Question

The concentration \(p\) (in \(\mu g/dL\) i.e., micrograms/deciliter) of a hormone, as a function of time \(t\) (in hours) is governed by the following differential equation for \(t \geq 0\)

\[ \frac{dp}{dt} = e^{-0.1t} - 0.1\, p \]

If \(p(0) = 20\ \mu g/dL\), then \(p(10) =\) \(\mu g/dL\). (Round off to one decimal place)

Hint:

This is a linear first-order ODE; use the integrating factor \(e^{0.1t}\) (or Laplace transforms) to solve for \(p(t)\), then plug in \(t=10\).

Answer 1

Correct Answer: 11

Step 1: Recognize the equation type.
The equation \(\dfrac{dp}{dt} = e^{-0.1t} - 0.1p\) can be rewritten as:
\[ \frac{dp}{dt} + 0.1p = e^{-0.1t} \]
This is a first-order linear ordinary differential equation of the standard form \(\dfrac{dp}{dt} + P(t)p = Q(t)\), with \(P(t) = 0.1\) (a constant) and \(Q(t) = e^{-0.1t}\). Such equations are solved using an integrating factor.

Step 2: Find the integrating factor.
The integrating factor is \(\mu(t) = e^{\int P(t)\, dt} = e^{\int 0.1\, dt} = e^{0.1t}\). Multiplying both sides of the equation by \(\mu(t)\) makes the left side an exact derivative:
\[ \frac{d}{dt}\left( p\, e^{0.1t} \right) = e^{0.1t} \times e^{-0.1t} = e^{0} = 1 \]

Step 3: Integrate both sides.
\[ p\, e^{0.1t} = \int 1 \, dt = t + C \]
\[ p(t) = (t + C)\, e^{-0.1t} \]

Step 4: Use the initial condition to find C.
At \(t = 0\), \(p(0) = 20\):
\[ 20 = (0 + C)\, e^{0} = C \implies C = 20 \]
So the full solution is:
\[ p(t) = (t + 20)\, e^{-0.1t} \]

Step 5: Evaluate at t = 10.
\[ p(10) = (10 + 20)\, e^{-0.1 \times 10} = 30\, e^{-1} \]
Since \(e^{-1} \approx 0.3679\):
\[ p(10) \approx 30 \times 0.3679 = 11.04 \]
Rounded to one decimal place, \(p(10) \approx 11.0\ \mu g/dL\).

Final Answer:
\(p(10) \approx 11.0\ \mu g/dL\). \[ \boxed{p(10) \approx 11.0\ \mu g/dL} \]