Variation of Parameters — Question 2

Question 2

Solve on $I=(-\pi/2,\pi/2)$: $$y''+y=\sec t,\qquad y(0)=y'(0)=0.$$ Use the homogeneous pair $\cos t,\sin t$.

Tasks

1. Derive the parameter derivatives and evaluate the integrals with lower endpoint zero.

2. Simplify the initial-value solution and verify its equation and both data.

3. Determine the limits of $y$ and $y'$ as $t\to(\pi/2)^-$. Can the solution extend as a $C^1$ function through that endpoint?

4. Explain why a finite limit of $y$ does not enlarge the interval on which this initial-value problem has a classical solution.

Question 2 – Solution

Strategy. The forcing is singular at the interval endpoints. Keep its domain visible even when the solution itself has a finite limit.

Step 1: Integrate the parameter system. Here $W=1$. The equations $u_1'\cos t+u_2'\sin t=0$ and $-u_1'\sin t+u_2'\cos t=\sec t$ give $$u_1'=-\tan t,\qquad u_2'=1.$$ Since $\cos t>0$ on $I$, integration from zero yields $u_1=\ln(\cos t)$, $u_2=t$.

Step 2: Verify the response. The zero lower endpoints fit both data, and $$\boxed{y=\cos t\ln(\cos t)+t\sin t.}$$ Differentiation simplifies to $$y'=-\sin t\ln(\cos t)+t\cos t, \quad y''=-\cos t\ln(\cos t)+\frac{\sin^2t}{\cos t}+\cos t-t\sin t.$$ Therefore $y''+y=(\sin^2t+\cos^2t)/\cos t=\sec t$, and both data vanish.

Step 3: Examine the endpoint. Using $x\ln x\to 0$ as $x\to 0^+$, we obtain $$\lim_{t\to(\pi/2)^-}y(t)=\pi/2,\qquad \lim_{t\to(\pi/2)^-}y'(t)=+\infty.$$ Thus even a $C^1$ extension is impossible. The graph approaches a finite height with an unbounded positive slope; the endpoint is excluded.

Step 4: Keep the equation’s domain. The right side $\sec t$ is undefined at $\pi/2$ and $-\pi/2$. A continuous assignment to $y$ at an endpoint would not define the differential equation there. The maximal open interval containing zero on which this IVP is classical is exactly $I$.