# 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...

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.

*print*Print*list*Cite

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.`**