91Ó°ÊÓ

Firstly, we are given a solution to the differential equation and initial value problem \(m(t) = \frac{I}{k}\left(1 - e^{-kt}\right)\) when \(m(0) = 0\). To verify this solution, we have to plug it into the given differential equation, and check if it satisfies the equation. The original differential equation is \(m^{\prime}(t) + k m(t) = I\). Now, we find the derivative of the given solution with respect to time, \(m^{\prime}(t)\). \(m^{\prime}(t) = \frac{d}{dt} \left(\frac{I}{k}\left(1 - e^{-kt}\right)\right) = \frac{I}{k} \frac{d}{dt} (1 - e^{-kt}) = \frac{I}{k} \cdot ke^{-kt}\) Now, we plug the given solution \(m(t)\) and its derivative \(m^{\prime}(t)\) into the differential equation: \(m^{\prime}(t) + k m(t) = \frac{I}{k} \cdot ke^{-kt} + k\left(\frac{I}{k}\left(1 - e^{-kt}\right)\right) = I \cdot e^{-kt} + I - Ik \cdot e^{-kt}\) As we can see, the terms \(I \cdot e^{-kt}\) and \(- Ik \cdot e^{-kt}\) cancel out, and we are left with just: \(I = I\) This means that the given solution \(m(t)\) is, in fact, correct. Now, we want to verify the initial condition \(m(0) = 0\): \(m(0) = \frac{I}{k}\left(1 - e^{-k(0)}\right) = \frac{I}{k}(1 - e^0) = \frac{I}{k}(1 - 1) = 0\) The initial condition is also satisfied.

We are asked to graph the solution \(m(t)=\frac{I}{k}\left(1-e^{-kt}\right)\) with the given values \(I=10 \mathrm{mg}/\mathrm{hr}\) and \(k=0.05\,\mathrm{hr}^{-1}\). Replacing these values gives: \(m(t) = \frac{10}{0.05} \left(1 - e^{-0.05t}\right) = 200\left(1 - e^{-0.05t}\right)\) To graph the solution, plot the function \(m(t) = 200\left(1 - e^{-0.05t}\right)\) over a suitable range of time values (e.g. \(t=0\) to \(t=50\,\mathrm{hr}\)). You should see the graph starting from zero and increasing until it reaches a steady-state value (which we'll find in the next step).

We are asked to find the steady-state drug level, which is the limit of the mass of the drug as time goes to infinity: \(\lim_{t \rightarrow \infty} m(t) = \lim_{t \rightarrow \infty} 200\left(1-e^{-0.05t}\right)\) As \(t \rightarrow \infty\), the exponent \(-0.05t \rightarrow -\infty\), so \(e^{-0.05t} \rightarrow 0\): \(\lim_{t \rightarrow \infty} 200\left(1 - e^{-0.05t}\right) = 200(1 - 0) = 200\,\mathrm{mg}\) This means the steady-state drug level is \(200\,\mathrm{mg}\). To verify this result using the graph from step 2, observe that the graph approaches a horizontal asymptote at \(m(t) = 200\,\mathrm{mg}\) as time goes to infinity.