Real & Distinct Roots — Question 8

Question 8 – Solution

Strategy. A mixture of two exponential modes need not have a constant effective decay rate; use its logarithmic derivative to quantify the change.

Step 1: Recover and check the equation. The roots are $-1,-3$, so the monic characteristic polynomial is $r^2+4r+3$. Hence $\boxed{y''+4y'+3y=0}$. Direct differentiation gives $y'=-e^{-t}-27e^{-3t}$, so $$\boxed{y(0)=10,\qquad y'(0)=-28}.$$ The response is positive because both mode terms are positive.

Step 2: Read the rate as a changing weighted average. Let $z=9e^{-2t}>0$. Then $$R=\frac{e^{-t}+27e^{-3t}}{e^{-t}+9e^{-3t}} =\frac{1+3z}{1+z}.$$ This averages 1 and 3 with positive weights $1,z$, so $1<R<3$. In particular, $$\boxed{R(0)=14/5},\qquad \boxed{R\to 1},\qquad R=2\ \Longleftrightarrow\ z=1\ \Longleftrightarrow\ \boxed{t=\ln 3}.$$

Step 3: Derive the rate equation. Using $y''=-4y'-3y$ and $y'/y=-R$, $$R'=-\frac{y''}{y}+(y'/y)^2 =R^2-4R+3=(R-1)(R-3)<0.$$ Thus $R$ is strictly decreasing; its long-time limit is the slower mode’s rate, rather than an unchanging average of the roots.

Step 4: Compare with a constant-rate prediction. Set $f=\ln y$. Then $f'=-R$ and $f''=-R'>0$. Since $f'$ is strictly increasing, integration on $[0,t]$ for $t>0$ gives $$f(t)>f(0)+t f'(0)=\ln 10-\frac{14}{5}t.$$ Exponentiation preserves this inequality, so $\boxed{y(t)>10e^{-(14/5)t}}$ for every $t>0$. The extrapolation matches the initial value and slope but decays too quickly afterward. The changing mode weights, not an error in the initial data, cause the discrepancy.