# Exponential growth differential equation

## Homework Statement

A biomedical company finds that a certain bacteria used for crop insect control will grow exponentially at the rate of 12% per hour. Starting with 1000 bacteria, how many will they have after 10h?

## Homework Equations

Topic is application of differential equations.
y=Ce^kt

## The Attempt at a Solution

This question is from the topic on application of differential equations.
I stumbled on the solution but i need to understand why it works and whether it is the correct method.

y+dy/dt=Ce^(0.12t)
at T=0, y=1000, dy/dt=0

1000+0=Ce^(0.12*0)
C=1000

Total T=y+dy/dt
T=1000e^(0.12t)

from here i just substitute in t=10 to get answer.

question is why exponential growth in this case uses base e and not other base? why does it work when C is initial value?

also is this the correct method when using differential equations to solve this problem? don't want the working to be affected just cause I know the formula is y=Ce^kt

You are making things more complicated than they have to be.

y=C exp(k t) is of course the right starting point.

y(t + 1 hr) = y(t) + 12% = y(t) * 1.12.

Plug that into above equation. C will drop out and you can solve for k.

Next, you have y(0) = 1000, which gives you C=1000 as you have already noted.

The final step is to calculate y(10 hrs).

Another way is to say that each hour the number is multiplied by 1.12, so after 10 hours,
you multiply by (1.12)^10.

Note that you get the wrong answer (you underestimate the number of bacteria) if you just say +12% per hour x 10 hours = +120%. This is linear growth instead of exponential.

i would use formula directly but the topic is on differential equations. Need to derive the formula from differential equations first. How do you do that?

Office_Shredder
Staff Emeritus
Gold Member
Is the question exactly as you worded it? The phrase 'will grow at a continuous rate of 12%' or 'will grow continuously with a relative rate of 12%' or something similar is often used to denote that the population satisfies the differential equation
dP/dt = kP where here k=.12. I strongly suspect that different courses will use slightly different terminology, so you should check carefully with how the words have been used in your class before

ok. now the problem is modified.

Number of bacteria is assumed to grow exponentially at the rate given by dN/dt = kN, where N is the number of bacteria at time t hour. The bacteria grow to double the original amount in 0.5 hour.

Find the equation for N as a function of t and the original amount N0.

dN/dt = kN

∫1/kN dN = ∫dt

ln (kN) = t+C

kN = et+C

kN = Cet

at t=0.5 , k=2

2N = Ce0.5

C = $\frac{2N}{e^0.5}$

substitute back C

N = 2N0et-0.5

don't know the correct answer, so i'm posting here if anyone can tell me if im correct... thanks!

Nope, check your second step, the integration. Should be ln(N)=kt+C. Then N=Cexp(kt), the standard expression for exponential growth or decay.