= Solution
Let $D=1+\omega^2$, with $\omega>0$. A particular solution of the <forced harmonic oscillator> is $y_p=e^{-t}(at+b)$. Substitution gives $Da=2$ and $Db-2a=0$, hence $a=2/D$ and $b=4/D^2$. Adding $A\cos\omega t+B\sin\omega t$ and imposing the two initial conditions gives $A=-4/D^2$ and $B=2(1-\omega^2)/(\omega D^2)$. Thus
$$
y(t)=e^{-t}\left(\frac{2t}{D}+\frac4{D^2}\right)-\frac4{D^2}\cos\omega t+\frac{2(1-\omega^2)}{\omega D^2}\sin\omega t.
$$
The first term tends to zero. The amplitude of the surviving sinusoid is
$$
\boxed{\sqrt{A^2+B^2}=\frac{2}{\omega(1+\omega^2)}.}
$$
As an independent interpretation, the <residual amplitude of an oscillator after an integrable force> is $\omega^{-1}|\int_0^\infty F(t)e^{-i\omega t}dt|$. Here that integral is $2/(1+i\omega)^2$, giving the same result.
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