Variation of Parameters — Question 9

Question 9 – Solution

Strategy. Variation of parameters supplies all initial-value responses; evaluating the integral at the second endpoint reveals the obstruction.

Step 1: Keep the free slope. The homogeneous pair is $\cos t,\sin t$, with $W=1$. Integrating the parameter derivatives from zero and imposing $y(0)=0$ gives every candidate: $$y(t)=A\sin t+\int_0^t\sin(t-s)f(s)\,ds, \qquad y'(0)=A.$$ Differentiating this formula verifies the equation and shows no candidate has been lost.

Step 2: Evaluate the second endpoint. Since $\sin\pi=0$ and $\sin(\pi-s)=\sin s$, the remaining condition is $$\boxed{\int_0^\pi\sin s\,f(s)\,ds=0.}$$ If it fails, no choice of $A$ helps. If it holds, every real $A$ works: there are infinitely many solutions, differing by $A\sin t$. Thus this endpoint pair never selects exactly one solution for a compatible forcing.

Step 3: Compare the two forcings. For $f=1$, the integral is $2$, so there is no solution. For $f=\cos t$, it is $\int_0^\pi\sin s\cos s\,ds=0$. Evaluation of the response integral, or a direct residual check, gives $$\boxed{y(t)=(A+t/2)\sin t,\quad A\in\mathbb R.}$$ Indeed, $(t\sin t/2)''+t\sin t/2=\cos t$, and both endpoints vanish.

Step 4: Apply the extra measurement. The needed integrals are $\int_0^\pi\sin t\,dt=2$ and, by integration by parts, $\int_0^\pi t\sin t\,dt=\pi$. Thus $$0=2A+\pi/2,\qquad A=-\pi/4, \quad \boxed{y(t)=(t/2-\pi/4)\sin t.}$$ The coefficient of $A$ in this measurement is nonzero, so it selects exactly one member of the compatible family. This does not contradict the earlier nonuniqueness: the added integral condition supplies new information.