Circuit with sinusoidal voltage source - transient analysis
I will attempt to solve the general solution here (in part because I am missing the resistance for R2). You can plug in your values wherever needed.

Before t=0

We first need to find the initial condition at t=0 in order to calculate the solution in the Laplace domain after t=0.

If we assume that the circuit is in steady-state before the switch is opened then the easiest method to find the steady-state solution would be to work with phasors (Fourier transform). We would get the following schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

Where

\$Rss = R_o + R_1//R_2\$

We can calculate the output voltage and current through L using these impedances:

\$\frac{Vout(j\omega)}{E(j\omega)}=\frac{j\omega L}{R_{ss}(1-LC\omega^2)+j\omega L}\$

\$\frac{i_L(j\omega)}{E(j\omega)}=\frac{1}{R_{ss}(1-LC\omega^2)+j\omega L}\$

The phasor at the input is given by using \$e(t)=A_o\ cos(\omega_o t+\phi)\$. At time equals zero, this phasor will be at (remember, the real part has to be the same):

\$E = A_o\ (\cos(\phi)+j\cdot sin(\phi))\$

All this information can now be used to find the steady-state solution by substituting all knowns.

\$Vout=E\cdot\frac{j\omega_o L}{R_{ss}(1-LC\omega_o^2)+j\omega_o L}\$

\$i_L=E\cdot\frac{1}{R_{ss}(1-LC\omega_o^2)+j\omega_o L}\$

It then suffices to take the real parts of \$Vout\$ and \$i_L\$ to find the voltage and current through the inductor at t=0:

\$v_C(t=0)=\Re(Vout)\$

\$i_L(t=0)=\Re(i_L)\$

After t=0

Now that we have the initial conditions, we can start calculating the solution using the Laplace transform.

You can prove that an inductor with an initial current is equivalent to an inductor with no initial current in parallel with a constant current source of that initial current. The same can be done for a capacitor, where a series voltage source is added. We can get the following equivalent circuit:







simulate this circuit

We can solve this circuit by writing the KCL equations (currents in = currents out):

\$\frac{E(s)-v_1}{R_o}=\frac{v_1-v_2}{R_1}+\frac{v_1-v_3}{R_2}\$

\$\frac{v_1-v_2}{R_1}=\frac{i_C(t=0)}{s}+\frac{v_2}{Ls}\$

\$\frac{v_1-v_3}{R_2}=\left(v_3-\frac{v_C(t=0)}{s}\right)\cdot Cs\$

This gives us three equations for three unknowns, which can be solved. I have to say that these equations are quite tedious though. I do not recommend solving these equations symbolically.

As an alternative, you can get the same equations without changing the circuit by staying in the time domain before transforming to the Laplace domain. We would get that:

\$\frac{e(t)-v_1}{R_o}=\frac{v_1-v_2}{R_1}+\frac{v_1-v_3}{R_2}\$

\$\frac{v_1-v_2}{R_1}=i_L(t)\$

\$v_2=L\cdot\frac{di_L(t)}{dt}\$

\$\frac{v_1-v_3}{R_2}=C\cdot\frac{dv_3}{dt}\$

After Laplace-transformation:

\$\frac{E(s)-v_1}{R_o}=\frac{v_1-v_2}{R_1}+\frac{v_1-v_3}{R_2}\$

\$\frac{v_1-v_2}{R_1}=I_L\$

\$v_2=L\cdot(I_Ls-i_L(t=0) \Rightarrow I_L=\frac{v2}{Ls}+\frac{i_L(t=0)}{s}\$

\$\frac{v_1-v_3}{R_2}=C\cdot(v_3s-v_C(t=0))\$

You will find that these equations are exactly the same.

I hope it helped!