RL circuit as a system of first-order ODEs

Question

The system is as follows:\begin{align}i_1&=i_2+i_3,\\50\sin t&=6i_1+i_2'+5i_2,\\50\sin t&=6i_1+i_3',\end{align} I have to find $i_2,i_3$.

This is my first circuit I'm trying to solve, but I don't know where to start. I tried differentiating the first equation and somehow plugging it into the other two but that lead me nowhere.

Answer 1

Because of the special structure of your equations, you could eliminate i1 to deal with 2 unknowns instead of 3. However, here I would pursue the general case. Notice that the equations could be re-written as $$\begin{array}{l}{i_1}' = {i_2}' + {i_3}'\\{i_2}' = 50\sin t - 6{i_{1,{\rm{H}}}} - 5{i_{2,{\rm{H}}}}\\{i_3}' = 50\sin t - 6{i_{1,{\rm{H}}}}\end{array}$$ now replacing from the second and third equations into the first one, after simplification, we could write $$\frac{d}{{dt}}\left[ {\begin{array}{*{20}{c}}{{i_1}}\\{{i_2}}\\{{i_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 12}&{ - 5}&0\\{ - 6}&{ - 5}&0\\{ - 6}&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{i_1}}\\{{i_2}}\\{{i_3}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{100\sin t}\\{50\sin t}\\{50\sin t}\end{array}} \right]$$now for the homogeneous part$$\frac{d}{{dt}}{{\bf{i}}_{\rm{H}}} = \left[ {\begin{array}{*{20}{c}}{ - 12}&{ - 5}&0\\{ - 6}&{ - 5}&0\\{ - 6}&0&0\end{array}} \right]{{\bf{i}}_{\rm{H}}}$$Decomposition of the above matrix yields $$\left[ {\begin{array}{*{20}{c}}{ - 12}&{ - 5}&0\\{ - 6}&{ - 5}&0\\{ - 6}&0&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&1&0\\3&{ - 2}&0\\2&3&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 15}&0&0\\0&{ - 2}&0\\0&0&0\end{array}} \right]{\left[ {\begin{array}{*{20}{c}}5&1&0\\3&{ - 2}&0\\2&3&1\end{array}} \right]^{ - 1}}$$so that by defining $${{\rm{j}}_{\rm{H}}} = {\left[ {\begin{array}{*{20}{c}}5&1&0\\3&{ - 2}&0\\2&3&1\end{array}} \right]^{ - 1}}{{\bf{i}}_{\rm{H}}}$$ we get $$\frac{d}{{dt}}{{\rm{j}}_{\rm{H}}} = \left[ {\begin{array}{*{20}{c}}{ - 15}&0&0\\0&{ - 2}&0\\0&0&0\end{array}} \right]{{\rm{j}}_{\rm{H}}}$$which are three uncoupled first order ODEs which yield $${{\rm{j}}_{\rm{H}}} = \left[ {\begin{array}{*{20}{c}}{{c_1}{e^{ - 15t}}}\\{{c_2}{e^{ - 2t}}}\\{{c_3}}\end{array}} \right]$$Also, for the particular part, according to the inhomogeneous part, we should have $${{\rm{i}}_{\rm{P}}} = \sin t\left[ {\begin{array}{*{20}{c}}{{A_1}}\\{{A_2}}\\{{A_3}}\end{array}} \right] + \cos t\left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_2}}\\{{B_3}}\end{array}} \right]$$Replacing the particular solution form into the original equation gives$$\sin t\left( {\left[ {\begin{array}{*{20}{c}}{ - 12}&{ - 5}&0\\{ - 6}&{ - 5}&0\\{ - 6}&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_1}}\\{{A_2}}\\{{A_3}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_2}}\\{{B_3}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{100}\\{50}\\{50}\end{array}} \right]} \right) + \cos t\left( {\left[ {\begin{array}{*{20}{c}}{{A_1}}\\{{A_2}}\\{{A_3}}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 12}&{ - 5}&0\\{ - 6}&{ - 5}&0\\{ - 6}&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_2}}\\{{B_3}}\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right] $$Now you should only summarize these steps to get the final result.

Answer 2

Now an alternative (non-matrix based) solution. After eliminating ${i_1}$ from the first equation we get $$\begin{array}{l}{i_2}' = - 11{i_2} - 6{i_3} + 50\sin t\\{i_3}' = - 6{i_2} - 6{i_3} + 50\sin t\end{array}$$. Decomposing into homogeneous and particular parts for the unknowns, for the first part we have $$\begin{array}{l}{i_{2,{\rm{H}}}}' = - 11{i_{2,{\rm{H}}}} - 6{i_{3,{\rm{H}}}}\\{i_{3,{\rm{H}}}}' = - 6{i_{2,{\rm{H}}}} - 6{i_{3,{\rm{H}}}}\end{array} $$Now we shall assume the solutions to have an exponential behavior. That is $$\begin{array}{l}{i_{2,{\rm{H}}}} = {A_2}{e^{rt}}\\{i_{3,{\rm{H}}}} = {A_3}{e^{rt}}\end{array}$$ Replacing into the equations, we get $$\begin{array}{l}(11 + r){A_2} + 6{A_3} = 0\\6{A_2} + (6 + r){A_3} = 0\end{array}$$ Assuming the determinant of the equation above to be zero, you could find two solutions for $r$ and the answer for your homogeneous part is their linear combination. For finding the particular solution, I suggest you use the Fourier method. That is essentially assuming the solutions to be a linear combination of $\sin50t$ and $\cos50t$ and finding their weight by replacement into the original equation.