 Reduction of order

Given a known solution $y_1$ of $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, reduction of order seeks another solution as $y_2=v{y}_1$. On an interval where $y_1\ne 0$, substitution cancels the terms containing $v$ and givesThis is first order in $v^{\prime }$, soIts Wronskian with $y_1$ is $e^{-\int p}$, nonzero, proving independence. An additive constant in the integral merely adds a multiple of $y_1$.

For $y^{\prime \prime }+y=0$ and $y_1=\cos x$, the formula gives $y_2=\cos x\tan x=\sin x$ where $\cos x\ne 0$. Although the integral factor has poles, the actual second solution continues smoothly through them. The reduction of order across a zero of the known solution explains why one continues the solution using the original regular equation rather than declaring the pole of an auxiliary integral a solution singularity.