Past exam of the mathematics course of the University of Cambridge 2011 ia Paper 2 8A b Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 2 5A Solution Created 2026-09-24 Updated 2026-10-07
Use a power-series solution of a differential equation, . Comparing coefficients givesIn particular , while and are free. The two prescribed derivative pairs select and . The next coefficients are , , , . HenceThese are the first three nonzero terms when the respective parameter is nonzero; if or , the corresponding solution vanishes identically.
For closed forms, set . Cancellation of the first-derivative terms reduces the equation, for , to . The analytic solutions extend across zero, and their hyperbolic cosine and hyperbolic sine expansions identifyThis is the hyperbolic reduction of a regular-singular differential equation: although the original leading coefficient vanishes at zero, both selected power-series solutions of a differential equation are entire.
Past exam of the mathematics course of the University of Cambridge 2022 ia Paper 2 5C a Solution Created 2026-09-24 Updated 2026-09-25
For a normalized equation is an ordinary point when are analytic there. It is a regular singular point when and are analytic there. These are the ordinary point criterion for a second-order equation and regular singular point criterion for a second-order equation.
Hereso is regular singular. Seek a power-series solution of a differential equationEquating the coefficient of givesand henceThereforeThe series terminates exactly when is a nonnegative integer: the factor with then makes . Thus the polynomial solutions occur for