Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 8

Away from zeros of the leading coefficient, the normal-form differential equation isThe only finite singular points are $z=0,i,-i$. At zero, $zP(z)=3/(1+z^2{)}^2$ and $z^{2}Q(z)=1/(1+z^2{)}^2$ are analytic, so zero is a regular singular point. At $z=\pm i$, the coefficient $P$ has a genuine double pole, so $(z\mp i)P$ is not analytic; these are irregular singular points. ThusIf infinity is included, set $t=1/z$ and $Y(t)=y(1/t)$. The transformed coefficients are $\tilde{P}=2/t-3t^3/(1+t^2{)}^2$ and $\tilde{Q}=t^2/(1+t^2{)}^2$. They satisfy the regular singular point at infinity criterion, so $\boxed{\infty \text{ is regular singular}}$ as well.

Seek a Frobenius solution $y=z^r{\sum }_{n\ge 0}{a}_n{z}^n$ with $a_0\ne 0$. Since $(z+z^3{)}^2=z^2+2z^4+z^6$, its lowest coefficient gives the indicial equationThus the repeated indicial root is $\boxed{r=-1}$. Equating the coefficient of $z^{n+r}$ before setting $r=-1$ givesFor $n\ge 1$ this becomes the linear recurrence relationEvery coefficient is fixed by $a_0$, since the denominator never vanishes for $n\ge 1$. In particular $a_1=0$ and then every odd coefficient is zero. Therefore there is $\boxed{\text{exactly one independent pure Frobenius series}}$, not two merely because the indicial root repeats. If "power series" means an ordinary Taylor series with nonnegative integer powers, the only such solution is zero: its coefficient equation starts with $a_0=0$ and inductively forces all coefficients to vanish.

A second independent local solution is a logarithmic solution from a repeated Frobenius exponent. Indeed, Abel's identity gives $W=C{z}^{-3}[1+O(z^2)]$. For the normalized first solution $y_1=z^{-1}[1+O(z^2)]$, reduction of order givesAfter choosing the harmless multiplicative constant in the numerator, its integrand is $1/s+O(s)$, so $y_2=y_1\log z+z^{-1}\times \text{an analytic series}$. The nonzero logarithmic coefficient proves that this second solution is not another pure Frobenius solution.

Normalize the free leading coefficient to $a_0=1$. The linear recurrence relation from part (b) givesConsequently the first four nonzero coefficients areand the single pure Frobenius solution isFor clarity about the second independent solution, a choice with no added multiple of $y_1$ isThese companion coefficients follow by reduction of order with the normalized Wronskian $W=z^{-3}(1+z^2{)}^{3/2}\exp [3z^2/(2(1+z^2))]$. The logarithm requires a local branch; it does not supply a second ordinary power series solution.