Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 17 / b / iv / Solution

Impose the five initial conditions at zero on the preceding general solution. They giveThe second and fourth equations give $C_1=C_2$ and $C_4=0$. The third and fifth then yield $C_1=C_2=1/4$ and $C_3=-1/2$, while the first gives $C_0=1$. ThusFor a direct check, the difference of its fifth and first derivatives is $x$. At zero its value is one; its first, second, third and fourth derivatives vanish. In particular the quadratic term cancels the otherwise nonzero second derivative of the hyperbolic-cosine and cosine combination.