Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 8

Away from $y=\pm 1$, divide by the coefficient of $x^{\prime \prime }$ to obtain the normal formThus $\boxed{y=1\text{ and }y=-1}$ are the only finite singular points. For a second-order linear differential equation, $y_0$ is a regular singular point if $(y-y_0)P(y)$ and $(y-y_0{)}^{2}Q(y)$ extend analytically to $y_0$. At $y_0=1$ these functions areand at $y_0=-1$ they areEach pair is analytic at its respective point. Therefore $\boxed{\text{both finite singular points are regular}}$.

Set $t=y-1$. Then $y^2-1=t(t+2)$, and a Frobenius solution has the formIn the differential equation, the lowest power is $t^{\sigma -1}$, whose coefficient isThe indicial equation and its indicial roots are consequentlyTheir difference is not an integer, so the Frobenius method produces two independent series without a logarithmic term:Choose a branch of $(y-1{)}^{1/2}$, or take $y>1$ for real solutions. A general local solution is $C_1{x}_1+C_2{x}_2$ by linear independence and the superposition principle. The analytic factors converge at least for $|y-1|<2$, up to the next singular point at $y=-1$.

In the shifted variable $t=y-1$, substitute the Frobenius solution $x={\sum }_{j\ge 0}{a}_j{t}^{j+\sigma }$ into $t(t+2)x^{\prime \prime }+x^{\prime }+x=0$. Equating the coefficient of $t^{j+\sigma }$ givesThus the linear recurrence relation, valid for $j=0,1,\dots$, isFor the indicial root $\sigma =0$, this becomesFor the indicial root $\sigma =1/2$, rename the coefficients $b_j$; thenWith $a_0=b_0=1$, the first terms of the two Frobenius solutions are