It's, arguably, easier to solve this in Laplace and, if the system were 1st order, the solution will be of the form: \$\small A(1-e^{-at})sin (\omega t+\phi)\$. You have an overdamped 2nd order system, so the final equation will have two transient exponentials.
A differential equation solution will obviously give the same answer, but will require integration by parts, probably a couple of times. But it's a nice exercise.
The following analysis has been added to the original answer
The differential equation relating current, \$\small I(t)\$, and applied voltage, \$\small v=V_msin(\omega t)\$, in a series RLC, circuit is:
$$\small \ddot I+\frac{R}{L}\:\dot I+\frac{1}{LC}\:I=\frac{1}{L}\:\dot v $$
evaluating \$\small \dot v\$, and writing the 2nd order term in standard form:
$$\small \ddot I+2\zeta\omega\:\dot I+\omega^2\:I=\frac{V_m}{L}\:\omega\:cos(\omega t) \:\:\:...\:(1)$$
where, in this case, the applied sinusoidal voltage is at the natural frequency, \$\small \omega=\frac{1}{\sqrt{LC}}\$
The particular integral (steady state solution), given a sinusoidal input, is:
$$\small I_{ss}= A\:sin(\omega t)+B\:cos(\omega t)$$
Differentiating \$\small I_{ss}\$ twice:
$$\small \dot I_{ss} =A\:\omega\:cos(\omega t)-B\:\omega\: sin(\omega t)$$
$$\small \ddot I_{ss} =-A\:\omega^2\:sin(\omega t)-B\:\omega^2\: cos(\omega t)$$
Substituting for \$\small I\$, \$\small \dot I\$, and \$\small \ddot I\$ in \$\small (1)\$:
$$\small -2\zeta\omega^2B\:sin(\omega t)+2\zeta\omega^2A\:cos(\omega t)=\frac{V_m}{L}\:\omega\:cos(\omega t) $$
Hence, comparing coefficients, $$\small B=0$$
$$\small A=\frac{V_m}{2\zeta\omega L}=\frac{V_m}{R}$$
and
$$ \small I_{ss}=\frac{V_m}{R}sin(\omega t)$$
The homogeneous solution (transient solution) is of the form:
$$\small I_{h}= e^{-\alpha t}\left ( D\:sin(\omega t)+E\:cos(\omega t)\right )$$
and the overall expression for \$\small I\$ is, thus:
$$\small I=I_h + I_{ss}= e^{-\alpha t}\left ( D\:sin(\omega t)+E\:cos(\omega t)\right )+\frac{V_m}{R}sin(\omega t)$$
By inspection, initial conditions are: \$\small t=0;\: I=0;\:\dot I=0\$
The first condition gives: \$\small E=0\$, and the second condition, obtained by evaluating \$\small \dot I\$, gives: $$\small D=-\frac{V_m}{R}$$
Therefore the final expression for \$\small I\$ is:
$$\small I=\frac{V_m}{R} \left (1-e^{-\alpha t}\right )\:sin(\omega t)$$
Best Answer
The best way would be to do it in the Laplace domain and then convert it back to the time domain.
Clearly the current supplied by the source is equal to the inductor current. Find the equivalent impedance and divide it by the supply voltage to get the current equation.
The equivalent impedance comes out to be -
The current equation then comes out to be -
Now you can take the inverse laplace of this to find the time domain expression.
Method 2 would be, as it has been pointed out in the earlier comments, is to find out the damping ratio and determine the nature of the system, but that would of course require the component values. I have provided you a general method to solve it.