Answer
$y=\frac{36}{1+8e^{-t}}$
Work Step by Step
Start by separating variables
$\frac{dy}{dt} = y(1-\frac{y}{36})$
$ \frac{1}{y(1-\frac{y}{36})} dy= dt$
Integrate both sides
$ \int\frac{1}{y(1-\frac{y}{36})} dy= \int dt$
before integrating the first part, we need to use partial fractions
$\frac{1}{y(1-\frac{y}{36})} = \frac{A}{y} + \frac{B}{1-\frac{y}{36}}$
Solving for A and B gives
$ A= 1, B=\frac{1}{36}$
Substituting A and B back into the Integration
$\int \frac{1}{y} + \frac{1}{36(1-\frac{y}{36})} = \int dt$
Integrate both sides to get
$\ln|y| - \ln|36-y| = t + C$
Now its time to isolate y
multiply both sides by -1
$-\ln|y| + \ln|36-y| = -t - C$
$\ln|\frac{36-y}{y}| = -t-C$
Exponentiate both sides
$|\frac{36-y}{y}| = e^{-t-C} $, Let $C=^+_-e^{-C}$ to get
$ \frac{36-y}{y}= Ce^{-t}$
$ 36-y= yCe^{-t} $
$36= y(1+Ce^{-t})$
$\frac{36}{y}= (1+Ce^{-t} )$
$y=\frac{36}{1+Ce^{-t} }$
Now lets find the particular solution at (0,4)
$4= \frac{36}{1+C}$
$1+C= 9$
$C=8$
Substitute back into the logistics function
$y=\frac{36}{1+8e^{-t}}$
