Solving transient circuit with serial RLC using Laplace Transform
I have tabulated and plotted my solution given below in Excel (fully general/universal solution, i.e. all 3 cases, see below; I can provide an interested person with it via e-mail; I'm Czech as the inquirer MightyPork most probably is) and have also simulated the circuit in PSpice (receiving identical results with Excel graph at the bottom). Instead of 'I with hat' I have used I(s) notation during my derivation.


http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&amp;id=7c762190486dfb47dca59a9a1f8cb1a8&amp;title=Inverse%20Laplace%20Transform%20Calculator&amp;theme=orange&amp;i0=(s-z0)/(s(s-p1)(s-p2))&amp;i1=s&amp;i2=t&amp;podSelect=&amp;includepodid=Input&amp;includepodid=Result




(plot added 2015-02-13)


Appendix: (added 2015-02-11)

MightyPork wrote:


  …There I got kinda stuck, I don't know how to proceed. Also, it's possible I've made some mistake.
  How do I get current in the time domain? It should be some kind of damped wave, looking at the poles, but I'm not sure how to do the inverse transform of this…



Yes, you've made a little mistake missing a minus sign in exponent (rather a sort of a "typing" error :) :





To do the inverse transform of that, then, if you want to use the "WolframAlpha Calculator" mentioned above (or some other similar tool, tabulated expressions etc.), you should find roots of the numerator (zeroes) and denominator (poles; you've already done it) and rewrite the right side as follows:




For our expression you can find the inverse Laplace transform as:



So, if we are interested just in our particular case (with complex poles), then we can write:



so, the final result (as rioraxe has already stated) is:





Looking for the voltage \$ u_C(t) \$, we have to integrate the calculated current as follows (we know that \$ u_C(0) = 0 \$):

(I have used a variable x instead of t within the function i(t) so as not to confuse the independent variable with the definite integral limits which finally will become the independent variable in the result (after integration and substituting them)



If we wanted to do it manually, it'd be better to use the above "exponential expression" because 

and the integration itself becomes quite easy, but let's do it using the WolframAlpha Calculator:

Requesting integration of a generic expression  'e^ax (c sin(bx) + d cos(bx))'

by entering the "integrate e^ax (c sin(bx) + d cos(bx))" command at

http://www.wolframalpha.com/calculators/integral-calculator/

we receive:



and for the following given values \$ \&gt; a \&gt; = \&gt; –4000; \&gt; b = \&gt; 3000; \&gt; c \&gt; = \&gt; \frac{1}{3}; \&gt; d \&gt; = \&gt; 1, \&gt; C \&gt; = \&gt; 0,4 \cdot 10^{–6} \&gt; \$ :





… astonishingly, the same result as earlier :)  (having used the Laplace transform till the end of calculation previously)