Question

The logistic growth model formula is given by
$ye^k^t +Ay=Le^k^t$


where y is the population at time t (t\geq 0) and A, k and L are positive constants. Use implicit differentiation to verify that
A)$\frac{dy}{dt} = \frac{k}{L} y(L-y)$
B)$\frac{d^2}{dt^2} =\frac{k^2}{L^2} y(L-y)(L-2y)$

Answer 1

Given :   $ye^{kt} + Ay = Le^{kt}$

To find:    $dy/dt = (ky/L)(L - y)$   , $d^2y/dt^2 = (k^2/L^2) y(L -y)( L - 2y)$  

Solution:

$ye^{kt} + Ay = Le^{kt}$

=> $y(e^{kt} + A) = Le^{kt}$

on Differentiating wrt x

$(dy/dt)(e^{kt} + A) + y(ke^{kt}) = kLe^{kt}$

=> $(dy/dt)(e^{kt} + A) = kLe^{kt} - y(ke^{kt})$

=> $(dy/dt)(e^{kt} + A) = ke^{kt} (L - y)$

$y(e^{kt} + A) = Le^{kt}$  

=> $(e^{kt} + A) = Le^{kt}/y$

Substituting  $(e^{kt} + A) = Le^{kt}/y$  

=>$(dy/dt)(Le^{kt}/y) = ke^{kt} (L - y)$  

Cancelling $e^{kt}$  from both sides

=> $(dy/dt)(L/y) = ke^{kt} (L - y)$  

=> $dy/dt = (ky/L)(L - y)$  

QED

Hence Proved

$dy/dt = (ky/L)(L - y)$  

=> $d^2y/dt^2 = (ky/L)(- dy/dt) + (k/L)(dy/dt)(L - y)$  

=> $d^2y/dt^2 = (k/L)(dy/dt) (-y + L - y)$  

=> $d^2y/dt^2 = (k/L)(dy/dt) ( L - 2y)$  

Substitute $dy/dt = (ky/L)(L - y)$  

=> $d^2y/dt^2 = (k/L) (ky/L)(L -y)( L - 2y)$  

=> $d^2y/dt^2 = (k^2/L^2) y(L -y)( L - 2y)$  

QED

Hence Proved

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