= Solution
The <complementary solution> comes from the characteristic equation $r^4-a^4=0$, whose roots are $a,-a,ia,-ia$. It is therefore
$$
y_h=C_1e^{ax}+C_2e^{-ax}+C_3\cos(ax)+C_4\sin(ax).
$$
For the <resonant exponential particular solution>, the forcing is already a homogeneous exponential, so the <method of undetermined coefficients> needs the resonant trial $y_p=Kxe^{-ax}$. If $P(r)=r^4-a^4$, the identity $P(D)[xe^{rx}]=P'(r)e^{rx}$ when $P(r)=0$ gives $P'(-a)=-4a^3$. Hence $K=-1/(4a^3)$ and
$$
y=C_1e^{ax}+C_2e^{-ax}+C_3\cos(ax)+C_4\sin(ax)-\frac{x}{4a^3}e^{-ax}.
$$
The decay condition first eliminates the growing exponential, $C_1=0$. A nonzero combination of the sine and cosine has a fixed nonzero amplitude and cannot tend to zero, so $C_3=C_4=0$ as well. Both remaining exponential terms decay because $a>0$. Finally $y(0)=C_2=1$, giving the unique answer
$$
\boxed{y(x)=\left(1-\frac{x}{4a^3}\right)e^{-ax}.}
$$
This is an instance of <resonance in a differential equation>: multiplication by $x$ supplies the <particular solution> when the forcing exponent is a simple characteristic root.
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