Undetermined Coefficients — Question 10

Question 10 – Solution

Strategy. Determine which forcing combination cancels the growing homogeneous mode after the initial data are fitted.

Step 1: Match and fit. The root $-1$ is simple, while $-2$ is not a root. Use $y_p=ate^{-t}+be^{-2t}$. Since $(D^2-1)(te^{-t})=-2e^{-t}$ and $(D^2-1)e^{-2t}=3e^{-2t}$, we obtain $a=-1/2$, $b=c/3$. Fitting the zero data gives $$\boxed{y=(\tfrac 14+\tfrac c6)e^t+(-\tfrac 14-\tfrac c2)e^{-t} -\tfrac 12te^{-t}+\tfrac c3e^{-2t}.}$$ The value is zero because the constant coefficients sum to zero; the slope is zero because the derivative contributions sum to zero.

Step 2: Cancel the unstable mode. Every term except the $e^t$ term tends to zero. Thus boundedness holds exactly when $1/4+c/6=0$, giving $\boxed{c=-3/2}$. The tuned response is $$\boxed{y_* =\tfrac 12e^{-t}(1-t-e^{-t}).}$$

Step 3: Prove sign and decay. For $t>0$, the supplied strict exponential inequality gives $1-t-e^{-t}<0$, so $y_*<0$. All its terms decay, hence $y_*\to 0$ from below. It has zero initial value and slope despite remaining strictly negative afterward.

Step 4: Examine imperfect cancellation. Replacing the tuned value by $-3/2+\delta$ makes the growing-mode coefficient $\delta/6$. More explicitly, the change in solution is $$\delta[e^t/6-e^{-t}/2+e^{-2t}/3].$$ Consequently $y(t)/e^t\to\delta/6$: the response tends to positive infinity for $\delta>0$ and negative infinity for $\delta<0$. The forcing still decays; the initial-data adjustment excites a growing homogeneous component unless the cancellation is exact.