Differential Equation: Application of D.E: Population Growth

Differential Equation: Application of D.E.: Population Growth
A bacterial population B is known to have a rate of growth proportional to (B + 25). Between noon and 2PM the population increases to 3000 and between 2PM and 3PM the population is increased by 1000 in culture. (a) Find an expression for the bacterial population B as a function of time. (b) What is the initial bacterial population in the culture? (c) What is the total bacterial population in the culture at 4:15PM?

$\dfrac{dB}{dt} = k(B + 25)$

$\dfrac{dB}{B + 25} = k \, dt$

$\displaystyle \int \dfrac{dB}{B + 25} = k \int dt$

$\ln (B + 25) = kt + C$
 

At 2:00PM, t = 2 and B = 3000
$\ln 3025 = 2k + C$   ←   eq. (1)
 

At 3:00PM, t = 3 and B = 4000
$\ln 4025 = 3k + C$   ←   eq. (2)
 

From eq. (1) and eq. (2)
$k = 0.2856$

$C = 7.4434$
 

Hence,
$\ln (B + 25) = 0.2856t + 7.4434$       answer for (a)
 

At noon, t = 0
$\ln (B + 25) = 7.4434$

$B = 1683$       answer for (b)
 

At 4:15PM, t = 4.25
$\ln (B + 25) = 0.2856(4.25) + 7.4434$

$B = 5726$       answer for (c)
 

Another solution (By Calculator - CASIO fx-991ES PLUS):
MODE 3 5

AC
$B + 25 = 0\hat{y}$

$B + 25 = 1708$

$B = 1683$       answer for (b)
 

$B + 25 = 4.25\hat{y}$

$B + 25 = 5751$

$B = 5726$       answer for (c)

Note:
$\hat{y}$ = SHIFT 1 5 5