Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 4

Divide by the leading coefficient for $x\ne 0,1$:The pole of $P$ makes $x=0$ a singular point of a differential equation. However,are real analytic at zero. ThereforeThe ordinary-point power-series theorem does not apply there, but the Frobenius method does.

For a Frobenius exponent $r$, put $p_0={\lim }_{x\to 0}xP(x)=1$ and $q_0={\lim }_{x\to 0}{x}^{2}Q(x)=0$. The indicial equation issoA fundamental pair has the repeated-root Frobenius formon a positive-$x$ interval. In this case a real analytic solution is particularly simple: $y_1=1-x$ satisfies the equation exactly. A convenient logarithmic companion, obtained below, is $y_2=x+(1-x)\ln [x/(1-x)]$ for $0<x<1$. It is independent of $y_1$ because of its logarithmic divergence at zero.

Let $A(x)={\sum }_{n=1}^{\infty }{a}_n{x}^n$ denote the analytic part of the prescribed ansatz. Its logarithmic part has derivativesAfter substitution, the full differential equation reduces toThe first terms are real analytic at zero and cannot cancel a nonzero logarithm. ThereforeThe cancellation of the apparent $1/x$ terms is already built into this calculation.

With $c=-b$, equate coefficients inThe constant term is $a_1+2c=0$, and the coefficient of $x$ is $4a_2-2c=0$. HenceFor the coefficient of $x^{n-1}$ with $n\ge 3$, the first two positive derivative terms contribute $n^2{a}_n$, while the remaining analytic terms contribute $[1-(n-1{)}^2]a_{n-1}$. ThusIteration gives $a_n=-b/[n(n-1)]$ for $n\ge 2$. Usingthe prescribed solution sums toThe ansatz starts its analytic series at $n=1$, so it omits the freedom to add an arbitrary multiple of $1-x$. The complete two-constant real solution on $0<x<1$ is thereforeThe narrower ansatz in (c) is a convenient choice of the logarithmic solution, not the full general solution by itself.

For nonzero $b$, the coefficients satisfy $a_n\sim -b/n^2$, so the analytic power series has radius of convergence $1$. It converges absolutely also at $x=\pm 1$, by comparison with $\sum n^{-2}$.

The real logarithm in the prescribed solution requires $x>0$. ConsequentlyThe series and the function itself have finite limits at $x=1$; in fact the logarithmic companion tends to $1$. Its derivatives are singular there, so convergence of its values at the endpoint does not make it a classical solution through that singular point of a differential equation. At $x=0$ the logarithmic branch diverges. The independent analytic solution $C(1-x)$ is a polynomial and has no such convergence restriction.