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

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.

Use the same general solution. The first two conditions now give $A+B=3$ and $-2A-6B+12=-2$, so $A=1$, $B=2$. HenceAt $t=\log \sqrt{2}$, its value is $1/2-3/4+2/8=0$, so the third condition is also satisfied. This is the unique consistent solution.

The repeated factor in the differential operator suggests $y=e^{-x}u(x)$. Direct differentiation gives $y^{\prime \prime }+2y^{\prime }+y=e^{-x}{u}^{\prime \prime }$, so $u^{\prime \prime }=4$ andThe first value gives $C=1$. The second gives $u(1)=ey(1)=1$, hence $2+B+1=1$ and $B=-2$. Therefore