Dimensionless Variables Help?

Question

Dynamics of human populations are sometimes modeled by the gompertz equation dy/dt= -ryln( k/y ) , where r and k are positive constants. If y(0) = y0 > 0 , formulate the initial value problem in terms of the dimensionless variables T=rt, and Y=y/y0. Use this formulation to distinguish and describe three qualitatively different regimes of growth or decay.

I am sorry that I have posted this question before but I made a slight error in the equation. I need as much help as possible on this question, but any sort of help would be greatly appreciated. Thank you very much in advance.

Answer 1

First substitute the dimensionless variables
Y = y / y₀ → dy = y₀· dY
T = r · t → dt = 1/r dT

The differential equation in terms of the dimensionless variables is:
dY/dT = - Y · ln{(k/y₀)/Y }
with the initial condition Y₀= 1

Solve the DE by separation of variables:
∫ -1/ [Y · ln{(k/y₀)/Y }] dY = ∫ dT

The integral on the right side can be solved with the substitution
u = ln{(k/y₀)/Y } <=> Y = k/y₀·exp{-u}
→ dY = -k/y₀·exp{-u} = -Y du
Hence:
∫ -1/ [Y · ln{(k/y₀)/Y }] dY
= ∫ 1/u du
= ln(u)
= ln(ln{(k/y₀)/Y })

Therefore the general solution of the DE is:
ln(ln{(k/y₀)/Y }) = T + C ( C is the constant of integration)
<=>
(ln{(k/y₀)/Y }) = C'·exp{T} (with C' = exp(C))
<=>
(k/y₀)/Y } = exp(-C'·exp{T})
<=>
Y = k/y₀· exp(C''·exp{T}) (with C'' = -C')

Apply initial condition to evaluate the constant:
1 = k/y₀· exp( C''·exp{0})
→
C'' = ln(y₀/k)

Hence the solution of the DE in dimensionless variables is:
Y = k/y₀· exp(ln(y₀/k) · exp{T})

The behavior of this function is determined by the value of the constant ln(y₀/k). There are three regimes:
(I)
y₀< k
→ ln(y₀/k) < 0
The solution is monotonically decreasing function with
Y→0 for T → ∞
(II)
y₀= k
→ ln(y₀/k) = 0
The solution is a constant function with Y=1
(III)
y₀> k
→ ln(y₀/k) > 0
The solution is monotonically increasing function with
Y→∞ for T → ∞

Answer 2

dy/dt= - r y ln( k/y )

yo Y = y T = r t

yo dY/dt = dy/dt and dT/dt = r
dY/dt = - r y ln( k/y )

dY/dT = - r y ln( k/y ) *1/r = - y ln( k/y )
dY/dT = - y ln (k / y ) putting y

dY/dT = - (yo Y) ln (k / Y yo)
dY/dT = (yo Y) [ - ln (k /yo) + ln Y]

(1/Y) dY/dT - yo ln Y = - yo ln (k /yo) ----(1)
substitution
z = ln Y so dz/dT = (1/Y) dY/dT

dz/dT - yo z = - yo ln (k /yo) ----(2)
it is 1st order linear differential equation
Integrating factor IF = exp ∫ P dT = exp ∫ -yo dT = exp(- yo T)
solution>>>
z * exp (- yo T) = - ∫ exp (- yo T) yo ln (k /yo) dT +C

z * exp (- yo T) = - yo ln (k /yo) ∫ exp (- yo T) dT +C
z * exp (- yo T) = yo ln (k /yo) [ exp (- yo T) / yo] +C

z * exp (- yo T) = ln (k /yo) [ exp (- yo T)] +C

z = ln (k /yo) + C exp (yo T) put back Y for z

ln Y = ln (k /yo) + C exp (yo T) -----(3)

given y(o)=y0 ie at t=0, y(o) = yo>>>
T=o, Y=y0/y0 =1>>>>>>>>>>>>>>>> this is catchy

anyway you email back>>>>>>>>>

ln 1 = ln (k /yo) + C exp (0) >>>> C = - ln (k /yo) put in (3)

ln Y = ln (k /yo) [1 - exp (yo T) -----(4)

this is relation between Y and T dimensionless variable

************************** you first confirm whether Y = 1 is a right approach then I will move forward. Email