we can use integration techniques, such as separation of variables or substitution, to find the function y in terms of x.

What is the significance of the constant of integration C in the equation dy/dt = ky?

The constant of integration C is added to the solution for y because when differentiating, the constant disappears. It represents the unknown initial value of y at the starting point of the function.

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