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...
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?