Question

Please help me on this calculus

In a model of epidemics, the number of infected individuals in a population at a time is a solution of the logistic differential equation dy/dt=0.6y-0.0002y^2 , where y is the number of infected individual in the community and t is the time in day.
1. Describe the population for this situation.
2. Assume that 10 people were infected at the initial time t=0 . Find the solution for the differential equation.
3. How many day will it take for half of the population to be infected?

Answer 1

So we have: \[ \large
\frac{dy}{dt} = 0.6y-0.0002y^2
\]

Answer 2

It's seperable so:\[
\large
\begin{array}{rcl}
\frac{dy}{dt} &=& 0.6y-0.0002y^2 \\
\frac{1}{ 0.6y-0.0002y^2}\frac{dy}{dt} &=& 1 \\
\int \frac{1}{ 0.6y-0.0002y^2}\frac{dy}{dt} dt &=& \int 1 dt \\
\end{array}
\]Remember that the differential is defined such that: \(dy = \frac{dy}{dt}dt\):\[
\int \frac{1}{ 0.6y-0.0002y^2} dy = \int 1 dt

\]Now we just need to set up our limit of integration given initial conditions.

Answer 3

Actually, scratch that, we'll just take the anti derivative of both and solve for the constants of integration later on.

Answer 4

We're gonna want to use partial fractions, I think: \[
\begin{array}{rcl}
\int \frac{1}{ 0.6y-0.0002y^2} dy &=& \int 1 dt \\
\int \frac{1}{ y(0.6-0.0002y)} dy &=& t+C_1

\end{array}
\]

Answer 5

We need to solve for \(A\) and \(B\):\[

\begin{array}{rcl}
\frac{1}{ y(0.6-0.0002y)} &=& \frac{A}{y} + \frac{B}{0.6-0.0002y} \\
\frac{1}{ y(0.6-0.0002y)} &=& \frac{A(0.6-0.0002y)}{y(0.6-0.0002y)} + \frac{B(y)}{y(0.6-0.0002y)} \\
1 &=& A(0.6-0.0002y) + B(y) \\
1 &=& 0.6A-0.0002Ay + By \\
(1) + (0)y &=& 0.6A+(B-0.0002A)y \\
\end{array}
\]This gives us a system of equations: \[
\begin{array}{rclrl}
1 &=& &0.6&A \\
0 &=& B&-0.0002&A
\end{array}
\]

Answer 6

@algradilla Think you can do the rest?

Answer 7

I'm still confused on how I'm supposed to use the info that 10 people are infected when t=0