Coupled differential equations

[plmokn]

Homework Statement

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}.

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{pmatrix}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.

 

[HallsofIvy]

Homework Statement

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}.

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{pmatrix}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]<br /> a second order non-homogeneous equation with constant coefficients.<br /> Since that is a second order differential equation, it will have two independent solutions involving two new constants. That&#039;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<br /> Q_1= \frac{C_1}{R}-\frac{L}{R} \frac{Q_2}{dt}

 

[plmokn]

Thank you.