Hi everybody!

You guys helped me a lot!. I figured out how to reach to the solution. I'll post it here just in case somebody need it!

From the original expression:

$\displaystyle \frac{dx}{dt}=\frac{(1-x)}{\tau}$

by separating the variables we have:

$\displaystyle \frac{dx}{(1-x)}=\frac{dt}{\tau}$

with substitution of $\displaystyle u = (1-x)$, I get that $\displaystyle dx = -du$

To obtain the following expression to be solved by direct integration:

$\displaystyle \int\frac{-du}{u}=\int\frac{dt}{\tau}$

whose solution is:

$\displaystyle -ln|u| = t/\tau + C$

for $\displaystyle C$ being the integration constant. Thanks

**Prove It** for your

excellent post about it!

Now it's simple to follow

**Soroban**:

$\displaystyle -ln|(1-x)| = t/\tau + C$

$\displaystyle ln|(1-x)| = -t/\tau - C$

$\displaystyle 1-x = \exp(-t/\tau - C)$, this was the difficult step for me :S

$\displaystyle 1-x = \exp(-t/\tau)*\exp(-C)$, now let $\displaystyle K=\exp(-C)$

$\displaystyle 1-x = \exp(-t/\tau)*K$

$\displaystyle -x = -1 + K*\exp(-t/\tau)$

$\displaystyle x = 1 - K*\exp(-t/\tau)$

Thanks so much !!!