Past exam of the year 1 of the computer science course of the University of Oxford / 2003 / Calculus, probability, linear algebra, discrete mathematics, logic and proof, and models of computation / 1 / a / ii / Solution

For $y_1=(1-x{)}^{-1}$, the derivatives are $y_1^{\prime }=(1-x{)}^{-2}$ and $y_1^{\prime \prime }=2(1-x{)}^{-3}$. Consequently the left-hand side of the ordinary differential equation iswhere $x\ne 1$.

Use reduction of order, writing $y=u/(1-x)$. A direct application of the product rule givesFor $v=u^{\prime }$ the resulting first-order ordinary differential equation is $v^{\prime }/v=2/(1-x)$ whenever $v\ne 0$. Integration yields $v=C(1-x{)}^{-2}$ and $u=A+B/(1-x)$; the case $v=0$ is already included. Henceon either interval $x<1$ or $x>1$. The two displayed solutions of a differential equation are linearly independent, so they span the solution space of a homogeneous linear differential equation on either interval. The point $x=1$ is a singular point of a differential equation and cannot be crossed by a nonzero member of this family that remains finite there.