Modeling — Question 2

Question 2 – Solution

Strategy. Distinguish retained mass from cumulative elimination; integrated compartment balances recover exposure without integrating long formulas.

Step 1: Write transfer and loss rates. Well-mixed compartments and constant, unsaturated first-order rate constants give $\boxed{x'=-2x+y,\ y'=x-y}$. The numerical coefficients have units h$^{-1}$. Adding gives $(x+y)'=-x$: internal transfers cancel, but central elimination removes drug. Amounts, rather than concentrations, are the states; no unspecified compartment volume is needed for these laws.

Step 2: Establish the physical state constraints. On $x=0,y\ge 0$, $x'=y\ge 0$; on $y=0,x\ge 0$, $y'=x\ge 0$. These inward boundary directions and uniqueness preserve the nonnegative quadrant. With $e'=x$, $e(0)=0$, $\boxed{x+y+e=D}$. Thus retained amounts lie between $0$ and $D$, and $e$ is nondecreasing. The third state records removal, not return flow.

Step 3: Resolve the two decay modes. The roots are $\lambda_\pm=(-3\pm\sqrt 5)/2$, both negative. Solving for the constants gives, with $r=\sqrt 5$, $$\boxed{x=D e^{-3t/2}\bigl(\cosh(rt/2)-r^{-1}\sinh(rt/2)\bigr), \quad y=\frac{2D}{r}e^{-3t/2}\sinh(rt/2).}$$ These give $x(0)=D,y(0)=0,x'(0)=-2D,y'(0)=D$. Differentiating the formulas gives $x'=-2x+y$ and $y'=x-y$ for all $t$. Since $r<3$, both modal exponentials decay: $x,y\to 0$ and $e\to D$.

Step 4: Integrate the balances. Decay makes both integrals finite. Integrating $(x+y)'=-x$ gives $\int_0^\infty x\,dt=D$ numerically, and integrating $y'=x-y$ then gives $\int_0^\infty y\,dt=D$. Restoring the rate units, $$\boxed{\int_0^\infty x(t)\,dt=\int_0^\infty y(t)\,dt=D\text{ mg}\cdot\text{h}.}$$ These are amount–time exposures in mg$\cdot$h, not eliminated masses. Their sum is $2D$ mg$\cdot$h of total residence exposure, while the eliminated mass is only $D$ mg. A molecule can contribute residence time in both compartments.