Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 14E / i / Solution

After division by $z$, the coefficient functions areBoth $zP(z)=1-z$ and $z^{2}Q(z)=\lambda z$ are analytic at zero, so the regular singular point criterion for a second-order equation shows that $z=0$ is regular singular. Substitution of $y=z^r$ into the leading terms gives the indicial equationThe exponent is repeated. Directly substituting $y={\sum }_{n\ge 0}{a}_n{z}^n$ givesThus $a_0$ determines the entire power series, so there is only one such solution $y_1$ up to scale. The logarithmic solution from a repeated Frobenius exponent predictsso its leading nonanalytic term is proportional to $\log z$.

For the contour ansatz, differentiation under the integral and one integration by parts giveThe integral vanishes whenwhose solution is the Laguerre contour-integral amplitudeThe contour and branches must make $f$ single-valued along the traversed path and must kill the endpoint term

Now suppose $\lambda <0$ is nonintegral. Near $t=0$, $f(t)=O(t^{-\lambda -1})$ is integrable and the endpoint factor is $O(t^{-\lambda })\to 0$. We may therefore choosea finite loop based at the branch point $0$ and avoiding $1$ except by encirclement. A second choice isalong one bank of the negative real axis, with a consistent branch. At $-\infty$, $e^{zt}$ kills the algebraic endpoint factor because $\mathrm{Re}z>0$.

The ${\gamma }_1$ integral is analytic in $z$ because its contour is finite, and it is a nonzero solution. By uniqueness of the analytic local solution, it is a constant multiple of $y_1$. This is the Finite Laguerre contour solution construction.

Solved by gpt-5.6-sol high.