Repeated Roots — Question 5

Question 5 – Solution

Strategy. Treat the zero root separately and distinguish decay in one time direction from boundedness in both directions.

Step 1: Retain both repeated-root freedoms. For every $r$, the family is $\boxed{y=(A+Bt)e^{rt}}$. At $r=0$ this becomes the affine family $y=A+Bt$ solving $y''=0$, not merely the constant family.

Step 2: Classify forward behavior. For $r<0$, both $e^{rt}$ and $te^{rt}$ tend to zero, so every solution decays and is forward bounded. For $r=0$, boundedness requires $B=0$, while decay to zero requires $A=B=0$. For $r>0$, only the zero solution is forward bounded or tends to zero. $$\begin{array}{c|l|l} r&\text{forward-bounded solutions}&\text{solutions tending to zero}\\\hline r<0&\text{all solutions}&\text{all solutions}\\ r=0&y=A&y=0\\ r>0&y=0&y=0 \end{array}$$ For negative $r$, boundedness follows from the finite limit and continuity on every finite initial interval.

Step 3: Inspect the other time direction. If $B\ne 0$, the magnitude of $A+Bt$ is at least $|B||t|/2$ for sufficiently large $|t|$. If $B=0$ but $A\ne 0$, it is a nonzero constant. Thus neither nonzero case can cancel the exponential in its growing direction.

When $r>0$, that direction is $t\to\infty$; when $r<0$, it is $t\to-\infty$. Hence the complete two-sided classification is $$\boxed{r\ne 0:\ y=0\text{ only}},\qquad \boxed{r=0:\ y=A\text{ for any constant }A}.$$

Step 4: Explain the exceptional repeated zero. A single zero root paired with a negative root gives a constant mode plus a decaying exponential, so every solution is forward bounded. A repeated zero root additionally gives the unbounded mode $t$.

For example, $\boxed{y(t)=t}$ solves $y''=0$, whose roots are both zero and therefore nonpositive, but is unbounded forward. Root signs alone are insufficient at a repeated zero; the polynomial multiplier must be included. For a strictly negative repeated root, the polynomial affects the rate and shape but not eventual decay.