= Solution
The <Fourier transform of a derivative> is $i\omega$ times the transform under the stated convention. Transforming the <linear ordinary differential equation> gives
$$
(-\omega^2+5i\omega+6)\widetilde x=\widetilde F,
\qquad\boxed{R(\omega)=\frac1{(i\omega+2)(i\omega+3)}.}
$$
The applied <force> is $F(t)=H(t)e^{-t}$, with $H$ the <Heaviside step function>, so $\widetilde F=(1+i\omega)^{-1}$. Partial fractions give
$$
\widetilde x=\frac{1/2}{1+i\omega}-\frac1{2+i\omega}+\frac{1/2}{3+i\omega}.
$$
Using the transform pair $H(t)e^{-ct}\leftrightarrow(c+i\omega)^{-1}$ yields
$$
\boxed{x(t)=H(t)\left(\tfrac12e^{-t}-e^{-2t}+\tfrac12e^{-3t}\right).}
$$
Its value and first <derivative> both vanish at $t=0^+$, so the joining introduces no spurious <Dirac delta> terms. Equivalently, this is the <convolution> of the <force> with the causal <impulse response> $H(t)(e^{-2t}-e^{-3t})$.
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