Laplace Transforms — Question 2

Question 2 – Solution

Strategy. Parameter differentiation resolves repeated quadratic poles while a Taylor check tests the high-order initial data.

Step 1: Transform the sixth-order problem. The equation is $y^{(6)}+3y^{(4)}+3y''+y=0$. Only $\mathcal L(y^{(6)})=s^6Y-1$ has a nonzero initial subtraction, so $$\boxed{Y(s)=\frac 1{(s^2+1)^3}.}$$

Step 2: Differentiate a transform family. Since $\partial_a(s^2+a^2)^{-m}=-2ma(s^2+a^2)^{-m-1}$, $$h_2(t)=\frac{\sin(at)-at\cos(at)}{2a^3},\qquad h_3(t)=-\frac 1{4a}\partial_a h_2(t) =\frac{3\sin(at)-3at\cos(at)-a^2t^2\sin(at)}{8a^5}.$$ Differentiation under the integral is valid for $a$ in a compact positive interval and $\operatorname{Re}s>0$, since the derivatives have polynomial growth. Hence $$\boxed{y(t)=\frac{3\sin t-3t\cos t-t^2\sin t}{8}.}$$

Step 3: Verify independently. Each term is the real or imaginary part of $e^{it}$ times a polynomial of degree at most two; $(D-i)^3$ annihilates these complex modes, and the conjugate factor annihilates their conjugates. Thus $(D^2+1)^3y=0$. Expanding the displayed trigonometric expression gives $$y(t)=\frac{t^5}{120}-\frac{t^7}{1680}+O(t^9).$$ This proves the five zero initial derivatives and $y^{(5)}(0)=1$.

Step 4: Distinguish oscillation from boundedness. There are triple poles at $s=\pm i$; $\operatorname{Re}s>0$ gives absolute convergence. Along $t_n=\pi/2+2\pi n$, $y(t_n)=(3-t_n^2)/8\to-\infty$, so the response is unbounded. The poles of $sY$ on the imaginary axis invalidate the final-value theorem. The formal value $\lim_{s\to 0}sY=0$ is not a time limit.