Reduction of Order — Question 4

Question 4 – Solution

Strategy. Separate a failure of division by the seed from a singularity of the original equation.

Step 1: Reduce on the valid interval. Substitution gives $$v''+2\cot t\,v'=0,\qquad v'=C\csc^2t,\qquad v=-C\cot t+D.$$ Thus $y=-C\cos t+D\sin t$; a second solution is $\cos t$.

Step 2: Extend the product. Although $\cos t/\sin t=\cot t$ has poles at multiples of $\pi$, its product with $\sin t$ is $\cos t$ on each valid interval. The latter is smooth and directly satisfies $y''+y=0$ on all of $\mathbb R$. The poles belong to the chosen quotient.

Step 3: Solve at the seed’s zero. In $y=A\cos t+B\sin t$, the data give $A=1$, $B=0$. Thus $\boxed{y=\cos t}$. Continuous coefficients and nonzero leading coefficient ensure global uniqueness for the data at zero.

Step 4: Match derivatives. Both one-sided values are $1$, so continuity holds for all real $a,b$. The one-sided slopes are $a,b$, so $C^1$ regularity requires $a=b$. In that case one smooth formula holds on both sides, and it is a classical solution. If $a\ne b$, a derivative at zero does not exist; matching only position is insufficient.