- #1
plmokn
Homework Statement
I need to find the general solution for:
[itex]R \frac{dQ_{1}}{dt} + \frac{1}{C}(Q_{1}-Q_{2})=0[/itex]
[itex]R\frac{dQ_{1}}{dt} + L\frac{d^2Q_{2}}{dt^2}}=0[/itex]
For the case [itex]R=1\times 10^3[/itex], [itex]L=4\times 10^{-3}[/itex], [itex]C=1 \times 10^{-9}[/itex].
The Attempt at a Solution
As an attempt I looked for solutions of the form, [itex]Q_{1}=q_{1}e^{at}, Q_{2}=q_{2}e^{at}[/itex] which then in matrix form gives:
[itex]\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}[/itex]
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): [itex]a=-\frac{1}{2RC}[/itex]. For this value of a we get a solution:
[itex]\begin{pmatrix}Q_{1}\\Q_{2}\end{pmatrix}=A \begin{pmatrix}2\\1\end{pmatrix} exp(\frac{-t}{2RC})[/itex]
where A is an arbitary constant.
But now I need to find another solution. I think it'll be of the form [itex](Bt+C)exp(\frac{-t}{2RC})[/itex] 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.
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