simplification of ODE

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Mar 7th 2010, 09:35 PM
wclayman

simplification of ODE

I saw this in a book:
$\frac{dy}{dt} = -\frac{y}{\tau} +f(t)$ ,
can be simplified to $y = f(t)\tau$ , if one is interested only in the time scale slower than $\tau$ ,

I don't know how this is got, could someone help me with it? Many thanks!
Mar 8th 2010, 12:06 PM
Coomast

Quote:

Originally Posted by wclayman

I saw this in a book:
$\frac{dy}{dt} = -\frac{y}{\tau} +f(t)$ ,
can be simplified to $y = f(t)\tau$ , if one is interested only in the time scale slower than $\tau$ ,

I don't know how this is got, could someone help me with it? Many thanks!

Presumably it means nothing more than that the change of y with respect to t is very small and in the approximation stated means equal to 0. So you have:

$\frac{dy}{dt}=0=-\frac{y}{\tau}+f(t)$

From which the relation [imath]y=f(t)\cdot \tau[/imath]

Coomast
Mar 9th 2010, 03:47 AM
wclayman

Thanks! I get it.
Mar 9th 2010, 10:23 AM
chisigma

The DE can be written as...

$y^{'} + \frac{y}{\tau} = f(t)$ (1)

... whose solution is...

$y(t) = y (0) \cdot e^{-\frac{t}{\tau}} + \varphi (t)$ (2)

... where...

$\varphi (t) = \int_{0}^{t} e^{-\frac{t-u}{\tau}}\cdot f(u)\cdot du$ (3)

If $t << \tau$ the exponential terms in (2) and (3) can be approximated at 1 so that is...

$y(t) \approx y(0) + \int_{0}^{t} f(u)\cdot du$ (4)

Kind regards

$\chi$ $\sigma$
Mar 13th 2010, 12:54 AM
wclayman

Quote:

Originally Posted by chisigma

The DE can be written as...

$y^{'} + \frac{y}{\tau} = f(t)$ (1)

... whose solution is...

$y(t) = y (0) \cdot e^{-\frac{t}{\tau}} + \varphi (t)$ (2)

... where...

$\varphi (t) = \int_{0}^{t} e^{-\frac{t-u}{\tau}}\cdot f(u)\cdot du$ (3)

If $t << \tau$ the exponential terms in (2) and (3) can be approximated at 1 so that is...

$y(t) \approx y(0) + \int_{0}^{t} f(u)\cdot du$ (4)

Kind regards

$\chi$ $\sigma$

I am confused now, first, the assumption is actually $t \gg \tau$ , and following this way, I can't arrive at the above-mentioned conclusion.
Mar 13th 2010, 01:43 AM
chisigma

What is [for me] difficult to understand is why Coomast sets in His post $\frac{dy}{dt}=0$ ... the same to say that $y(t)$ is a constant (Nerd)...

Kind regards

$\chi$ $\sigma$
Mar 13th 2010, 06:25 AM
wclayman

I'm not sure now, ...
The author of that book is coming to visit the day after tomorrow, I'll take that chance and consult him directly. After that I will post the what he says to me.
I strongly suspect he had some typo or something! That solution is wrong.