Separable Equations — Question 7

Question 7 – Solution

Strategy. Compare passage-time integrals on two concentration intervals related by scaling; the unknown constant $k$ then cancels.

Step 1: Write the two passage times. Since $y>0$ and $k>0$, the concentration decreases strictly. Separation gives $$1=\frac 1k\int_8^{16}s^{-p}\,ds,\qquad 1=\frac 1k\int_4^8s^{-p}\,ds.$$ These integrals are positive and finite for every real $p$.

Step 2: Recover the exponent. In the first integral put $s=2u$: $$\int_8^{16}s^{-p}\,ds=2^{1-p}\int_4^8u^{-p}\,du.$$ The observed equal passage times force $2^{1-p}=1$, hence $\boxed{p=1}$. Conversely, $p=1$ makes the two integrals equal. No division by $1-p$ was used, so the logarithmic case was retained throughout.

Step 3: Determine the rate and prediction. For $p=1$, $$k=\int_8^{16}\frac{ds}{s}=\ln 2,\qquad \ln\frac{y}{16}=-t\ln 2.$$ Therefore $\boxed{y(t)=16\,2^{-t}}$. It gives exactly $16,8,4$ at the measured times, and differentiating gives $y'=-(\ln 2)y$. It is positive for every real $t$. The value $1$ is first reached at $\boxed{t=4}$ because the solution is strictly decreasing.

Step 4: Identify the information supplied by the third observation. With only $y(0)=16$ and $y(1)=8$, each real $p$ could be paired with $$k(p)=\int_8^{16}s^{-p}\,ds>0.$$ The corresponding separated solution decreases from $16$ to $8$ in one unit of time and stays positive during that interval. Thus infinitely many exponents fit the first two data; the second measured halving time identifies the exponent within this model family.