Solving IVP’s with Laplace Transforms — Question 5

Question 5 – Solution

Strategy. Initial data control the coefficient of the unstable mode. Exact cancellation can give a bounded trajectory but does not remove sensitivity to an initial error.

Step 1: Transform and resolve the modes. The transformed equation is $$(s^2-1)Y=s+v+\frac 1{s+2}.$$ Partial fractions give $$Y=\frac{v/2+2/3}{s-1}+\frac{-v/2}{s+1}+\frac{1/3}{s+2},$$ $$\boxed{y(t)=(v/2+2/3)e^t-(v/2)e^{-t}+\tfrac 13e^{-2t}.}$$ Recombining the fractions reproduces the transformed equation.

Step 2: Cancel the growing coefficient. Boundedness requires $v/2+2/3=0$, so the unique choice is $$\boxed{v=-4/3,\qquad y_b(t)=\tfrac 23e^{-t}+\tfrac 13e^{-2t}.}$$ Then $y_b(0)=1$, $y_b'(0)=-2/3-2/3=-4/3$, and $y_b''-y_b=e^{-2t}$. If the growing coefficient is nonzero, it eventually dominates both decaying terms, so no other velocity works.

Step 3: Quantify the initial-error tolerance. Replacing $v$ by $-4/3+\delta$ changes the solution by $$\boxed{y-y_b=\tfrac\delta 2(e^t-e^{-t})=\delta\sinh t.}$$ Since $\sinh t$ increases on $t\ge 0$, the maximum absolute error on $[0,5]$ is $|\delta|\sinh 5$. The exact, necessary and sufficient tolerance is $$\boxed{|\delta|\le\frac{0.01}{\sinh 5}\approx 1.3477\times 10^{-4}.}$$ Equality is permitted and attains the error limit at time five. The graph compares the bounded choice with a small positive velocity error.

Step 4: Check the actual tails. For the bounded choice, the nonzero slow tail is $(2/3)e^{-t}$, so the exact domain is $s>-1$. For every other $v$, a nonzero $e^t$ term gives exact domain $s>1$. The operator factor $s-1$ cancels from the selected response numerator. Its presence in the operator does not mean it survives in every solution; the initial data determine the residue.