Model for spread of infection, with vaccination I'm trying to solve following problem:


*

*$N = 10^6$ ... number of people

*$ir = 8\% $ ... infection rate

*time unit - 1 day


And when there are 3% of population infected, vaccination begins. Its effect is constant - 600 people is healed in one day.
When will be the population completely "clean"?

My attempt


*

*I(t) ... number of infected people in one day


I tried to put the equation together like this:
$$
\frac{dI(t)}{dt} = ir \cdot I(t) - 600
$$
I understand it that the change in the amount of infected people is determined by new infected people (the former amount increased by 8%) minus 600 healed people.
With specific values:
$$
\frac{d(I(t))}{dt} = 1.08 \cdot I(t) - 600
$$
And I know, that I'm not beginning from zero, but the 3% of population are already infected, so:
$$
I(0) = 0.03 \cdot N = 30 000
$$

I'm not sure if my attempt is even partially correct. Could you, please, correct me?

Edit 1
The equation is supposed to be like this:
$$
\frac{dI(t)}{dt} = ri \cdot \left( N - I(t) \right) - 600,
$$
because the remaining healthy people can be infected, not the already infected ones (as I've written in my previous equation).

Edit 2
I was trying to solve the equation, unfortunately, I'm stuck at the very beginning.
At first I separated the homogeneous equation:
$$
\frac{1}{ir \cdot (N - I(t))} dI = 1 dt
$$
Then I tried to integrate both its sides:
$$
\int \frac{1}{ir \cdot (N - I(t))} dI = t + C_1, C_1 \in \mathbb{R}
$$
Here I'm not sure about the integral on the left side:
\begin{align}
-\frac{1}{ir} \cdot ln(N - I(t)) &= t + C_1\\
ln(N - I(t))^{-\frac{1}{ir}} &= t+ C_1\\
(N-I(t))^{-\frac{1}{ir}} = C_1 \cdot e^{t}
\end{align}
Because now, when I want to substitute numbers I'll get this:
$$
(10^6 - I(t))^{\frac{1}{1.08}} = \frac{1}{C_1 \cdot e^t}
$$
And how am I supposed to get rid of the fraction in the exponent? As far as I know, I can't compute the root with decimal number...
 A: Hint: I would prefer the following differential equation:
$$\frac{dI(t)}{dt}=k\cdot (1,000,000-I(t))-600$$
The more people have been infected the less new infections can happen. 
A: Let's repeat the results so far (Eenoku & callculus):
$$
N = 1000000 \quad ; \quad k = 0.08 \quad ; \quad I_0 = 3000 \quad ; \quad H = 600 \\
\frac{dI(t)}{dt}=k \left[N-I(t)\right]-H
$$
Ansatz (suppose the solution has the following form):
$$
I(t) = C e^{-k\, t} + D
$$
Initial condition:
$$
I(0) = C + D = I_0 \quad \Longrightarrow \quad C = I_0 - D
\quad \Longrightarrow \\I(t) = (I_0 - D) e^{-k\, t} + D = I_0 + D\left[1 - e^{-k\, t}\right]
$$
Substitute into the differential equation:
$$
\frac{dI(t)}{dt}=k \left[N-I(t)\right]-H \quad \Longrightarrow \quad
-k\,C e^{-k\cdot t} = k\,N - k\,C e^{-k\, t} - k\,D - H
\\ \Longrightarrow \quad k\,N - k\,D - H = 0 \quad \Longrightarrow \quad D = N - H/k
$$
Plugging in the numbers:
$$
I(t) = 3000 + 992500 \times \left[ 1 - \exp(-0.08 \times t) \right]
$$
Seems a bit of an undesirable outcome to me. Something wrong with the medication?
