Final Thoughts — Question 5

Question 5 – Solution

Strategy. A trajectory tests a model only along one curve in the $(x,y)$-plane. Compare predictions away from that curve.

Step 1: Agreement on the observed curve. Since $p'=-e^{-x}=-p$, model (A) holds. Along $y=p(x)$, the added term in (B) is $(p-e^{-x})^2=0$, so (B) also gives $p'=-p$. Both match the entire observation and its initial value exactly.

Step 2: Different off-curve predictions. At $x=0$, $e^{-x}=1$. Therefore $$\begin{array}{c|cc} \text{Point}&\text{Slope from (A)}&\text{Slope from (B)}\\\hline (0,0)&0&1\\ (0,2)&-2&-1 \end{array}$$ Thus $\boxed{\text{the models agree on }p\text{ but are different laws.}}$

Step 3: An equilibrium distinguishes them. For $y\equiv 0$, the derivative is zero. Model (A) assigns zero, so this is an equilibrium. Model (B) assigns $e^{-2x}>0$, so the same constant function is not a solution of (B).

Step 4: General non-identifiability. For any smooth $H$, the model $$\boxed{y'=-y+H(x,y)(y-e^{-x})^2}$$ admits $p=e^{-x}$, because the added term vanishes identically along that curve. Many choices of $H$ change slopes elsewhere. Additional experiments with different initial data or restrictions on the allowed model class are needed to distinguish such laws. Complete data on one trajectory constrain a curve, not the whole plane.