Past exam of the differential equations 2 course of the University of Oxford / 2017 / 3 / a

Move the expansion point to the origin with $z=x+1$. The transformed ordinary differential equation isAfter division by $z$, the functions $z(1-1/z)=z-1$ and $z^2(c/z)=cz$ are analytic functions at zero, so this is a regular singular point. For a Frobenius solution with leading term $a_0{z}^{\rho }$, $a_0\ne 0$, the lowest power is $z^{\rho -1}$ and its coefficient isTherefore the indicial equation and its indicial exponents areThe exponent $\rho$ denotes the quantity called $\alpha$ in the paper, and avoids confusion with other uses of that letter.

Substitute a Frobenius series $y={\sum }_{n\ge 0}{a}_n{z}^{n+\rho }$ into the transformed equation. For $n\ge 1$, equating the coefficient of $z^{n+\rho -1}$ gives the recurrence relationAt the larger indicial root $\rho =2$, the coefficient of $a_n$ is $n(n+2)\ne 0$, so a Frobenius solution always exists. At the smaller indicial root $\rho =0$, the first two equations areSince $a_0\ne 0$, the exceptional equation is consistent exactly whenThese are both admissible values, including the zero value; the singular wording of the question does not remove either. This is the undetermined coefficient at Frobenius resonance: at either value $a_2$ is free, whereas otherwise the second independent solution requires a Logarithmic Frobenius solution.

Explicit pairs demonstrate sufficiency and linear independence. In the variable $z=x+1$ one may takeBoth displayed $u_2$ have leading term $z^2/2$, while $u_0(0)=1$. They therefore give independent Frobenius solutions with the two indicial exponents, and direct differentiation verifies the equation. The point is an apparent regular singular point in these two cases.

For the requested closed forms, particularly short choices areIn addition, $(x+2)e^{-x}$ solves the zero-parameter equation and $e^{-x}$ solves the equation with $c=-1$.