how from ln[y/y-1] = kt+C to y = Ce^kt / (Ce^kt-1).??? If y(0)=a, what is the value of c??

Answer 1

e raised to each side gives you:

y / (y-1) = e ^ (kt+C)

Multiply both sides by (y-1)

y = (y-1) * e ^ (kt+C)

Distribute e ^ (kt+C) on the right hand side

y = y * e ^ (kt+C) - e ^ (kt+C)

Move all terms with y to one side and factor out y

y - y * e ^ (kt+C) = - e ^ (kt+C)

y (1-e ^ (kt+C)) = - e ^ (kt+C)

Divide

y = - e ^ (kt+C) / (1- e ^ (kt+C))

Now you can multiply through top the bottom by -1 to get

y = e ^ (kt+C) / (e ^ (kt+C) -1)

e ^ (kt+C) = e^(kt) * e^C and what they are doing here is replacing e^C by C as the new coefficient of e^(kt).

So you get y = Ce^kt / (Ce^kt-1) and now let t=0 since you are looking for y(0)=a.

So y = Ce^k0 / (Ce^k*0-1) = C / (C-1) = a

C / (C-1) =a so mutiply both sides by C-1

C = (C-1)a

C = Ca - a

a = Ca-C

a = C(a-1)

C = a/(a-1)

Answer 2

I'll start with what I had before, and I figure out the other half.
dy/dt = ky(1-y)
separation of variables
(1/y*(1-y)) dy = k dt
integrate
-ln ((y-1)/y) + c = kt +c
move c's to right side and apply both sides as power of e
(y/(y-1)) = e ^ (kt)
drop down c since e ^ (kt) = c*e ^ (kt)
y / (y - 1) = c*e ^ (kt)
multiply by y-1
y = (y - 1)*c*e ^ (kt)
distribute
y = y*c*e^(kt) - c*e ^ (kt)
move y*c*e^(kt) to left
y - y*c*e ^ (kt) = -c*e ^ (kt)
factor out y on left side
y (c*e ^ (kt) - 1) = c*e ^ (kt)
divide (e ^ (kt) - 1) to move to right side
y = c*e ^ (kt) / (c*e ^ (kt) - 1)

Yay!

So, for y(0) = a

a = c*e ^ (k(0)) / (c*e ^ (k(0)) - 1)
e^0 = 1 so,
a = c / (c-1)
multiply by (c-1)
a*(c-1) = c
I'll flip it around so it becomes better looking
c= a*(c-1)
expand it out
c = a*c - a
subtract a*c to move it to the left
c-a*c = -a
factor out the 1-a
c(1-a)=-a
divide and make this!!
c = a / (a-1)

Ok, now it's done. Hooray!!

Answer 3

You have:
ln[y/y-1] = kt+C
then apply exp function to both sides:
y/y-1 = exp(kt+C)
Then get y:
y = (y-1)exp(kt+C)
y - y*exp(kt+C)= -exp(kt+C)
y(1-exp(kt+C)) =-exp(kt+C)
y= exp(kt+C)/(exp(kt+C)-1)
y= exp(kt)*exp(C)/(exp(kt)*exp(C)-1)
Now y(0) assuming t is the variable:
y(0) = exp(0k + C)/(exp(0k + C) -1)
y(0) = exp(C)/(exp(C)-1) = a
Now you get C:
exp(C)= a(exp(C)-1)
exp(C)= a exp(C) -a
exp(C)-a exp(C)=-a
exp(C)(1-a) = -a
exp(C)= a/(a-1)
C= ln(a/(a-1))

Observer that since exp(C) is a constant you can redefine exp(C)= C, which is a common practice when solving differential equations, but is confusing. So I'd rather to call exp(C)=D, and thus previous analysis becomes:
y= exp(kt)*D/(exp(kt)*D-1)
y= D*exp(0t)/(D*exp(0t)-1)
y= D/(D-1) = a
Now you get D:
D= a*(D-1)
D= a*D - a
D-a*D=-a
D(1-a)=-a
D=a/(a-1)

Which is the solution for D in y= D*exp(kt)/(D*exp(kt)-1), and you can see that is the same if you keep D=exp(C) in your equation:
y= exp(kt)*exp(C)/(exp(kt)*exp(C)-1)
since when you replace C=ln(a/(a-1)) you get the same result than replacing D=a/(a-1) in y= D*exp(kt)/(D*exp(kt)-1).
Also you can see that we have conserved the relation D=exp(C):
a/(a-1) = exp [ ln(a/(a-1)]

Answer 4

First,

y / (y - 1) = e^(kt + C)
y = (y - 1)e^(kt + C)
y = ye^(kt + C) - e^(kt + C)
y - ye^(kt + C) = -e^(kt + C)
y (e^(kt + C) - 1) = e^(kt + C)
y = e^(kt + C) / (e^(kt + C) - 1)