$\displaystyle m_{t+h}-m_{t}=hm_t$,

so $\displaystyle \frac{dm_t}{dt}=m_t$

I am told that you use differential equations to the get the differentiation above, but I don't see how. Can anybody see?

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- Feb 19th 2006, 04:45 AMemterics90differential equation
$\displaystyle m_{t+h}-m_{t}=hm_t$,

so $\displaystyle \frac{dm_t}{dt}=m_t$

I am told that you use differential equations to the get the differentiation above, but I don't see how. Can anybody see? - Feb 19th 2006, 06:48 AMCaptainBlackQuote:

Originally Posted by**emterics90**

$\displaystyle

m(t+h)-m(t)=hm(t)

$,

so:

$\displaystyle \frac{dm}{dt}=m(t)$

So you are looking for a function which is its own derivative, so

something like:

$\displaystyle m(t)=Ae^t$

should do.

RonL