Solving a series RLC DC step response circuit
Well, we have the following circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

Using KCL, we can write (I'm doing the analysis in the s-domain):

$$
\begin{cases}
\text{I}_{\text{R}_1}=\text{I}_{\text{R}_3}+\text{I}_\text{C}\\
\\
\text{I}_{\text{R}_2}+\text{I}_{\text{R}_3}+\text{I}_2=0\\
\\
\text{I}_1=\text{I}_{\text{R}_1}+\text{I}_{\text{R}_2}\\
\\
\text{I}_\text{C}=\text{I}_1+\text{I}_2
\end{cases}\tag1
$$

Using KVL, we can write:

$$
\begin{cases}
\text{I}_{\text{R}_1}=\frac{\text{V}_1-\text{V}_3}{\text{R}_1}\\
\\
\text{I}_{\text{R}_2}=\frac{\text{V}_1-\text{V}_4}{\text{R}_2}\\
\\
\text{I}_{\text{R}_3}=\frac{\text{V}_3-\text{V}_4}{\text{R}_3}\\
\\
\text{I}_2=\frac{\text{V}_2-\text{V}_4}{\text{sL}}\\
\\
\text{I}_\text{C}=\frac{\text{V}_3}{\left(\frac{1}{\text{sC}}\right)}=\text{V}_3\text{sC}
\end{cases}\tag2
$$

Using your given values we get:

$$\text{i}\left(t\right)=-\text{i}_2\left(t\right)=-\mathcal{L}_\text{s}^{-1}\left[-\frac{10\left(12+\text{s}\right)}{\text{s}\left(\text{s}+4\right)^2}\right]_{\left(t\right)}=\frac{5\left(3-\exp\left(-4t\right)\left(3+8t\right)\right)}{2}\tag3$$