## Coupled differential equations

1. The problem statement, all variables and given/known data
I need to find the general solution for:
$R \frac{dQ_{1}}{dt} + \frac{1}{C}(Q_{1}-Q_{2})=0$
$R\frac{dQ_{1}}{dt} + L\frac{d^2Q_{2}}{dt^2}}=0$
For the case $R=1\times 10^3$, $L=4\times 10^{-3}$, $C=1 \times 10^{-9}$.

3. The attempt at a solution
As an attempt I looked for solutions of the form, $Q_{1}=q_{1}e^{at}, Q_{2}=q_{2}e^{at}$ which then in matrix form gives:
$\begin{pmatrix} Ra+\frac{1}{C} & -\frac{1}{C} \\Ra & La^2\end{pmatrix}\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}=\begin{pmatri x}0\\0\end{pmatrix}$

Setting the determinant of the matrix equal to zero, leads to a cubic. There is going to be a constant term corrisponding to the solution a=0, and the remaining quadratic gives a repeated root (using the values of L,C,R given): $a=-\frac{1}{2RC}$. For this value of a we get a solution:
$\begin{pmatrix}Q_{1}\\Q_{2}\end{pmatrix}=A \begin{pmatrix}2\\1\end{pmatrix} exp(\frac{-t}{2RC})$
where A is an arbitary constant.

But now I need to find another solution. I think it'll be of the form $(Bt+C)exp(\frac{-t}{2RC})$ where B is related to A but I'm not sure how to go about finding it? Most of the results google gives seem to be for a slightly different style of question where there is only one first order derivative in each equation.

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 Quote by plmokn 1. The problem statement, all variables and given/known data I need to find the general solution for: $R \frac{dQ_{1}}{dt} + \frac{1}{C}(Q_{1}-Q_{2})=0$ $R\frac{dQ_{1}}{dt} + L\frac{d^2Q_{2}}{dt^2}}=0$ For the case $R=1\times 10^3$, $L=4\times 10^{-3}$, $C=1 \times 10^{-9}$. 3. The attempt at a solution As an attempt I looked for solutions of the form, $Q_{1}=q_{1}e^{at}, Q_{2}=q_{2}e^{at}$ which then in matrix form gives: $\begin{pmatrix} Ra+\frac{1}{C} & -\frac{1}{C} \\Ra & La^2\end{pmatrix}\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}=\begin{pmatri x}0\\0\end{pmatrix}$ Setting the determinant of the matrix equal to zero, leads to a cubic. There is going to be a constant term corrisponding to the solution a=0, and the remaining quadratic gives a repeated root (using the values of L,C,R given): $a=-\frac{1}{2RC}$. For this value of a we get a solution: $\begin{pmatrix}Q_{1}\\Q_{2}\end{pmatrix}=A \begin{pmatrix}2\\1\end{pmatrix} exp(\frac{-t}{2RC})$ where A is an arbitary constant. But now I need to find another solution. I think it'll be of the form $(Bt+C)exp(\frac{-t}{2RC})$ where B is related to A but I'm not sure how to go about finding it? Most of the results google gives seem to be for a slightly different style of question where there is only one first order derivative in each equation. Thanks in advance.
I would have done it in a slightly different way- not using matrices. First integrate the second equation to get
$$RQ_1+ L\frac{dQ_2}{dt}= C_1$$
so
$$Q_1= \frac{C_1}{R}-\frac{L}{R} \frac{Q_2}{dt}$$
Also from the second equation,
$$R\frac{dQ_1}{dt}= -L\frac{d^2Q_2}{dt^2}$$
so we can replace both instances of Q1 in the first equation:
$$-L\frac{d^2Q_2}{dt^2}+\frac{1}{C}(\frac{C_1}{R}- \frac{L}{R}\frac{dQ_2}{dt}-Q2)= 0$$
or
$$-L\frac{d^2Q_2}{dt^2}-\frac{L}{CR}\frac{dQ_2}{dt}- \frac{1}{C}Q_2= -\frac{C_1}{CR}[/itex] a second order non-homogeneous equation with constant coefficients. Since that is a second order differential equation, it will have two independent solutions involving two new constants. That's as it should be since you orginally had first and second order equations to solve. After you have solved for Q2(t) you can immediately get Q1 from [tex]Q_1= \frac{C_1}{R}-\frac{L}{R} \frac{Q_2}{dt}$$
 Thank you.