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

[UrbanXrisis]

\frac{dy}{dt}=ky

I need to solve for y

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

is this correct? also, I've always wondered why the integral of dy goes away? the integration of dx 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?

 

[Zurtex]

No, you need to add the constant of integration as soon as you have integrated, you will see why.

 

[dextercioby]

Could u reformulate your question...?I don't understand "the 'dy' goes away"...

It's justa plain simple integration,

Daniel.

 

[whozum]

\int dy

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

\infty * \left(\frac{y}{\infty}\right) = y

 

[UrbanXrisis]

okay.. here's the question that's bothering me...

k=constant

\int k dx=kx
\int x^2 dx = \frac{1}{3} x^3

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

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

I was never explained why...

\int x^n dx = \frac{x^{n+1}}{n+1}
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?

 

[UrbanXrisis]

No, you need to add the constant of integration as soon as you have integrated, you will see why.

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


also.. for \int\frac{dy}{y}=k \int dt

why doesn't it become
yln(y)=kt+C

since \int dy = y just as \int dt = t

 

[whozum]

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

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

\sum_{i=1}^{\infty} f(x_i)\Delta x_i

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 \Delta x turn the dx into x.

I was never explained why...

\int x^n dx = \frac{x^{n+1}}{n+1}
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?

x^{n+1} = x*x^n

Do you see why?

Take f(x) = x^n [/tex]<br /> <br /> \int f(x) dx = \sum_{i=1}^{\infty}(x_i^n)\Delta x_i<br /> <br /> The \Delta x_i turn into x, and when multiplied by x^{n} \mbox{ gives } x^{n+1} I don&#039;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.

 

[dextercioby]

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.

 

[UrbanXrisis]

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)

 

[Jameson]

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

This is correct. Anytime you see something in the form of \frac{dy}{dt} = ky you should spot that this will integrate to the form of y = Ce^{kt}. This is the general model for exponential growth / decay.

also.. for \int\frac{dy}{y}=k \int dt

why doesn't it become
yln(y)=kt+C

since \int dy = y just as \int dt = t

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

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

Jameson

 

[Data]

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

y=e^{kt}*e^C
y=Ce^{kt}

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 C_1, for example).

As well, keep in mind that

\int \frac{1}{x} \ dx = \ln x+C

is not quite true. It's really

\int \frac{1}{x} \ dx = \ln |x|+C,

and this changes things sometimes.

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

\int f(x) \ dx,

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

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

is true precisely because the right side is the most general form of an antiderivative of x^n (read: every antiderivative of x^n has this form for some C, 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.

 

[UrbanXrisis]

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

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

Jameson

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

okay.. here's the question that's bothering me...

k=constant

\int k dx=kx
\int x^2 dx = \frac{1}{3} x^3

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

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:

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

\sum_{i=1}^{\infty} f(x_i)\Delta x_i

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 \Delta x turn the dx into x.

why does the dx turn into a x and the dy goes away?

 

[whozum]

edit: Come to think of it, there's a lot more to this than I can properly explain so I'll abstain.

 

[dfollett76]

also, I've always wondered why the integral of dy goes away? the integration of dx 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 \int\dx. 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. \int1dx

 

[UrbanXrisis]

okay, that makes more sense now...

 

[UrbanXrisis]

so...

\int dy = y?

 

[dfollett76]

Yeah,

\int d(insert anything here) = (insert anything here)

 

[Hippo]

Wrong.

\int \ d y = y + K

 

[Jameson]

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:

\int ax^ndx = \frac{ax^n}{n+1}*x + C

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 \int \frac{1}{y} dy

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

Does that make sense? I hope this helps.

Jameson

 

[Data]

well, a simpler example is just

\int e^x \ dx = e^x + C.

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, e^x - which is precisely what an indefinite integral is defined to be!