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Question

A population of bacteria given by y(t) grows according to the equation dy/dt=ky, where k is a constant and t is measured in minutes. If y(10)=10 and y(30)=25, what is the value of K?

Answer 1

dy/dt = ky is a differential equation with separable variables. Therefore, dy/y = k dt. Integrating both sides, we get ln(y) = kt + C, where C is the constant integration. So, y = e^(kt + C) = e^C e^kt = K e^(kt), where K = e^C is also a constant

According to the conditions,

y(10) = Ke(10k) = 10
y(30) = Ke(30k) = 25. Since K >0. it follows that

e(30k)/e^(10k) = e^(20k) = 2.5 => 20k = ln(2.5) => k = ln(2.5)/20 =~ 0.045814537

Answer 2

Solve the differential equation:

dy/dt = ky
dy = ky dt
(1/y) dy = (k) dt
∫(1/y) dy = ∫(k) dt
Ln(y) + c = kt + c

Solve for y, redefining c as "some other constant" along the way:
Ln(y) = kt + c
y = e^(kt + c)
y = e^(kt) e^(c)
y = c e^(kt)

You're given two sets of values that satisfy the equation, so you should be able to use these to solve for c and k.

Answer 3

The best way to do this problem would be by using the exponential growth formula (y = y(sub 0) + e^(kt)).

y(sub 0) = the y-value when t=0
e = Euler's number
k = the constant
t = time (independent variable)

Since you know two points on the line, (10,10) & (25,30), you can plug each in to deduce the k value.

(Step #1) 10 = y(sub 0) + e^(10k)

(Step #2) Then solve for y(sub 0) --> y(sub 0) = 10 - e^(10k)

(Step #3) Now plug in the other point (25,30) --> 30 = y(sub 0) + e^(25k)

(Step #4) Plug in the value for y(sub 0) found in Step #2.
30 = 10 - e^(10t) + e^(25k)

(Step #5) Now solve for k.
20 = e^(15k).
ln 20 = 15k
k = (ln 20)/15

And there's your answer! Hope my explanation was helpful!