How to caculate the step response of the second order system with nonzero initial condition,$$ \ddot{y}+ 2ξω_n\dot y +2y = 0, y(0) = 0, \dot y(0) = c$$ (where $c$ is a contant)
Question: how to calculate the transfer function of this equation? since the input function is $0$? so no way to calculate the step response of this equation.
We have a linear differential equation with constant coefficients:
$$L[y]=y''(t) + 2 \xi \omega _n y'(t)+2 y(t)$$
In order to solve $L[y]=0$ we try using $y(t)=\exp (\lambda \cdot t)$ and get the characteristic polynomial
$$p_{L}(\lambda)=\lambda ^2+2 \xi \omega_{n} \cdot \lambda _n+2$$
Then the transfer function is
$$H(s)=\frac{1}{p_{L}(s)}=\frac{1}{s^2+2 \xi \omega_{n} \cdot s+2}$$
For the step response we use the Heaviside $\theta(t)$ unit step function and it's Laplace transform:
$$\mathcal{L}_t[\theta (t)](s)=\frac{1}{s}$$
The step response using inverse Laplace transform then is
$$\sigma(t)=\mathcal{L}_s^{-1}\left[H(s)\cdot\frac{1}{s}\right](t)=\mathcal{L}_s^{-1}\left[\frac{1}{s \left(s^2+2 \xi \omega _n \cdot s +2\right)}\right](t)$$
The general solution is:
$$\sigma(t)=\frac{\xi \omega _n e^{t \left(-\sqrt{\xi ^2 \omega _n^2-2}-\xi \omega _n\right)}}{4 \sqrt{\xi ^2 \omega _n^2-2}}-\frac{\xi \omega _n e^{t \left(\sqrt{\xi ^2 \omega _n^2-2}-\xi \omega _n\right)}}{4 \sqrt{\xi ^2 \omega _n^2-2}}-\frac{1}{4} e^{t \left(-\sqrt{\xi ^2 \omega _n^2-2}-\xi \omega _n\right)}-\frac{1}{4} e^{t \left(\sqrt{\xi ^2 \omega _n^2-2}-\xi \omega _n\right)}+\frac{1}{2}$$
Example with $\xi=1$ and $\omega_{n}=\frac{1}{2}$:
$$\sigma(t)=-\frac{e^{-\frac{t}{2}} \sin \left(\frac{\sqrt{7} t}{2}\right)}{2 \sqrt{7}}-\frac{1}{2} e^{-\frac{t}{2}} \cos \left(\frac{\sqrt{7} t}{2}\right)+\frac{1}{2}$$
Visualize example:
See also Transfer function