Jtdylktuy

I want to solve the following equations for $x_1$, $x_2$, and $x_3$

$$ K_0, x_1(t) + K_1, frac{dx_1(t)}{dt} + M_0, frac{dx_2(t)}{dt} + x_3(t) =A_0,sin (alpha_0 t) tag{1} $$

$$ M_0, frac{dx_1(t)}{dt} + L_0,x_2(t) + K_1, frac{dx_2(t)}{dt}+L_1,int x_2(t), dt =0 tag{2} $$

$$ x_1(t) -alpha_1,B_1,cos(alpha_1,t + phi) , x_3(t)- big[B_0+B_1,sin(alpha_1,t + phi) big] frac{dx_3(t)}{dt} =0 tag{3} $$

Here is my try using Laplace transform:

I can assume $x_1(0)=0$, and define $x_2(0)=x_{20}$, and $x_3(0)=x_{30}$

$$(1) Rightarrow K_0, X_1(s) + K_1, sX_1(s)+ M_0, sX_2(s)+ X_3(s) + M_0, x_{20} = frac{A_0alpha_0}{s^2+alpha_0^2} tag{4} $$

$$(2) Rightarrow M_0, X_1(s) + L_0, X_2(s) +K_1, sX_2(s)+ L_1, frac{X_2(s)}{s}+ X_3(s)+ K_1, x_{20} =0 tag{5} $$
Next step brings me the trouble. I cannot simply take the Laplace transform of (3)

Update 1 (as suggested by @Cesareo)

Using the Euler's expansion of sine and cosine (3) can be transformed to

$$ X_1(s) - B_0 s X_3(s) -\
frac{B_1 (alpha_1 +1) s}{2} big(e^{i phi} X_3(s-i alpha_1) + e^{-i phi} X_3(s+i alpha_1)big)+x_{30} (B_0 - B_1 sin phi)=0 tag{6}$$

Now, my question is:
How can I move forward from here? How can I handle $X_3(s pm i alpha_1)$ ifI want to solve for $X_3(s)$?

Hint.

$$
mathcal{L}(cos(alpha t)x(t)) = mathcal{L}(frac 12(e^{ialpha t}+e^{-ialpha t})x(t))
$$

and

$$
int_0^{infty}e^{ialpha t}x(t)e^{-s}dt = int_0^{infty}e^{-(s-ialpha)t}x(t) dt = X(s-ialpha)
$$

so after the Laplace transformation you should handle a system of difference equations ...