Homework Help: How they say that it is the solution

1. Apr 19, 2010

nhrock3

they said that the solution of

$$-dn=-\frac{1}{\tau}ndt\\$$

is

$$n=n_0e^{-\frac{t}{\tau}}$$

i got a totaly different answer

$$-dn=-\frac{1}{\tau}ndt\\$$
$$\int -dn=\int -\frac{1}{\tau}ndt\\$$
$$\int \frac{-dn}{n}=\int -\frac{1}{\tau}dt\\$$
$$-\ln{n}=-\frac{t}{\tau}\\$$
$$\ln{n^{-1}}=-\frac{t}{\tau}\\$$
$$e^{-\frac{t}{\tau}{={n^{-1}}$$

2. Apr 19, 2010

vela

Staff Emeritus
Are you sure the negative sign in front of dn in the original equation is supposed to be there? You also forgot the constant of integration.