Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 6A (Differential Equations)

For the stated order of the two solutions, define their Wronskian byTaking a derivative cancels the two mixed products: $W^{\prime }=y_1{y}_2^{\prime \prime }-y_1^{\prime \prime }{y}_2$. Substituting the second-order linear differential equation for each second derivative cancels the $q$ terms and leavesThis is the Abel identity. On an interval with continuous coefficients, linear independence means $W(x_0)\ne 0$; hence the Wronskian never vanishes anywhere on that interval.

The defining Wronskian identity gives the first-order linear differential equationWork first on an interval where $y_1\ne 0$. Dividing by $y_1$ recognizes a derivative:This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant $C$ only adds a multiple of $y_1$. If $y_1$ has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.

For real parameters, work on $x>0$ so arbitrary real powers are unambiguous. Taking derivatives givesThe residual after substitution isWhen $\gamma \ne 0$, $x^{\gamma }$ ranges over $(0,\infty )$. Evaluating at the zeros of the cosine first forces the sine coefficient to vanish; then the cosine coefficient vanishes identically. If $\gamma =0$, the nonzero constant $\cos 1$ cannot satisfy the equation. Thus the necessary and sufficient conditions areUnder these conditions the normalized coefficient is $p=(1-\gamma )/x$. The Abel identity gives $W=C{x}^{\gamma -1}$. Choose $C=\gamma$ and apply reduction of order on an interval where $\cos (x^{\gamma })\ne 0$:The substitution $u=x^{\gamma }$ evaluates the integral. After continuation across the cosine zeros,This is a power substitution for an oscillatory second-order equation: in the variable $u=x^{\gamma }$, the equation reduces to the harmonic oscillator equation.