Solution of current in series RLC circuit for a forcing function
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:

$$\small I=\frac{V_m}{R} \left (1-e^{-\alpha t}\right )\:sin(\omega t)$$