Solving IVP’s with Laplace Transforms — Question 4

Question 4 – Solution

Strategy. Matching forcing and natural frequencies produce repeated transform poles. Explicit values and the work identity distinguish real growth from a mere upper bound.

Step 1: Transform and invert. Zero initial data give $$(s^2+4)Y=\frac 2{s^2+4},\qquad Y=\frac 2{(s^2+4)^2}.$$ Combining the sine pair with the supplied time-cosine pair yields $$\boxed{y(t)=\frac{\sin 2t-2t\cos 2t}{8}.}$$ Indeed the forward numerator is $(s^2+4)/4-(s^2-4)/4=2$.

Step 2: Verify the IVP. Direct differentiation gives $$y'=\tfrac t2\sin 2t,\qquad y''=\tfrac 12\sin 2t+t\cos 2t.$$ Adding $4y=\tfrac 12\sin 2t-t\cos 2t$ gives $\sin 2t$. Both initial values are zero, proving the claimed solution by uniqueness.

Step 3: Exhibit unbounded values and the domain. At $t=n\pi$, $$\boxed{y(n\pi)=-n\pi/4\longrightarrow-\infty.}$$ The actual response is therefore unbounded; a growing envelope alone would not prove that. Its linear envelope gives absolute transform convergence for $s>0$. At zero, its finite integral is $[1-\cos 2R-R\sin 2R]/8$, which has no limit. Below zero, intervals around successive peaks of the $t\cos 2t$ term violate the Cauchy criterion. Hence the exact domain is $\boxed{s>0}$.

Step 4: Account for the energy input. Multiplying the equation by $y'$ gives $$E'=y'(y''+4y)=y'\sin 2t=\tfrac t2\sin^2 2t\ge 0.$$ At $n\pi$, the velocity is zero, so $$\boxed{E(n\pi)=2y(n\pi)^2=n^2\pi^2/8.}$$ Although the forcing amplitude is at most one, its work accumulates over an increasing time interval. Resonance makes that accumulated work unbounded. The graph shows the growing displacement oscillations, not the energy itself.