# simplification of ODE

• Mar 7th 2010, 08:35 PM
wclayman
simplification of ODE
I saw this in a book:
$\displaystyle \frac{dy}{dt} = -\frac{y}{\tau} +f(t)$,
can be simplified to $\displaystyle y = f(t)\tau$, if one is interested only in the time scale slower than $\displaystyle \tau$,

I don't know how this is got, could someone help me with it? Many thanks!
• Mar 8th 2010, 11:06 AM
Coomast
Quote:

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

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:

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

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

Coomast
• Mar 9th 2010, 02:47 AM
wclayman
Thanks! I get it.
• Mar 9th 2010, 09:23 AM
chisigma
The DE can be written as...

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

... whose solution is...

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

... where...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Mar 12th 2010, 11:54 PM
wclayman
Quote:

Originally Posted by chisigma
The DE can be written as...

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

... whose solution is...

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

... where...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

I am confused now, first, the assumption is actually $\displaystyle t \gg \tau$, and following this way, I can't arrive at the above-mentioned conclusion.
• Mar 13th 2010, 12:43 AM
chisigma
What is [for me] difficult to understand is why Coomast sets in His post $\displaystyle \frac{dy}{dt}=0$... the same to say that $\displaystyle y(t)$ is a constant (Nerd)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Mar 13th 2010, 05:25 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.