Past exam of the mathematical methods I course of the University of Oxford / 2016 resit / 8 / a / 2 / Solution

Here $m=2$, so $z=1/y$ obeys $z^{\prime }-2z\tan x=-{\tan }^{2}x$. An integrating factor is ${\cos }^{2}x$, givingThe nonzero solutions are thereforeThe additional solution is $\boxed{y=0}$. Each formula is interpreted on an interval where $\tan x$ is defined and its displayed denominator is nonzero.