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

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 $x=0$.

The logarithm requires $y>0$, and $x>0$ is given. Set $z=\log (y/x)$, so $y=x{e}^z$ and the chain rule yields $x{y}^{\prime }/y=1+x{z}^{\prime }$. Dividing the ordinary differential equation by $y$ therefore givesSince $(z/x{)}^{\prime }=(x{z}^{\prime }-z)/x^2=0$, we obtain $z=Cx$, including the case $z=0$. Thus the general positive solution is $y=x{e}^{Cx}$. The initial condition $y(1)=e$ gives $C=1$:For the separate decay requirement, $x{e}^{Cx}\to 0$ precisely when $C<0$: if $C=0$, the result is $x$, and if $C>0$, it grows. Consequently the full family with the stated limit isThis latter family does not also satisfy $y(1)=e$; the solution selected by that initial condition grows. The identically zero function is excluded by the original logarithm.