Solving a source free series RLC circuit
All that matter here are the capacitor and the dependent current source. Regardless of the rest of the circuit, the current source adjusts its voltage to guarantee that the current is \$3v_c\$.

Note that, with the polarities given, the current source discharges the capacitor from an initial \$10V\$, and thus we may write:

$$v_c=10-\frac{1}{C}\int 3v_c\:dt\:=10-300\int v_c\:dt$$

Where the minus sign indicates that the current flow is discharging \$C\$.

Differentiating this equation:

$$\frac{dv_c}{dt}=-300v_c$$

This is a first order differential equation and, clearly, the solution is an exponential of the form:

$$v_c=Ae^{\alpha t}$$

By inspection, \$\alpha =-300\$, and using the initial condition: \$t=0, \:v_c=10\$, gives \$A=10\$. Hence:
$$v_c=10e^{-300t}$$

From the circuit, we see that \$i_L=-3v_c\$, therefore:

$$i_L=-30e^{-300t}$$