- #1

Kolika28

- 146

- 28

- Homework Statement
- A rumor is spread by a student 00.00 night to 12 of December at party. Two days later half of the students know the rumor. According to one model, the spread rate at time t (in hours after discovery) is proportional to the product of three factors

(1) the number of students knowing the rumor at time t, [i.e. how many that can disseminate information]

(2) the number of students who do not know the rumor at time t [i.e. How many who can receive the rumor] and

(3) 1 - sin (πt / 12) [students are least active at 06 in the morning].

We ignore that the number of students is an integer. We describe the process of a derivative function f (t) which describes the number of students who know the solution at time t (in hours). There are A> 1 students.

(a) Set up the initial value problem that describes the situation.

(b) Solve the initial value problem.

- Relevant Equations
- Differential equation

So this is what I have done:

##f'(t)=k*f(t)*(A-f(t))*(1-sin(\frac{pi*x}{12}))##

##\frac{1}{f(t)*(A-f(t))}=k*(1-sin(\frac{pi*x}{12}))##

I see that the left can be written as this (using partial fractions):

##1/A(\frac{1}{f(t)}-\frac{1}{A-f(t)})## And then I take the integral on both sides and get

##1/A*(ln(f(t))-ln(A-f(t))=k(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)##

##ln(\frac{f(t)}{A-f(t)})=kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)##

##\frac{f(t)}{A-f(t)}=e^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C}=Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}##

##f(t)=Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}*(A-f(t))##

##f(t)=A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}-f(t)*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}##

##f(t)+f(t)*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}=f(t)*(1+Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})})=A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}##

##f(t)=\frac{A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}}{(1+Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})})}##

I don't feel like I'm doing this right. I'm struggling with finding both C and k. What have I done wrong?

##f'(t)=k*f(t)*(A-f(t))*(1-sin(\frac{pi*x}{12}))##

##\frac{1}{f(t)*(A-f(t))}=k*(1-sin(\frac{pi*x}{12}))##

I see that the left can be written as this (using partial fractions):

##1/A(\frac{1}{f(t)}-\frac{1}{A-f(t)})## And then I take the integral on both sides and get

##1/A*(ln(f(t))-ln(A-f(t))=k(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)##

##ln(\frac{f(t)}{A-f(t)})=kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)##

##\frac{f(t)}{A-f(t)}=e^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C}=Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}##

##f(t)=Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}*(A-f(t))##

##f(t)=A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}-f(t)*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}##

##f(t)+f(t)*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}=f(t)*(1+Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})})=A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}##

##f(t)=\frac{A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}}{(1+Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})})}##

I don't feel like I'm doing this right. I'm struggling with finding both C and k. What have I done wrong?