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

The repeated root of the characteristic equation suggests factoring out $e^x$. Set $y=e^{x}u$. The product rule givesThus the ordinary differential equation reduces to $u^{\prime \prime }=1$, with general solution $u=x^2/2+Ax+B$. The first initial condition gives $B=1$. Since $y^{\prime }(0)=A+B=2$, the other initial condition gives $A=1$. ThereforeSubstitution into the reduced ordinary differential equation and the two initial conditions verifies the result.

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.

Multiplying the ordinary differential equation by two suggests using the dependent variable $Y=y^2$, whose derivative is $Y^{\prime }=2y{y}^{\prime }$. On the interval through $x=1$, this gives the linear ordinary differential equationAfter integration, $Y=x^2+Cx$. The initial condition $Y(1)=4$ gives $C=3$. As $y(1)>0$, continuity selects the positive branch:The interval $x>0$ is the maximal real interval through the given initial condition: at $x=0$ the formula approaches $0$, but its derivative is unbounded, so it does not extend through zero as a differentiable solution of a differential equation.