Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 3 / a

The homogeneous linear differential equation has solutions $\cos x$ and $\sin x$. Since the forcing is already a homogeneous solution, seek a particular solution proportional to $x\cos x$. Direct differentiation givesThus the general solution of the linear ordinary differential equation isThe first initial condition gives $A=0$. Alsoso the second initial condition gives $B-\frac{1}{2}=-\frac{1}{2}$, hence $B=0$. Therefore

For $y_1=e^x/x$, on an interval avoiding zero,Substitution into the ordinary differential equation givesTo find an independent solution, put $v=xy$. The product rule gives $v^{\prime \prime }=x{y}^{\prime \prime }+2y^{\prime }$, so the ordinary differential equation becomes $v^{\prime \prime }-v=0$. Its independent solutions are $e^x$ and $e^{-x}$. HenceThe two displayed solutions are linearly independent, since their ratio $e^{2x}$ is nonconstant on any open interval. Constants can be chosen separately on the two sides of the singular point $x=0$.