Differential equations?

Question

Paramecia in a pond sample are growing according to the principle of Malthusian growth (i.e. unlimited growth, with no food or space limitations). Initially, there are 1100 Paramecia. Four hours later, the population has 1400 individuals. Find the population of Paramecia as a function of time.

I tried exponential growth, but it doesn't seem to work. Please help!

Matt · Accepted Answer

Malthusian growth differential equation
dP/dt = k*P

1/P dP = k dt

∫ (1/P) dP = ∫ (k) dt

ln(P) = k*t + c

e^ln(P) = e^(k*t + c)

P = e^(k*t) * e^(c)

P = e^(k*t) * C ............ [ C is equal to e^c - it's a new arbitrary constant (for now) ]

P = C * e^(k*t)



When t = 0:
1100 = C * e^(k*0)

1100 = C * e^0 = C * 1 = C

1100 = C



Substituting back for C into the original equation:
P = 1100 * e^(k*t)



When t=4 and P = 1400:
1400 = 1100 * e^(k*4)

1400/1100 = e^(k*4)

ln( 1400/1100 ) = ln( e^(k*4) )

ln( 1400/1100 ) = k*4

ln( 1400/1100 ) / 4 = k



Substituting back for k in the original equation: 
P = 1100 * e^( ( ln(1400/1100) / 4 ) * t)

And done. Reducing 1400/1100 yields the final answer of:
P = 1100 * e^( ( ln(14/11) / 4 ) * t  )

? · Answer

(0,1100)
(4,1400)
=(300/4)
=75
1100=75^(x-a)
ln(1100)=-a ln(75)
a=-ln(1100)/ln(75)
=-1.62202310108
y=75^(x+1.622023)

? · Answer

123 you are incorrect, I don't see a single differential equation in your work.

Connor · Answer

http://imageshack.us/f/341/yahooo.png/

Differential equations?

4 respuestas