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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)*(Af(t))*(1sin(\frac{pi*x}{12}))##
##\frac{1}{f(t)*(Af(t))}=k*(1sin(\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}{Af(t)})## And then I take the integral on both sides and get
##1/A*(ln(f(t))ln(Af(t))=k(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)##
##ln(\frac{f(t)}{Af(t)})=kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)##
##\frac{f(t)}{Af(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})}*(Af(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)*(Af(t))*(1sin(\frac{pi*x}{12}))##
##\frac{1}{f(t)*(Af(t))}=k*(1sin(\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}{Af(t)})## And then I take the integral on both sides and get
##1/A*(ln(f(t))ln(Af(t))=k(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)##
##ln(\frac{f(t)}{Af(t)})=kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)##
##\frac{f(t)}{Af(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})}*(Af(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?