Is there a way to find the transfer function from only your input and the steady state response?
Clearly, no. Steady state response means assentially the 0 frequency response. Obviously systems can have the same 0 frequency (DC) response but various responses to other frequencies.
For example, consider a simple R-C low pass filter. The DC response is simply 1 (output = input), but that is not true of higher frequencies. Different values of R and C will have different frequency profiles, but all have the same response to DC.
Or consider a simple R-C high pass filter. The DC response is 0, but obviously that is not true of other frequencies.
Added:
As Roger C mentioned in a comment, perhaps you meant steady state response to a fixed input waveform, not necessarily a steady (DC) one. If that is the case, it is possible to get a good idea of the overall transfer function if you do the test at lots of frequencies. However, a good idea of the transfer function is not the same as the actual transfer function, so the answer is still no in theory.
One way to look at this is that by putting in a particular pure frequency (sine signal) and measuring the steady state output (amplitude and phase), you measure one point of the transfer function in frequency space. By point-sampling the transfer function in frequency space, you can infer the continuous frequency response. The inverse Fourier transform of that gives you the impulse response, which is what you are looking for.
The problem with this method is that it only point-samples the freqency response, which never guarantees what the response is between the samples. If you know something about your system, then point sampling it at strategic frequencies could be good enough in practice, but this does not work in the general case. For example, it is quite common to measure a audio amplifier this way, since you know there aren't supposed to be sharp resonant peaks or the like in the frequency response.
"Vin is 5V, so Vout should be 50,000V."
Why? The OpAmp amplifies the the difference between the + and - inputs, not just the value on the + input!
OK, you might start with: the output is at 0V, and the input (connected to the + input) is 5V. What you have done is apply a 5V step to the input.
Now what happens is that the OpAmp starts to rise the voltage on the output. It can't do this at once, so it will rise 'slowly' (for some rather fast value of slowly, which has a technical name in OpAmp world: the slew rate, which is an importnat charactreistic of a real OpAmp). When it reaches 5V, this is fed back to the negative input, at which time it compensates the 5V at the + input, so the OpAmp no longer tries to rise its output level. (To be really accurate: this happens a little bit earlier, when the difference is 5V/10k.)
Depending on timing characteristics, the output might 'slowly' settle to 5V, or overshoot the 5V, drop below 5V, etc (oscillate towards 5V). If the circuit is designed badly the oscillation might increase (and never end).
Best Answer
If you replace \$s\$ by \$j\omega\$ you get the system's frequency response \$H(j\omega)\$, which you'll need later on. First you have to compute the Fourier series of the periodic input signal:
$$v_b(t)=\sum_{n=-\infty}^{\infty}c_ne^{jn\omega_0t},\quad \omega_0=\frac{2\pi}{T}$$
where \$T=6\$ms is the period of \$v_b(t)\$. The Fourier coefficients \$c_n\$ are given by
$$c_n=\frac{1}{T}\int_0^Tv_b(t)e^{-jn\omega_0t}dt\tag{1}$$
I haven't evaluated the integral, but it should be pretty straightforward because of all the straight lines in \$v_b(t)\$. Once you have the coefficients \$c_n\$ you need to realize that the response of the system to an exponential input \$e^{j\omega_0t}\$ is simply \$H(j\omega_0)e^{j\omega_0t}\$ (because the system is linear and time-invariant). So you finally get for the output signal
$$v_o(t)=\sum_{n=-\infty}^{\infty}c_nH(jn\omega_0)e^{jn\omega_0t}$$
EDIT: In order to compute the Fourier coefficients you need to write down the piecewise definition of the input signal \$v_b(t)\$:
$$v_b(t)=\left\{\begin{array}{rc}3t-4,& 1\le t<2\\ -3t+8,&2\le t<3\\ -1,&3\le t<7\end{array}\right.$$
Then you split the integral (1) into three intervals:
$$c_n=\frac{1}{6}\left\{\int_1^2(3t-4)e^{-jn\omega_0t}dt+ \int_2^3(-3t+8)e^{-jn\omega_0t}dt- \int_3^7e^{-jn\omega_0t}dt \right\}$$
I guess you can take it from here.