Inverse Laplace Transforms — Question 4

Question 4 – Solution

Strategy. An irreducible quadratic requires a linear numerator. The transient and persistent parts then have different long-time roles.

Step 1: Resolve the real factors. Use $$F=\frac A{s+1}+\frac{Bs+C}{s^2+4}.$$ Clearing denominators gives $5=A(s^2+4)+(Bs+C)(s+1)$, so $A+B=0$, $B+C=0$, $4A+C=5$. Hence $A=1$, $B=-1$, $C=1$.

Step 2: Invert each numerator term. Since the sine pair has numerator two, $$\boxed{f(t)=e^{-t}-\cos 2t+\tfrac 12\sin 2t.}$$ The forward numerator is $(s^2+4)-s(s+1)+(s+1)=5$, giving the required transform. A constant numerator over the quadratic alone would have omitted the cosine contribution.

Step 3: Check the onset cancellations. Direct differentiation gives $$f'=-e^{-t}+2\sin 2t+\cos 2t,\qquad f''=e^{-t}+4\cos 2t-2\sin 2t.$$ Therefore $\boxed{f(0)=0,\ f'(0)=0,\ f''(0)=5}$. Also $F(s)=5/s^3+O(s^{-4})$: the $1/s$ and $1/s^2$ terms vanish. Repeated integration by parts connects precisely these coefficients to the displayed initial values. In time, the first nonzero Taylor term is $5t^2/2$.

Step 4: Identify the persistent oscillation. The exponential term decays, but $-\cos 2t+\tfrac 12\sin 2t$ does not. In fact, $$f(n\pi)\longrightarrow-1,\qquad f(n\pi+\pi/2)\longrightarrow 1.$$ Thus $f$ has no limit at infinity, although it is bounded. Boundedness gives absolute convergence for $s>0$. At $s=0$, the primitive contains the nonconstant periodic term $-\sin 2R/2+(1-\cos 2R)/4$, so it has no limit. For $s<0$, the persistent sinusoid has positive fixed-length intervals with exponentially growing weighted integrals; the decaying time term cannot cancel it on late such intervals. Consequently the exact domain is $\boxed{s>0}$.