Undetermined Coefficients — Question 5

Question 5 – Solution

Strategy. Resonance adds a polynomial factor, but the sign of the exponential rate still controls long-time decay.

Step 1: Identify the resonance. The roots are $-1\pm 2i$. Thus the exponential-trigonometric forcing matches a simple complex root pair, and a suitable trial is $te^{-t}(A\cos 2t+B\sin 2t)$. Differentiation gives $$(e^{-t}F)''+2(e^{-t}F)'+5e^{-t}F=e^{-t}(F''+4F).$$

Step 2: Match and fit. For $F=t(A\cos 2t+B\sin 2t)$, the transformed residual is $-4A\sin 2t+4B\cos 2t$. Hence $A=0$, $B=1/4$. The particular solution and its first derivative vanish at zero, so $$\boxed{y_0=\tfrac 14te^{-t}\sin 2t.}$$ The full family is $e^{-t}[C\cos 2t+D\sin 2t+(t/4)\sin 2t]$. Zero initial data force $C=D=0$.

Step 3: Prove decay for every solution. For fixed $C,D$ and $t\ge 0$, its absolute value is at most $e^{-t}(|C|+|D|+t/4)$, which tends to zero. The resonant factor $t$ does not overcome the negative exponential rate.

Step 4: Distinguish an envelope bound from an attained maximum. We have $|y_0(t)|\le te^{-t}/4$. The envelope attains its unique maximum at $t=1$, since its derivative is $e^{-t}(1-t)/4$. Thus $$\boxed{|y_0(t)|\le 1/(4e).}$$ Equality would require both $t=1$ and $|\sin 2t|=1$. But $|\sin 2|<1$, so the response never attains this bound. It is a valid envelope estimate, not the exact largest displacement.