Table Of Laplace Transforms — Question 8

Question 8 – Solution

Strategy. Divide an improper rational expression before consulting ordinary-function rows. Its polynomial part may represent an impulse.

Step 1: Identify the input class. For an ordinary candidate continuous near zero and of exponential order, its Laplace transform tends to zero as $s\to\infty$: the bounded initial piece has integral $O(1/s)$ and the exponentially bounded tail tends to zero. But $s/(s+1)\to 1$. Polynomial division gives $$F=1-\frac 1{s+1},\qquad \boxed{f=\delta(t)-e^{-t}}.$$ The impulse supplies the nonvanishing constant term.

Step 2: Transform the equation and account for the jump. With the stated pre-impact convention, $(s+2)Y=F$, so $$Y=\frac{s}{(s+1)(s+2)}=-\frac 1{s+1}+\frac 2{s+2}, \qquad \boxed{y=2e^{-2t}-e^{-t}\quad(t\ge 0).}$$ Write $[y]_0=y(0^+)-y(0^-)$. Integrating across zero gives $[y]_0=1$, consistent with $y(0^+)=1$. The negative exponential input has no impulse mass.

Step 3: Find the sign changes and area. The unique positive zero is $t=\ln 2$. The derivative is $y'=e^{-t}-4e^{-2t}$, negative before $\ln 4$ and positive afterward. Thus the global minimum is $\boxed{y(\ln 4)=-1/8}$. The area is $2/2-1=\boxed{0}$: positive and negative parts cancel. For comparison, the signed input area is $1-1=0$ when the full initial impulse is included.

Step 4: Verify the ordinary restart and convergence. For $t>0$, direct substitution gives $y'+2y=-e^{-t}$. The same response solves this ordinary equation with $y(0^+)=1$. Adding another initial impulse to that restart would count it twice. The tail is asymptotic to $-e^{-t}$, so the exact real transform domain is $\boxed{s>-1}$. The figure shows only the post-impact branch on $t\ge 0$; the filled initial point records the prescribed right-hand value.