help (1 Viewer)

[Arithela]

if dy/dt = ky, show that y = Ae^kt


i took the reciprocal

dt/dy = 1/ky
y = 1/k integral 1y
= 1/k lny + c

im stuck here

help plz!

 

[pLuvia]

dt=dy/ky
t=(lny)/k+C where C is a constant
kt-C=lny
y=e^kt-C
=Ae^kt where A=e^-C

 

[minijumbuk]

dy/dt = ky
1/(dt/dy) = ky
dt/dy= 1/ky
dt= dy/ky
t = ∫dy/ky
t = 1/k ∫dy/y
t = 1/k x ln y + C
kt = ln y + C (doesn't matter what you do to C, it is still a constant)
ln y = kt + C
e^{kt + C} = y
y= e^kt x e^C
But e^C is just a constant, so we call it A
y= e^kt x A
y = Ae^kt

 

[Arithela]

thanks i get it now

this question follows on from the one above:

Given that dM/dt = -kM (where k is a positive constant) and when t = 0, M = M_o, show that M=M_oe^-kt is a solution of this equation.

2. The population of a town is increasing continously at a rate proportional to the existing population. If the present population is 10000 and the population 5 years ago was 8000, determine the population in 5 years time.

 

[minijumbuk]

1. dM/dt = -kM_oe^-kt
But M= M_oe^-kt (from the question)
Therefore dM/dt = -kM
Therefore M= M_oe^-kt is a solution for dM/dt = -kM

2. The first sentence tells you this: dP/dt = kP, and then from that, you get P=P_oe^kt
Take 5 years ago as t=0
8000 = P_oe^-k(0)
Therefore P_o = 8000

Present is when t=5

10000 = 8000e^5k
k = ln (10000/8000) / 5
= some number, press it on your calculator

In 5 years time from the present, it means 10 years time from t=0, therefore t=10

P= 8000e^10k
Remember that you found k in the last step, so just sub it into here
Then you have P when t=10, then you're done.