# 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.