I have the following system $$\ddot{x}+w^2 x=0,$$ with the following initial conditions: $\dot{x}(0)=0$ and $x(0)=x_o$, the solution reads: $$x(t)=x_o cos(t).$$ Now I want to include the fluctuations due to thermal fluctuations (in frictionless system), so the equation of motion is modified as $$\ddot{x}+w^2 x=\lambda \zeta(t),$$ with $\dot{x}(0)=0$ and $x(0)=0$ solution reads: $$x(t)=\lambda \int_0^{t}cos(t-t')\zeta(t')dt'$$ 1- I would like to know how to include the initial condition $\dot{x}(0)=0$ and $x(0)=x_o$ in the solution?
2- how to obtain the frequency spectrum due to these fluctuations?
You have to give the correlation function for the noise. I guess it is Gaussian being thermal noise and set $$ \langle\zeta(t)\zeta(t')\rangle=\sigma^2\delta(t-t') $$ where $\sigma$ is a constant. The general solution of the forced harmonic oscillator can be written down, with the given initial conditions, $$ x(t)=x_0\cos(\omega_0 t)+\frac{\lambda}{\omega_0}\int_0^t\sin(\omega_0(t-t'))\zeta(t')dt', $$ where I have slightly changed the OP notation by setting $w\rightarrow\omega_0$. This will yield for the correlation function $$ \langle x(t)x(t')\rangle=x_0^2\cos^2(\omega_0t)+\frac{\lambda^2}{\omega_0^2} \int_0^t dt'\int_0^t dt''\sin(\omega_0(t-t'))\sin(\omega_0(t-t'')) \langle\zeta(t')\zeta(t'')\rangle, $$ that is $$ \langle x(t)x(t')\rangle=x_0^2\cos^2(\omega_0t)+\frac{\lambda^2}{\omega_0^2}\sigma^2 \int_0^t dt'\sin^2(\omega_0(t-t')). $$ I think you can go on from here.