How they say that it is the solution

[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}}

 

[vela]

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.