# Thread: Help with physics word problems

1. ## Help with physics word problems

Hi!

I am having a lot of trouble with these word problems in the modeling part of my differential equations books. Here is one I am just lost on. I think the wording is throwing me off. Shouldn't I approach this as a homogenous first order ODE, with an initial value???

Bacteria is being cultured for the production of medication. Without harvesting the bacteria, the rate of change of the population is proportional to its current population, with a proportionality constant of .2 per hour. Also, the bateria are being harvested at a rate of 1000 per hour. If there are initially 8000 bacteria in the culture, solve the initial value problem:

$\displaystyle (dN/dt) = .2N - 1000$

for the number N of bacteria as a function of time and find the time it takes for the population to double its initial number.

Thanks for any guidance suggested.

2. Originally Posted by kpizle
Hi!

I am having a lot of trouble with these word problems in the modeling part of my differential equations books. Here is one I am just lost on. I think the wording is throwing me off. Shouldn't I approach this as a homogenous first order ODE, with an initial value???

Bacteria is being cultured for the production of medication. Without harvesting the bacteria, the rate of change of the population is proportional to its current population, with a proportionality constant of .2 per hour. Also, the bateria are being harvested at a rate of 1000 per hour. If there are initially 8000 bacteria in the culture, solve the initial value problem:

$\displaystyle (dN/dt) = .2N - 1000$

for the number N of bacteria as a function of time and find the time it takes for the population to double its initial number.

Thanks for any guidance suggested.
This is a first order seperable ODE so

$\displaystyle \frac{dN}{.2N-1000}=dt \iff \frac{dN}{N-5000}=.2tdt$ Now just integrate to get the equation for $\displaystyle N(t)$

Use the inital condition $\displaystyle N(0)=8000$ to find the arbitary constant then you want to use the equation to find $\displaystyle t$ such that

$\displaystyle N(t)=16000$

3. Originally Posted by TheEmptySet
This is a first order seperable ODE so

$\displaystyle \frac{dN}{.2N-1000}=dt \iff \frac{dN}{N-5000}=.2tdt$ Now just integrate to get the equation for $\displaystyle N(t)$

Use the inital condition $\displaystyle N(0)=8000$ to find the arbitary constant then you want to use the equation to find $\displaystyle t$ such that

$\displaystyle N(t)=16000$

I finally came up with $\displaystyle N(t) = 5000 + 3000 e^(.2*t)$ as my equation and that t = 6.496 hours to get the population to double.