Taking the Laplace transform of the system
\\[
\begin{aligned}
2 i_{1}^{\prime}+50 i_{2} &=60 \\
0.005 i_{2}^{\prime}+i_{2}-i_{1} &=0
\end{aligned}
\\]
gives
\\[
2 s \mathscr{L}\left\\{i_{1}\right\\}+50
\mathscr{L}\left\\{i_{2}\right\\}=\frac{60}{s}
\\]
\\[
-200 \mathscr{L}\left\\{i_{1}\right\\}+(s+200) \mathscr{L}\left\\{i_{2}\right\\}=0
\\]
so that
\\[
\begin{aligned}
\mathscr{L}\left\\{i_{2}\right\\} &=\frac{6,000}{s\left(s^{2}+200
s+5,000\right)} \\
&=\frac{6}{5} \frac{1}{s}-\frac{6}{5} \frac{s+100}{(s+100)^{2}-(50
\sqrt{2})^{2}}-\frac{6 \sqrt{2}}{5} \frac{50 \sqrt{2}}{(s+100)^{2}-(50
\sqrt{2})^{2}}
\end{aligned}
\\]
Then
\\[
i_{2}=\frac{6}{5}-\frac{6}{5} e^{-100 t} \cosh 50 \sqrt{2} t-\frac{6
\sqrt{2}}{5} e^{-100 t} \sinh 50 \sqrt{2} t
\\]
and
\\[
i_{1}=0.005 i_{2}^{\prime}+i_{2}=\frac{6}{5}-\frac{6}{5} e^{-100 t} \cosh 50
\sqrt{2} t-\frac{9 \sqrt{2}}{10} e^{-100 t} \sinh 50 \sqrt{2} t
\\]