I don't think I can give you all the details, but this may help you get started. Fortunately, there are lots of information on the Web for this. A typical way of reverse transforming an expression like yours is to look up a table of Laplace transforms, rewrite the expression such that the components match one or more of the forms in the table. With your expression, it is going to fit into these two forms (copied from Wikipedia):
Now, rewrite your expression such that \$I(s) = AP(s) + BQ(s), A,B\$ are constants.
The first step of rewriting would probably be rewriting the denominator by "completing the square". And you would be factoring out constants during the rewrite.
After you get P(s) and Q(s) to match the forms in the table exactly, then, \$i(t) = Ap(t) + Bq(t)\$ by using the original functions given by the table.
New edit: I was curious, so finished the reverse transform as below
$$
I=\frac{8 \times 10^{-5}s + 0.4}{4 \times 10^{-3} s^2 + 32s +10^5}
=\frac{0.02s + 100}{s^2 + 8000s + 25000000} $$
$$
=0.02 (\frac{s + 5000}{s^2 + 8000s + 25000000})
=0.02 (\frac{s + 5000}{(s+4000)^2 + 3000^2}) $$
\$\alpha=4000\$, \$\omega=3000\$
These match the roots you calculated for the poles.
$$
I = 0.02 (\frac{s + 4000 - 4000 + 5000}{(s+4000)^2 + 3000^2})
= 0.02 (\frac{s + 4000}{(s+4000)^2 + 3000^2}+\frac{1000}{(s+4000)^2 + 3000^2})
= 0.02 (\frac{s + 4000}{(s+4000)^2 + 3000^2}+\frac13\frac{3000}{(s+4000)^2 + 3000^2})$$
$$
i(t) = 0.02(e^{-4000t}cos(3000t) + \frac13e^{-4000t}sin(3000t))$$
What got me curious was I tried to get the reverse transform from WolframAlpha also, and got some really complicated answer in real form. My guess is that it uses a mechanical method that produces an answer too complicated for it to reduce. So if one just plug numbers into a computer to get answers, sometimes simpler relationships may stay hidden.
Thevenin Equiv circuit for v1 goes from 60V to 40% or 24V with Rs=30//20=12 ohms. The steady state v(t) across cap is then +24-30= -6Vdc
Then we know resonance ω= √(L/C)= v(0.5) = 0.707 and Q=XL/R = 2 π 0.5/12 which is <1 and thus overdamped and the initial voltage across cap will be 0V.
Does this method make it easier than Laplace?
Best Answer
It's easy to solve simple filter circuits in the time domain but there becomes a tipping-point where most engineers would prefer to solve problems in the frequency domain and apply (for example) a step-function. Finding the inverse Laplace is fairly straightforward because of Laplace tables.
As an example of an RLC low pass filter, engineers become accustomed to the transfer function: -
$$H(s) = \dfrac{\omega_n^2}{s^2+2\zeta\omega_n^2 s + \omega_n^2}$$
And, applying (for example) a step function is as simple as multiplying by 1/s: -
$$\dfrac{1}{s}\cdot\dfrac{\omega_n^2}{s^2+2\zeta\omega_n^2 s + \omega_n^2}$$
Engineers who are familiar with this get to recognize that this converts to a standard form and the next step is just using the tables to derive the transient response.
In addition, the term \$\omega_n\$ (the natural resonant frequency) can be is factored out a lot of the time so that solving the step-function formula for \$\omega_n=1\$ begins with: -
$$\dfrac{1}{s}\cdot\dfrac{1}{s^2+2\zeta s + 1}$$
This is rearranged to a standard form such as this (underdamped resonance): -
$$\dfrac{1}{s[(s+a)^2 + b^2]}$$
Where \$a =\zeta\$ and \$b=\sqrt{1-\zeta^2}\$
The Laplace tables give us this: -
$$H(t) = 1+\dfrac{1}{\sqrt{1-\zeta^2}}\cdot e^{-\zeta t}\cdot \sin(t\cdot\sqrt{1-\zeta^2}+\phi)$$
Where \$\phi = \arccos(\zeta)\$
But you will probably only be convinced when you are dealing with slightly more complex situations or really do need to analyse the frequency spectrum.