Repeated Roots — Question 3

Question 3 – Solution

Strategy. Reduce both sign and monotonicity to affine functions of time after factoring out a positive exponential.

Step 1: Resolve the data. The root is $-\lambda$ twice. Writing $y=(A+Bt)e^{-\lambda t}$ gives $A=a$, $B=b+\lambda a$. Thus $$\boxed{y=(a+(b+\lambda a)t)e^{-\lambda t}}.$$ For brevity let $B=b+\lambda a$.

Step 2: Classify nonnegativity. The multiplier starts at $a>0$. It stays positive at every finite nonnegative time exactly when $B\ge 0$; if $B<0$, it crosses zero and becomes negative. Therefore $$\boxed{y(t)\ge 0\text{ for all }t\ge 0\quad\Longleftrightarrow\quad b\ge-\lambda a}.$$ At equality, the polynomial factor is constant and $y=ae^{-\lambda t}$.

Step 3: Impose nonincrease as well. Differentiation gives $y'=e^{-\lambda t}(b-\lambda Bt)$. With $B\ge 0$, this derivative is nonpositive for all $t\ge 0$ exactly when $b\le 0$. Hence $$\boxed{-\lambda a\le b\le 0}.$$ At $b=-\lambda a$, the response is a decreasing pure exponential. At $b=0$, it has zero initial slope but negative derivative for every $t>0$ because $B=\lambda a>0$.

Step 4: Classify the exterior ranges. If $b<-\lambda a$, then $B<0$, and the unique zero is $$\boxed{t_z=-\frac{a}{b+\lambda a}>0}.$$ The response crosses from positive to negative and tends to zero from below. If $b>0$, then $B>0$ and the unique maximum occurs at $$\boxed{t_m=\frac{b}{\lambda(b+\lambda a)}>0}.$$ Here $y'$ changes from positive to negative, and the response remains positive. The repeated root produces an affine multiplier; the zeros of that multiplier and its differentiated counterpart govern these thresholds. All cases still decay to zero.