Initial Conditions and Parallel Resonant Circuit Problem
I'll present here the theoretical solution. For \$- \infty &lt; t &lt; 0\$ only the voltage source is present in this RLC parallel circuit, which establishes an initial inductor current of \$i(0-)=i(0+)= 30 \space V / 10 \space \Omega = 3 \space A\$. Note also that \$v(0-)=v(0+)= 0 \space V\$, since the inductor can be seen as a short circuit for \$t &lt; 0\$. So, the circuit can be converted as shown on figure below:

Applying KCL on upper node:
$$ -I + \frac{v(t)}{R} + C\frac{d}{dt}v(t) + i(t) = 0 $$
Replacing \$v(t) = L\frac{d}{dt}i(t)\$
$$ LC\frac{d^2}{dt^2}i(t) + \frac{L}{R}\frac{d}{dt}i(t) + i(t) = I $$
The characteristic polynomial is:
$$ LCs^2 + \frac{L}{R}s + 1 = 0 $$
With roots
$$ s_{1,2} = -\frac{1}{2RC} \space \pm \sqrt{\left ( \frac{1}{2RC}\right )^2 - \frac{1}{LC}} $$
As \$\left ( \frac{1}{2RC}\right )^2 &gt; \frac{1}{LC}\$, the system is overdamped, with two real and distinct roots \$s_1 = -2.5\$ and \$s_2 = -10\$ for the current case.
The complete response has the form
$$ i(t) = i_f + A_1e^{-s_1t} + A_2e^{-s_2t} $$
where \$i_f\$ is the forced response (in this case, \$9 \space A\$). The constants \$A_1\$ and \$A_2\$ can be determined from the initial conditions.
$$
\left\{\begin{matrix}
 i(0) = I + A1 + A2 &amp; (1)\\ 
 \frac{d}{dt}i(0) = -s_1A_1 -s_2A_2 &amp; (2)\\
\end{matrix}\right.
$$
Note that \$v(0) = L\frac{d}{dt}i(0)\$. As \$v(0) = 0 \space V\$ then, \$\frac{d}{dt}i(0) = 0 \space A/s\$
Therefore
$$
\left\{\begin{matrix}
 3 = 9 + A1 + A2 &amp; (1)\\ 
 0 = -2.5A_1 -10A_2 &amp; (2)\\
\end{matrix}\right.
$$
Resolving, \$A_1 = -8\$ e \$A_2 = 2\$
Finally, the inductor current in Ampere is
$$ i(t) = 9 -8e^{-2.5t} + 2e^{-10t}$$
With graph