I'll stay completely in the time domain.
In my own way of writing:
$$\begin{align*}R_1\,i+\frac1{C_1}\int i\:\text{d}t+L_1\frac{\text{d}}{\text{d} t}i&=V_0\sin\left(\omega\:t\right)\tag{1}\label{eq:1}\\\\\frac{\text{d}}{\text{d}t}\left\{R_1\,i+\frac1{C_1}\int i\:\text{d}t+L_1\frac{\text{d}}{\text{d} t}i\right\}&=\frac{\text{d}}{\text{d}t}\left\{V_0\sin\left(\omega\:t\right)\vphantom{\int}\right\}\\\\R_1\frac{\text{d}}{\text{d} t}i+\frac1{C_1}i+L_1\frac{\text{d}^2}{\text{d} t^2}i&=V_0\:\omega\:\cos\left(\omega\:t\right)\tag{2}\label{eq:2}\\\\\left[\frac{\text{d}^2}{\text{d} t^2}+\frac{R_1}{L_1}\frac{\text{d}}{\text{d} t}+\frac1{L_1\,C_1}\right]i&=\frac{V_0}{L_1}\,\omega\:\cos\left(\omega\:t\right)\\\\\left[\left(\frac{\text{d}}{\text{d} t}-\alpha_{_1}\right)^2+\beta_{_1}^2\right]i&=\frac{V_0}{L_1}\,\omega\:\cos\left(\omega\:t\right)\tag{3}\label{eq:3}\end{align*}$$
where \$\alpha_{_1}=-\frac12\frac{R_1}{L_1}\$ and \$\beta_{_1}=\pm\frac12\sqrt{4\frac1{L_1\,C_1}-\left(\frac{R_1}{L_1}\right)^2}\$. As \$\beta_{_1}\$ is real, the result will be under-damped.
But the above is inhomogeneous. So annihilate the right side:
$$\begin{align*}\left[\left(\frac{\text{d}}{\text{d} t}-0\right)^2+\omega^2\right]\left[\left(\frac{\text{d}}{\text{d} t}-\alpha_{_1}\right)^2+\beta_{_1}^2\right]i&=\left[\left(\frac{\text{d}}{\text{d} t}-0\right)^2+\omega^2\right]\frac{V_0}{L_1}\,\omega\:\cos\left(\omega\:t\right)\\\\\left[\left(\frac{\text{d}}{\text{d} t}-0\right)^2+\omega^2\right]\left[\left(\frac{\text{d}}{\text{d} t}-\alpha_{_1}\right)^2+\beta_{_1}^2\right]i&=0\end{align*}$$
All solutions to the above homogeneous equation must be of the following general form:
$$ i=A_1\cos\left(\omega\,t\right)+A_2\sin\left(\omega\,t\right)+A_3\exp\left(\alpha_{_1}\,t\right)\cos\left(\beta_{_1}\,t\right)+A_4\exp\left(\alpha_{_1}\,t\right)\sin\left(\beta_{_1}\,t\right)\tag{4}\label{eq:4}$$
(Had \$\beta_{_1}\$ been imaginary, the \$\cos\$ and \$\sin\$ functions for the latter two terms would have been converted to \$\cosh\$ and \$\sinh\$ functions.)
There are two initial conditions that you have specified in comments:
- the inductor current is \$i=0\:\text{A}\implies\:R_1\,i=0\:\text{V}\$ at \$t=0\$; and,
- the voltage across \$C_1\$ is \$0\:\text{V}\implies \frac1{C_1}\int i\:\text{d}t=0\:\text{V}\$ at \$t=0\$.
From the above two results. and using Eq.\$\ref{eq:1}\$, then find \$\frac{\text{d}}{\text{d}t}i=0\$. And using that information plus Eq.\$\ref{eq:2}\$, then find \$\frac{\text{d}^2}{\text{d}t^2}i=\frac{V_0}{L_1}\omega\$.
The above results, and Eq.\$\ref{eq:4}\$, lead to these simultaneous equations:
- \$A_1+A_3=0\$
- \$A_2\:\omega-A_1\:\alpha_{_1}+A_4\:\beta_{_1}=0\$
- \$-A_1\:\omega^2 + A_3\left(\alpha_1^2 - \beta_1^2\right) + 2\,A_4\:\alpha_1\:\beta_1=\frac{V_0}{L_1}\omega\$
Also, at \$t\to\infty\$, the square of the rotating magnitude is (noting that the cosine and sine functions are relatively orthogonal):
- \$A_1^2+A_2^2=\frac{V_0^2}{\,\left[\left(R_1\right)^2+\left(\omega\:L_1-\frac1{\omega\,C_1}\right)^2\right]}\$
Which completes the need for four equations to find the four unknowns.
You already know that \$\omega=2\pi\:24\:\text{kHz}\approx 150796.44737231\$.
With your part values find:
- \$\alpha_{_1}=-\frac12\frac{10\:\text{m}\Omega}{40\:\mu\text{H}}=-125\$
- \$\beta_{_1}=\frac12\sqrt{\frac4{40\:\mu\text{H}\,\cdot\,1.25\:\mu\text{F}}-\left(\frac{10\:\text{m}\Omega}{40\:\mu\text{H}}\right)^2}\approx 141421.301\$
- \$A_1=-687.918062062593\$
- \$A_2=9.46736242762291\$
- \$A_3=687.918062062593\$
- \$A_4=-9.48600876867154\$
If you substitute into Eq.\$\ref{eq:4}\$ and plot using Desmos:
it will look just the same as it does under LTspice:
Just as an additional quick sanity test, set \$R_1=200\:\text{m}\Omega\$ and try again:
- \$\alpha_{_1}=-\frac12\frac{200\:\text{m}\Omega}{40\:\mu\text{H}}=-2500\$
- \$\beta_{_1}=\frac12\sqrt{\frac4{40\:\mu\text{H}\,\cdot\,1.25\:\mu\text{F}}-\left(\frac{200\:\text{m}\Omega}{40\:\mu\text{H}}\right)^2}\approx 141399.257423793\$
- \$A_1=-639.601304243966\$
- \$A_2=176.030646801803\$
- \$A_3=639.601304243966\$
- \$A_4=-176.030646801803\$
Again substitute into Eq.\$\ref{eq:4}\$ and plot using Desmos:
And here's LTspice:
Entirely done in the time domain. No Laplace.