# Math Help - population growth with differential

1. ## population growth with differential

Population y grows according to the equation dy/dt=ky where k is a constant and t is measured in years. if the population doubles every 10 years then the value of k is?

2. Originally Posted by lbj2322
Population y grows according to the equation dy/dt=ky where k is a constant and t is measured in years. if the population doubles every 10 years then the value of k is?
Separate and integrate:

$\int\frac{dy}{y} = \int k\,dt \implies \ln(y)=kt+C \implies y=c_0e^{kt}$

We are looking for the value of $k$ that makes $e^{10k}=2$, so $k=\frac{\ln 2}{10}\approx0.0693$.

3. thanks again for the help

when i did the differential i did come upon y=Ce^kt but i wasnt sure how to represent a population doubling every 10 years.
now i understand
thanks again