Solving the Hormone Concentration Differential Equation

The given differential equation is $\frac{dp}{dt} = e^{-0.1t} - 0.1p$. This is a first-order linear differential equation. We rewrite it in the standard form $\frac{dp}{dt} + P(t)p = Q(t)$.

Rearranging the equation gives: $ \frac{dp}{dt} + 0.1p = e^{-0.1t} $ Here, $P(t) = 0.1$ and $Q(t) = e^{-0.1t}$.

Calculating the Integrating Factor

The integrating factor (IF) is calculated using the formula $IF = e^{\int P(t) dt}$.

For this equation, the integrating factor is: $ IF = e^{\int 0.1 dt} = e^{0.1t} $

Applying the Integrating Factor

Multiply the standard form equation by the integrating factor $e^{0.1t}$: $ e^{0.1t} \left( \frac{dp}{dt} + 0.1p \right) = e^{0.1t} \cdot e^{-0.1t} $ The left side is the derivative of the product $(p \cdot IF)$: $ \frac{d}{dt} (p e^{0.1t}) = e^{0} $ $ \frac{d}{dt} (p e^{0.1t}) = 1 $

Integrate both sides with respect to $t$: $ \int \frac{d}{dt} (p e^{0.1t}) dt = \int 1 dt $ $ p e^{0.1t} = t + C $ where $C$ is the constant of integration.

Solving for $p(t)$: $ p(t) = (t + C)e^{-0.1t} $

Applying Initial Conditions

We are given the initial condition $p(0) = 20\ \mu$g/dL. Substitute $t=0$ and $p=20$ into the equation for $p(t)$ to find $C$.

$ 20 = (0 + C)e^{-0.1 \times 0} $ $ 20 = C e^0 $ $ 20 = C $

So, the particular solution is: $ p(t) = (t + 20)e^{-0.1t} $

Calculating Concentration at t=10

Now, we need to find the concentration $p$ at $t=10$ hours. Substitute $t=10$ into the particular solution.

$ p(10) = (10 + 20)e^{-0.1 \times 10} $ $ p(10) = (30)e^{-1} $ $ p(10) = \frac{30}{e} $

Using the approximate value of $e \approx 2.71828$: $ p(10) \approx \frac{30}{2.71828} \approx 11.036 $

Rounding the result to one decimal place, we get $p(10) \approx 11.0\ \mu$g/dL. This value lies within the given correct answer range of 10.8 to 11.2.