= Solution
Impose the five <initial conditions> at zero on the preceding <general solution>. They give
$$
\begin{aligned}
C_0+C_1+C_2+C_3&=1,\\
C_1-C_2+C_4&=0,\\
C_1+C_2-C_3-1&=0,\\
C_1-C_2-C_4&=0,\\
C_1+C_2+C_3&=0.
\end{aligned}
$$
The 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 $C_0=1$. Thus
$$
\boxed{y(x)=1+\frac14(e^x+e^{-x})-\frac12\cos x-\frac{x^2}{2}
=1+\frac12(\cosh x-\cos x)-\frac{x^2}{2}.}
$$
For 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.
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