Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 6 / Solution

Differentiating the logarithm givesThe real logarithm is valid where its argument is positive; $\log |\sec t+\tan t|$ gives the same derivative on the other nonsingular intervals.

For the first-order linear differential equation, an integrating factor on the interval containing $t=0$ isMultiplying the equation by it yields $(x{\sec }^{3}t{)}^{\prime }=\sec t$. Integration and $x(0)=2$ therefore giveThe interval ends at singularities of the differential equation's coefficient $\tan t$; a finite limiting value of the expression would not remove those coefficient singularities.