Write the solution of the initial value problem and use it to find the population when t=20 ? (dP)/(dt)=0.1P(1-P/2000) P(0)=100 When does the population reach 1200? Please explain as thorough as possible, thank you.

This is a separable differential equation. This means we can completely separate dependent and independent variables into two expressions. General form if such equation is
dy/dx=f(x)g(y)
and the solution is obtained by solving the following integrals
int dy/g(y)=int f(x)dx+C
Let us now return to the problem at hand.
dy/dt=0.1P(1-P/2000)
Now we need to put everything containing P on the left and everything containing t to the right side.
(dP)/(0.1P-P^2/20000)=dt
Let us first simplify the expression on the left.
(dP)/((2000P-P^2)/20000)=dt
(20000dP)/(P(2000-P))=dt
We shall write the term on the left using partial fractions to make integration easier.
(20000dP)/(P(2000-P))=A/P+B/(2000-P)
2000A=20000 
B-A=0
A=B=10
(2000dP)/P(2000-P)=10/P+10/(2000-P)
We can now integrate the equation.
int 10/P dP+int 10/(2000-P)dP=int dt
10ln P-10ln(2000-P)+10ln C=t

In the line above we have written the constant term as 10ln C in stead of just C. This is often used to make the expression easier to manipulate.
ln P-ln(2000-P)+ln C=t/10
Use formulae for logarithm of product and quotient:
log_a (xy)=log_a x+log_a y
log_a(x/y)=log_a x-log_a y
ln((CP)/(2000-P))=t/10
Take antilogarithm.
(CP)/(2000-P)=e^(t/10)
CP=2000e^(t/10)-e^(t/10)P
P(C+e^(t/10))=2000e^(t/10)
P=(2000e^(t/10))/(C+e^(t/10))
We can now calculate C by using the initial value.
P(0)=100
(2000e^(0/10))/(C+e^(0/10))=100
Since e^0=1, we have
2000/(C+1)=100
C+1=2000/100
C=20-1
C=19
The solution to the initial value problem is
P(t)=(2000e^(t/10))/(19+e^(t/10))   
We can now calculate population when t=20.
P(20)=(2000e^2)/(19+e^2)approx560
Population is approximately 560 at time t=20.
 
To find when the population reaches 1200, we need to solve the following equation
(2000e^(t/10))/(19+e^(t/10))=1200
Multiply by the denominator.
2000e^(t/10)=22800+1200e^(t/10)
800e^(t/10)=22800
e^(t/10)=28.5 
Take logarithm.
t/10=ln28.5
t=10ln28.5approx33.499  
The population will reach 1200 at time t=33.499.