# Modeling the Course of a Viral Illness

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## Homework Statement

When certain viral particles enter the body, they replicate to 160% every four hours and the immune system eliminates these particular viral particles at the rate of 50000 viral particles per hour. Find an equation modeling this viral growth.

N/A

## The Attempt at a Solution

Let $n$ be the function modeling this viral growth, then clearly

$$n_0(1.6^{t/4}) - 50000t \geq n(t) \geq (n_0 - 50000t)(1.6^{t/4})$$

since the first expression assumes that every particle replicates first and then 50000t particles are eliminated and the last expression assumes that all 50000t particles are eliminated and then they replicate. In reality, the particles are always replicating and being eliminated simultaneously, so $n(t)$ must be between these two extremes. Using this same train of thought, we know that . . .

$$(n_0 - 50000t)(1.6^{t/4}) \leq n(t)$$

$$[(n_0 - 50000t/2)(1.6^{t/8}) - 50000t/2](1.6^{t/8}) \leq n(t)$$

$$\vdots$$​

$$n_0(1.6^{t/4}) - 50000t \sum_{i=0}^{n-1} \frac{1.6^{\frac{(n-i)x}{n}}}{n} \leq n(t)$$

and

$$n_0(1.6^{t/4}) - 50000t \geq n(t)$$

$$(n_0(1.6^{t/8}) - 50000t/2)(1.6^{t/8}) - 50000t/2 \geq n(t)$$

$$\vdots$$​

$$n_0(1.6^{t/4}) - 50000t \sum_{i=1}^{n} \frac{1.6^{\frac{(n-i)x}{n}}}{n} \geq n(t)$$

Since $1.6^{t/4}$ is an integrable function and because the two sums above represent the upper and lower Darboux sums for $1.6^{t/4}$, it follows that

$$n(t) = n_0(1.6^{t/4}) - 50000t \int_0^t 1.6^{x/4} dx$$

I'm sorry that my work isn't perfectly clear at the moment, but seeing as I'm fairly certain that it's wrong, I'm not sure that it matters much. I would appreciate help actually starting this problem correctly (which I'm fairly certain that I haven't done). Any help is appreciated. Thanks!