Solution
= Solution
In the homogeneous equation, substitute the <exponential function> $e^{kx}$. The <characteristic equation> is
$$
k^5-k=k(k^4-1)=k(k-1)(k+1)(k^2+1)=0.
$$
Its distinct roots are $0,1,-1,i,-i$. Combining the complex-conjugate pair into real <sine> and <cosine> gives the <complementary function>
$$
\boxed{y_c=C_0+C_1e^x+C_2e^{-x}+C_3\cos x+C_4\sin x.}
$$
Five linearly independent terms supply the required five homogeneous degrees of freedom.