RL circuit as a system of first-order ODEs 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.
 A: 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.
