Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 7A (Differential Equations)

The Wronskian isOn an interval where $p,q$ are continuous, the solutions are linearly independent exactly when $W(x_0)\ne 0$ at one point, equivalently when the Wronskian is nonzero throughout the interval. Differentiation and the homogeneous linear differential equation giveHence Abel identity reads $W(x)=W(x_0)\exp (-{\int }_{x_0}^{x}p(s)ds)$. If $W(x_0)=0$, the two initial-data vectors are dependent; the uniqueness theorem for ordinary differential equations then makes the corresponding linear combination identically zero. This proves the converse as well as the necessity.

Direct calculation givesWhere $1+\cos x\ne 0$, Abel identity gives $p=-W^{\prime }/W=\sin x/(1+\cos x)$, and substitution of either solution gives $q=1/(1+\cos x)$. ThusThese coefficients have singularities at odd multiples of $\pi$. This does not contradict the Wronskian criterion: its continuous-coefficient interval hypothesis fails at those points.

On $0\le x<\pi$, the initial conditions uniquely select $y=\sin x$. The normalized differential equation is not defined on the whole interval $0\le x<\infty$, so the usual global uniqueness theorem for ordinary differential equations cannot give the global initial-value assertion as posed. No choice of finite coefficients at $\pi$ can repair the normalized equation while retaining both displayed solutions: at that point $1+\cos x$ and its first derivative vanish, but its second derivative equals one.

The continuation through a zero of the leading ODE coefficient is a different problem: consider the multiplied, degenerate equationFor this different equation, a globally $C^2$ solution with the given initial data is uniquely $\sin x$. On each regular interval every solution is $A_j(1+\cos x)+B_j\sin x$; continuity of $y^{\prime }$ and $y^{\prime \prime }$ at an odd multiple of $\pi$ forces both constants to agree across it. If only $C^1$ piecewise solutions away from the singular points are required, the $A_j$ may change and uniqueness fails. Thus the coefficient singularity and the intended solution regularity must be distinguished.

Both functions satisfy the constant-coefficient homogeneous linear differential equationIts characteristic polynomial is $\lambda ({\lambda }^2+1)$, so the general real solution is $y=A+B\cos x+C\sin x$. The three initial conditions give $A+B=0$, $C=1$, and $-B=0$, henceHere the equation has continuous coefficients and nonzero highest-derivative coefficient everywhere. The uniqueness theorem for ordinary differential equations, applied to its three-component first-order system, gives a unique solution on the entire half-line.