Hello again derick,
Well im very sorry to say this but something still isnt right. The output at the first instant of time has to be zero (0.000000) because all the voltage appears across the inductor and the initial voltage of the capacitor is zero and the initial inductor current is zero.
Again this is using the following values which we had been using in order to make comparison of our separate results easier:
L=100uH
C=22uF
R=10
You may wish to note that the response is an exponentially damped sinusoidal with R>=1.066 ohms. With R less than that we have to consider another type of response, but R=10 is a good choice for now so we'll stick with the damped sinusoidal.
Since your G(s) is correct, i can only imagine that you must be making a mistake in the transformation from G(s) to what you are calling the time function: Y(t).
Y(t)=1-e^(-at)*[A*sin(wt)+B*cos(wt)]
and you end up with A=B, but in real life we almost never end up with A=B as A is almost always different than B. It's true that they are both negative, but still they are not equal. I didnt want to yell out the result yet but just for a hint they are different by almost a factor of 10.
Also, when we do a simple circuit like this analytically, we never get results that are off by as much as 0.01 or -0.01. If they are off they would be off by a tiny amount like 1e-14 or -1e-14 on a typical PC computer using 16 digit floating point math. That's because the equations are so incredibly exact that the only error can come in during the numerical evaluation, and that's usually very accurate for small simple equations like this resulting time equation.
Just for reference, the numerical result for this problem is slightly greater than +200uv at t=1us.
You might also be interested to know that you can always check your results with a numerical simulator program like LT Spice.
Just to add an additional note, we can do this problem in a purely numerical environment (numerical calculations only) and that will back up our analytical results. This is also very interesting as im sure you will find. We can use a standard approach for ODE's to get there so it wont be a waste of time where we can only do this on this one application but can do this for many different applications too