Real & Distinct Roots — Question 7

Question 7 – Solution

Strategy. Track the two root signs and retain the distinction between a mode being available and its coefficient actually being present.

Step 1: Keep the root separation visible. The characteristic polynomial is $(r+\alpha)^2-4$, so $$\boxed{r_+=2-\alpha,\quad r_-=-2-\alpha},\qquad r_+-r_-=4.$$ Consequently $y=Ae^{(2-\alpha)t}+Be^{(-2-\alpha)t}$ for every real $\alpha$; no repeated or nonreal roots occur.

Step 2: Require both modes to decay. Every solution tends to zero exactly when both roots are negative. Since $r_+$ is the larger root, this is equivalent to $$\boxed{\alpha>2}.$$ If $\alpha\le 2$, choosing $A=1$, $B=0$ gives a constant or growing mode, disproving universal decay.

Step 3: Allow a constant mode for boundedness. Every solution is forward bounded exactly when both roots are nonpositive, or $\boxed{\alpha\ge 2}$. At $\alpha=2$, $$y=A+Be^{-4t}\longrightarrow A.$$ All solutions are bounded, but only those with $A=0$ tend to zero. A zero root supplies a constant mode rather than a decaying one.

Step 4: Classify the selected bounded solutions. Using the coefficients $A,B$ from Step 1, the full classification is $$\begin{array}{c|l|l} \alpha&\text{forward-bounded solutions}&\text{their limits}\\\hline \alpha<-2&y=0&0\\ \alpha=-2&y=B\quad(A=0)&B\\ -2<\alpha<2&y=Be^{(-2-\alpha)t}\quad(A=0)&0\\ \alpha=2&y=A+Be^{-4t}&A\\ \alpha>2&\text{all solutions}&0 \end{array}$$ For $\alpha<-2$, both roots are positive. If $A\ne 0$, the faster mode forces unbounded magnitude; if $A=0$ and $B\ne 0$, the remaining positive mode still grows. Distinct growth rates cannot cancel for all large times.

At $\alpha=-2$ the roots are 4 and 0, so only constant solutions are bounded. For $-2<\alpha<2$, one positive mode must be removed, leaving a decaying mode. These boundary cases show why a classification based only on whether roots are real would be incomplete.