Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 15 / a / i / Solution

The characteristic equation has roots $-2,-6$, and a trial multiple of $e^{-4t}$ gives the particular solution $-3e^{-4t}$. ThusThe first two conditions give $A+B=3$ and $-2A-6B+12=0$, hence $A=B=3/2$. At $t_{\ast }=\log \sqrt{2}$, the exponentials $e^{-2t_{\ast }},e^{-4t_{\ast }},e^{-6t_{\ast }}$ equal $1/2,1/4,1/8$. ThereforeNo solution satisfies all three conditions. The first two initial data already determine a unique solution of the second-order linear differential equation; the extra condition is inconsistent with it.