Since this has been here a while and I just cannot resist:
dr/dt= kr(1000- r)
dr/[r(1000-r)]= kdt
To integrate on the left, separate using "partial fractions"- find constant A and B such that 1/[r(1000- r)]= A/r+ B/(1000- r)
Multiplying on both sides by r(1000- r)
1= A(1000- r)+ Br
Let r= 0: 1= 1000A so A= 1/1000.
Let r= 1000: 1= 1000B so B= 1/1000
1000 dr/r+ 1000 dr/(1000- r)= kdt
The integral of 1000 dr/r is 1000 ln(|r|).
To integrate 1000 dr/(1000- r) let u= 1000- r so that du= -dr.
then dr/(1000- r)= -du/u.
The integral is -1000 ln(|u|)= -1000 ln(|1000- r|)
We have 1000 ln(|r|)-1000 ln(|1000- r|)= kt+ C.
$1000 ln\left(|\frac{r}{1000- r}|\right)= kt+ C$
$ln\left(\left(\frac{r}{1000- r}\right)^{1000}\right)= kt+ C$
(I have dropped the absolute value since this is to an even power.)
Taking the exponential of both sides
$\left(\frac{r}{1000- r}\right)^{1000}= C' e^{kt}$
where $C'= e^C$.
Now, we are given that r(0)= 5 and r(1)= 10.
Setting t= 0, r= 5
$\frac{5}{9995}= \frac{1}{1999}=C'$
Setting t= 1, r= 10
$\frac{10}{9990}= \frac{1}{999}= \frac{e^k}{1999}$
$e^k= \frac{1999}{999}$
so $k= ln\left(\frac{1999}{999}\right)$
and $e^{kt}= (e^k)^t= \left(\frac{1999}{999}\right)^t$
$r(t)= \frac{\left(\frac{1999}{999}\right)^t}{1999}$.