Laplace Transforms — Question 1

Question 1 – Solution

Strategy. Collect every initial contribution before exploiting the repeated shifted pole.

Step 1: Transform all derivatives. The derivative transforms are $s^4Y-s^3$, $s^3Y-s^2$, $s^2Y-s$, and $sY-1$. Thus $$(s+1)^4Y=s^3+4s^2+6s+4.$$ The student omitted the initial terms from $4y'''+6y''+4y'$.

Step 2: Use shifted powers. Writing $z=s+1$ makes the numerator $z^3+z^2+z+1$. Therefore $$Y=\frac 1z+\frac 1{z^2}+\frac 1{z^3}+\frac 1{z^4},\qquad \boxed{y(t)=e^{-t}\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}\right).}$$

Step 3: Check the state and shape. For $y=e^{-t}p(t)$, $(D+1)^4y=e^{-t}p^{(4)}=0$. Moreover $y'=-t^3e^{-t}/6$, so $y'(0)=y''(0)=y'''(0)=0$ and $y(0)=1$. Every term in $p$ is positive on $t\geq 0$; hence $y>0$ and $y'<0$ for $t>0$.

Step 4: Check convergence and limits. The transform converges absolutely for $\operatorname{Re}s>-1$. Also $\lim_{s\to\infty}sY=1$ and $\lim_{s\to 0}sY=0$. The final-value theorem applies because all poles of $sY$ lie at $-1$. Polynomial growth from multiplicity is dominated by $e^{-t}$, so $y(t)\to 0$.