Real & Distinct Roots — Question 4

Question 4 – Solution

Strategy. Factor out a positive exponential and reduce sign questions to affine expressions in $z=e^{-t}\in(0,1]$.

Step 1: Fit the real-root solution. The roots are $-1,-2$. Solving $A+B=a$, $-A-2B=b$ gives $$\boxed{y=(2a+b)e^{-t}-(a+b)e^{-2t}}.$$ Write $A=2a+b$, $B=-(a+b)$, so $A+B=a$.

Step 2: Classify nonnegativity. Since $y=e^{-t}(A+Bz)$, its sign is the sign of an affine function with values $A$ at $z=0$ and $a>0$ at $z=1$. If $A\ge 0$, it is positive for all $0<z\le 1$. If $A<0$, it is negative for sufficiently small positive $z$. Therefore $$\boxed{y(t)\ge 0\text{ for all }t\ge 0\quad\Longleftrightarrow\quad b\ge-2a}.$$ At $b=-2a$, the solution is the strictly positive $ae^{-2t}$ at every finite time.

Step 3: Impose monotonicity as well. We have $y'=-e^{-t}(A+2Bz)$. Its bracket has endpoint values $A$ and $A+2B=-b$. Thus it is nonnegative on $0<z\le 1$ exactly when $A\ge 0$ and $b\le 0$. Combining the conditions yields $$\boxed{-2a\le b\le 0}.$$ At $b=0$, the derivative vanishes initially but is negative for every $t>0$. Both boundary slopes are included.

Step 4: Resolve the two remaining regimes. If $b<-2a$, solving $A+B e^{-t}=0$ gives the unique zero $$\boxed{t_z=\ln\frac{a+b}{2a+b}>0}.$$ Both numerator and denominator are negative, and their ratio exceeds 1. The solution crosses from positive to negative and tends to zero from below.

If $b>0$, it remains positive but initially rises. Solving $y'=0$ gives the unique positive-time maximum at $$\boxed{t_m=\ln\frac{2(a+b)}{2a+b}>0}.$$ The derivative changes from positive to negative there. These cases exhaust all real $b$; every solution still tends to zero because both roots are negative.