Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics III / 1

The integrating factor for this linear ordinary differential equation is $\mu (x)=e^{x^3/3}$. Multiplication gives an exact derivative:Consequently $y=1+C{e}^{-x^3/3}$. The initial condition $y(1)=2$ gives $C=e^{1/3}$, so the initial value problem has solutionThe integrating factor calculation determines every solution, and the initial condition fixes its remaining constant of integration uniquely.

For the given matrix $A$, the characteristic polynomial isFor $\lambda =4$, the eigenvector equation gives $v_2=v_1$; for $\lambda =-1$, it gives $3v_1+2v_2=0$. Thus the eigenvalues and their corresponding eigenspaces areFor any eigenvector $v$ with eigenvalue $\lambda$, the vector-valued function $e^{\lambda x}v$ satisfies $Y^{\prime }=AY$, since its derivative is $\lambda e^{\lambda x}v=e^{\lambda x}Av$. The two displayed eigenvectors are linearly independent, so the general solution of the linear system of ordinary differential equations is

The function $y_1=x$ solves the linear ordinary differential equation. Apply reduction of order by writing $y=xv(x)$. Substituting $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$ and cancelling the terms in $v$ givesFor $w=v^{\prime }$, this becomes a first-order linear ordinary differential equation:A partial fraction decomposition provides the required antiderivative:ThereforeThe logarithm has positive argument on the required interval. The Wronskian of $x$ and $1+x\log ((x-1)/x)$ is $1/(x-1)\ne 0$, proving their linear independence and hence completeness of this general solution.