- #1
nhrock3
- 415
- 0
they said that the solution of
[tex]-dn=-\frac{1}{\tau}ndt\\[/tex]
is
[tex]n=n_0e^{-\frac{t}{\tau}}[/tex]
i got a totaly different answer
[tex]-dn=-\frac{1}{\tau}ndt\\[/tex]
[tex]\int -dn=\int -\frac{1}{\tau}ndt\\[/tex]
[tex]\int \frac{-dn}{n}=\int -\frac{1}{\tau}dt\\[/tex]
[tex]-\ln{n}=-\frac{t}{\tau}\\[/tex]
[tex]\ln{n^{-1}}=-\frac{t}{\tau}\\[/tex]
[tex]e^{-\frac{t}{\tau}{={n^{-1}}[/tex]
[tex]-dn=-\frac{1}{\tau}ndt\\[/tex]
is
[tex]n=n_0e^{-\frac{t}{\tau}}[/tex]
i got a totaly different answer
[tex]-dn=-\frac{1}{\tau}ndt\\[/tex]
[tex]\int -dn=\int -\frac{1}{\tau}ndt\\[/tex]
[tex]\int \frac{-dn}{n}=\int -\frac{1}{\tau}dt\\[/tex]
[tex]-\ln{n}=-\frac{t}{\tau}\\[/tex]
[tex]\ln{n^{-1}}=-\frac{t}{\tau}\\[/tex]
[tex]e^{-\frac{t}{\tau}{={n^{-1}}[/tex]