Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 8

For the normalized equation, $x_0$ is an ordinary point when both $p$ and $q$ are analytic near $x_0$. A singular point is a regular singular point when $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$; equivalently $p$ has at most a simple pole and $q$ at most a double pole there.

Thus ordinary points admit ordinary power-series solutions, while the Frobenius method applies at a regular singular point. The definitions refer to the coefficients after dividing by the leading coefficient of $y^{\prime \prime }$.

After normalization, $p=-6x^2/(1-x^3)$ and $q=-6x/(1-x^3)$. Both are analytic at zero, so $\boxed{x=0\text{ is ordinary}}$. At one,are analytic, whereas $p,q$ themselves have poles. Hence $\boxed{x=1\text{ is a regular singular point}}$.

Insert $y={\sum }_{n\ge 0}{a}_n{x}^n$. The coefficient of $x^0$ gives $a_2=0$; for $k\ge 1$ the coefficient of $x^k$ givesThus $a_n=a_{n-3}$ for $n\ge 3$. The initial data give $a_0=1$, $a_1=0$, soThe radius is one because the three cube roots of unity are poles of this solution. A direct structural check observes that the differential equation is exactly $[(1-x^3)y{]}^{\prime \prime }=0$. Thus $(1-x^3)y=A+Bx$, and the initial data select $A=1,B=0$. This is the product-derivative reduction of a second-order differential equation.

Use a Frobenius solution $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$. The lowest power $x^{\sigma -1}$ has coefficient $2\sigma (2\sigma -1)a_0$, giving the indicial equationFor $n\ge 1$, matching the coefficient of $x^{n+\sigma -1}$ givesFor both candidate exponents the cancelled factor is nonzero in this range, so the recurrence isWith $a_0=1$, the exponent-zero solution isFor $\sigma =1/2$, the coefficients are $a_n=1/(2n+1)!!=2^{n}n!/(2n+1)!$, givingThe ratio test gives infinite radius of convergence for both series factors, since the successive ratios in magnitude are $|x|/(2n+2)$ and $|x|/(2n+3)$. The second full solution still needs a square-root branch in the complex plane; on $x>0$ it is a real solution. The distinct leading behaviors $1$ and $\sqrt{x}$ prove linear independence.

For an independent closed-form check on the structure, put $z=x/2$. The equation becomes $z{Y}^{\prime \prime }+(1/2-z)Y^{\prime }-(1/2)Y=0$, a Kummer differential equation with equal parameters. Its exponential reduction of a confluent hypergeometric equation with equal parameters explains the elementary first solution and the branched companion.