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)

Question

Find the general solution of the differential equation: $ \frac{dy}{dx} + \frac{y}{x} = x^2 $

Answer 1

An alternate way to solve this initial value problem is with the Laplace transform, which turns the differential equation into an algebraic one. Let $P(s)$ denote the Laplace transform of $p(t)$.

Taking the Laplace transform of both sides of $\dfrac{dp}{dt} + 0.1p = e^{-0.1t}$, and using $\mathcal{L}\{dp/dt\} = sP(s) - p(0)$ and $\mathcal{L}\{e^{-0.1t}\} = \dfrac{1}{s+0.1}$:

\[ sP(s) - p(0) + 0.1P(s) = \frac{1}{s+0.1} \]

Substitute $p(0) = 20$ and group the $P(s)$ terms:

\[ (s + 0.1)P(s) = 20 + \frac{1}{s+0.1} \]

Divide through by $(s+0.1)$ to isolate $P(s)$:

\[ P(s) = \frac{20}{s+0.1} + \frac{1}{(s+0.1)^2} \]

This is now a sum of two standard Laplace transform pairs. Recall that $\mathcal{L}^{-1}\left\{ \dfrac{1}{s+a} \right\} = e^{-at}$ and $\mathcal{L}^{-1}\left\{ \dfrac{1}{(s+a)^2} \right\} = t\,e^{-at}$. Applying these with $a = 0.1$:

\[ p(t) = 20\,e^{-0.1t} + t\,e^{-0.1t} = (t + 20)\,e^{-0.1t} \]

This matches the general solution obtained by other methods, confirming it is correct. Now evaluate at $t = 10$:

\[ p(10) = (10 + 20)\,e^{-1} = 30\,e^{-1} \approx 30 \times 0.3679 = 11.04 \]

Let's summarize:

The Laplace transform turns a linear ODE with an initial condition directly into an algebraic equation for $P(s)$.
Splitting $P(s)$ into recognizable transform pairs, $1/(s+a)$ and $1/(s+a)^2$, avoids doing the time-domain integration by hand.
Both the integrating-factor method and the Laplace method land on the same formula, $p(t) = (t+20)e^{-0.1t}$.

$p(10) \approx 11.0\ \mu g/dL$.

Answer 2