Differential equations + exponenetial models

• May 16th 2009, 06:28 AM
gamboo
Differential equations + exponenetial models
7a

A differential equation is given by $\displaystyle \frac{dx}{dt}=-kte^\frac{x}{2}$, where k is a positive constant. Note: The power to e is x/2, tried making it bigger but still suck at using latex.

(i) Solve the differential equation.
(ii) Hence, given that x=6 when t=0, show that $\displaystyle x=-2\ln(\frac{kt^2}{4}+\frac{1}{e^3})$.

7b The population of a colony of insects is decreasing according to the model $\displaystyle \frac{dx}{dt}=-kte^\frac{x}{2}$, where x thousands is the number of insects in the colony after time t minutes. Initially, there were 6000 insects in the colony.

Given that k=0.004 , find:

(i) the population of the colony after 10 minutes, giving your answer to the nearest hundred.

(ii) the time after which there will be no insects left in the colony, giving your answer to the nearest 0.1 of a minute.
• May 16th 2009, 06:49 AM
e^(i*pi)
Quote:

Originally Posted by gamboo
7a

A differential equation is given by $\displaystyle \frac{dx}{dt}=-kte^\frac{x}{2}$, where k is a positive constant. Note: The power to e is x/2, tried making it bigger but still suck at using latex.

(i) Solve the differential equation.
(ii) Hence, given that x=6 when t=0, show that $\displaystyle x=-2\ln(\frac{kt^2}{4}+\frac{1}{e^3})$.

7b The population of a colony of insects is decreasing according to the model $\displaystyle \frac{dx}{dt}=-kte^\frac{x}{2}$, where x thousands is the number of insects in the colony after time t minutes. Initially, there were 6000 insects in the colony.

Given that k=0.004 , find:

(i) the population of the colony after 10 minutes, giving your answer to the nearest hundred.

(ii) the time after which there will be no insects left in the colony, giving your answer to the nearest 0.1 of a minute.

Separate the variables:

$\displaystyle \frac{dx}{dt}=-kte^\frac{x}{2}$

$\displaystyle \frac{dx}{e^\frac{x}{2}} = -kt dt$

Note that $\displaystyle \frac{1}{e^\frac{x}{2}} = e^{-\frac{x}{2}}$

$\displaystyle -2e^{-\frac{x}{2}} = -\frac{k}{2}t^2 +C$ where C is the constant of integration

Now we need to use a data point to solve for C. In this case we have x=6 and t=0

$\displaystyle -2e^{-\frac{6}{2}} = -2e^{-3}= C$ which makes the equation:

$\displaystyle -2e^{-\frac{x}{2}} = -\frac{k}{2}t^2 - 2e^{-3}$
• May 16th 2009, 06:59 AM
gamboo
Thanks, matches mine.

for b I got (i) 3800 insects
(ii) 30.8 minutes

can anyone confirm this for me
• May 16th 2009, 07:09 AM
e^(i*pi)
Quote:

Originally Posted by gamboo
Thanks, matches mine.

for b I got (i) 3800 insects
(ii) 30.8 minutes

can anyone confirm this for me

I got those two answers too