How to solve coupled linear ODE?

Question

I wan to solve the following ODE's:- $$L_1 q''(t)+R_1q'(t)+\frac 1C_1 q(t)-Mq_2''(t)=V\sin(\omega t)$$ $$L_2 q_2''(t)+R_2q_2'(t)+\frac 1C_2 q_2(t)-Mq''(t)=V\sin(\omega t)$$ $L,C,R,V>0$, I already know how to solve linear non-homogenous and non-coupled ODE, using the homogenous solution plus the particular solution found by using a trial function.

How can I reduce the given equations to non=coupled form so as to apply similar methodologies to solve them?

I just know the basic methods to solve simple ODE's. I am comfortable with solving simple coupled ODE such as $$y_1'(t)=Ay_2(t)\text{ and } y_2'(t)=B+Cy_1(t)$$ by simple substitution.
(The physics tag is added because these equations describe the behaviour of coupled RLC circuit with an application in metal detectors)

Answer 1

From the first equation, we get $$q_2''=Aq''+Bq'+Cq+D\sin(\omega t)\tag1$$ for some constants $A,B,C,D$. Then also $$q_2'''=Aq'''+Bq''+Cq'+D\omega\cos(\omega t)\tag2$$ and $$q_2''''=Aq''''+Bq'''+Cq''-D\omega^2\sin(\omega t)\tag3$$

Differentiating the second equation twice gives $$Eq_2''''+Fq_2'''+Gq_2''+Hq''''=I\omega^2\sin(\omega t)\tag4$$ for some constants $E,F,G,H,I$. Now replace $q_2''$, $q_2'''$, and $q_2''''$ in (4) by using (1), (2), and (3), respectively, and voila! equations uncoupled.

Answer 2

You can also use the operator method (a bit advanced) to find the solution of the system. First write the system as an algebraic equations their coefficients are second order differential operators: $$\big (L_1 D^2+R_1 D+\frac{1}{C_1}\big)q-MD^2 q_2=V\sin(\omega t)$$ $$-M D^2 q+\big(L_2 D^2+R_2 D+\frac{1}{C_2}\big)q_2=V\sin(\omega t).$$ Now we can get fourth order differential equations for $q$ and $q_2$, respectively by a simple eliminaton (Cramer's rule is convenient). However, this way will also be cumbersome because you need to find the roots of a fourth degree polynomial (characteristic equation) with the coefficients depending on $C_1, C_2, L_1, L_2, R_1$, and $R_2$. Notice that finding the roots of this polynomial is required for the decoupled equation found in the first answer too. As a result, whatever method you use, you will eventually meet this fourth degree polynomial.