What is the Integration Process for a Logistic Equation?

[nhmllr]

Can somebody explain to me what this is called
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/logistic_37.gif
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/logistic_38.gif

I mean what's happening on the left side of this equation
Why does the T turn into y(T) and 0 into y(0)?

Thanks

 

[JG89]

Could you give some context behind the equations?

 

[nhmllr]

I mean I can give the link to the lesson where itt's mentioned
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/logistic.html
but I don't think that the context gives any new insight
It seems to me that it's just some rule that I never learned

 

[tiny-tim]

hi nhmllr!

… what's happening on the left side of this equation
Why does the T turn into y(T) and 0 into y(0)?

the variable of integration is changing from t to y

the original limit was 0 < t < T

y is a function of t

so that's the same as y(0) < y(t) < y(T)

(you need the same limit, written in the new variable)

 

[nhmllr]

hi nhmllr!


the variable of integration is changing from t to y

the original limit was 0 < t < T

y is a function of t

so that's the same as y(0) < y(t) < y(T)

(you need the same limit, written in the new variable)

Ohhh... I think I see, Thanks

 

[Curious3141]

Can somebody explain to me what this is called
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/logistic_37.gif
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/logistic_38.gif

I mean what's happening on the left side of this equation
Why does the T turn into y(T) and 0 into y(0)?

Thanks

It's basically a change of variable, or if you prefer to think about it this way, a reversal of the usual method of substitution.

If you're working out \int f(y)dy, where y is dependent on t, i.e. y = g(t), then you can state:

\int f(y)dy = \int f(g(t))dy = \int f(g(t))\frac{dy}{dt} dt = \int f(y)\frac{dy}{dt}dt.

What happened there is I made a change of variables from y to t. dy = \frac{dy}{dt}dt. You should be able to recognise that as the basis for substitution.

In the example, they're just going in reverse.

The reason the bounds change is that the bounds must follow the variable of integration. So if the integration is wrt t, the bounds will be [0,T]. If the integration is wrt y, the bounds will be y_0, y_T where the lower bound refers to the y-value at t = 0 and the upper bound refers to the y-value at t=T.

 

[electronicscm]

The limit of integration is from 0 to T and you're integrating with respect to the varible t, as shown by dt.