Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 8

Divide by $2x^2$ to put the ordinary differential equation into normalized form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$:The coefficients are singular at zero, butare both analytic there. This is precisely the regular singular point criterion for a second-order equation, so $x=0$ is a regular singular point.

Apply the Frobenius method. For $y=x^{\sigma }{\sum }_{n=0}^{\infty }{a}_n{x}^n$, the coefficient of the lowest power $x^{\sigma }$ comes only from $2x^2{y}^{\prime \prime }-x{y}^{\prime }+y$; the remaining term $xy$ starts one power higher. As $a_0\ne 0$, this lowest coefficient gives the indicial equationTherefore the two possible characteristic exponents at a regular singular point areTheir difference is not an integer, so the two corresponding Frobenius solutions do not require a logarithmic correction.

For $n\ge 1$, the coefficient of $x^{n+\sigma }$ isFactoring the bracket gives the recurrence relationFor $\sigma =1/2$ this is $a_n=-a_{n-1}/[n(2n-1)]$. Iteration and $(2n)!=2^{n}n!(2n-1)!!$ yieldFor $\sigma =1$, the recurrence is $a_n=-a_{n-1}/[n(2n+1)]$, givingHere the double factorial products are one at $n=0$, consistent with the chosen normalization. Recognizing the cosine and sine power series, for positive $x$ one may also write the two Frobenius solutions asThe constants multiplying the two independent solutions may be chosen separately.

For either exponent, the recurrence from part (c) impliesfor every finite complex $x$. The ratio test therefore givesThe factor multiplying $x^{\sigma }$ is an entire function in each case. For $\sigma =1/2$, this does not make the full Frobenius solution single-valued at zero: its square-root prefactor still requires a branch. The infinite radius refers to $\sum a_n{x}^n$, not to a Taylor expansion of the branched full solution.