Homogenous differential equation of the first order I cannot solve this question... excuse my bad english!
Boiling water cools down proportionally to the temperature the water has at the specific point of time. What temperature does the water have after 8 hours if it has a temperature of 60 degrees Celsius after 4 hours.
This is how far I have got:
$T'(t) = -kT$.   $k$ is a constant, $T$ is the temperature after $t$ hours.
$T'(t) + kT = 0$
$T = Ce^{-kt}$     general solution, $C$ is a constant
$T(4) = Ce^{-4k}$
$60 = Ce^{-4k}$
What is $T(8)$???
Thanks in advance!
 A: As pointed out in the comments above, there is one piece of information you haven't used, yet: that the water was boiling initially. From this, we have $100=T(0)=C,$ and knowing $C$ allows us to then solve for $k$ by the work you've done so far, whence we can find $T(8).$
Note that we can (and should) find an exact value of $k$, instead of a numerical approximation, if we want to get the exact (numerical) value of $T(8).$
A: As has been stated the first condition is $100 = T(0)$. From the solution of the first order equation it is known that $T(t) = c_{0} \, e^{-k \, t}$. Now, $T(0) = 100 = c_{0}$ for which $T(t) = 100 \, e^{- k \, t}$. Now using the the given time and value it is determined that $T(4) = 60 = 100 \, e^{- 4 k}$ which provides $e^{4k} = \frac{5}{3}$ and taking the logarithm of both sides yields $k = \frac{1}{4} \, \ln(\frac{5}{3})$. The solution now becomes:
$$T(t) = 100 \, e^{- \frac{t}{4} \, \ln\left(\frac{5}{3}\right)}.$$
After 8 hours then
$$T(8) = 100 \, e^{-2 \, \ln(\frac{5}{3})} = 100 \, e^{\ln\left(\frac{9}{25}\right)} = 100 \cdot \frac{9}{25} = 36$$
