= Solution
The <integrating factor> is $e^{kx}$, giving $(e^{kx}y)'=ae^{kx}\sin(mx)$. The required <antiderivative> is
$$
\int e^{kx}\sin(mx)\,dx
=\frac{e^{kx}}{k^2+m^2}[k\sin(mx)-m\cos(mx)].
$$
After dividing by $e^{kx}$, the <general solution> is the oscillatory term plus $Ce^{-kx}$. The <initial condition> fixes $C=1+am/(k^2+m^2)$, so
$$
\boxed{y(x)=\left(1+\frac{am}{k^2+m^2}\right)e^{-kx}
+\frac{a[k\sin(mx)-m\cos(mx)]}{k^2+m^2}.}
$$
The real, nonzero parameters ensure $k^2+m^2>0$; direct substitution verifies both the equation and $y(0)=1$.
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