Variation of Parameters — Question 2

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 $I$.