Find a system of differential equations and initial conditions for the currents in the networks given in the schematic diagrams (Figures \({\bf{5}}{\bf{.39 - 5}}{\bf{.42}}\) on pages \({\bf{294 - 295}}\)). Assume that all initial currents are zero. Solve for the currents in each branch of the network.

One has that\(L = 0.5H,C = 0.5\;F,R = 1\Omega \), and\(E\left( t \right) = \cos 3t\;V\).

Applying Kirchhoff's voltage law on the first loop one will get\(0 = L\frac{{d{I_1}}}{{dt}} + R{I_2}\)and from the second loop we have that\(E\left( t \right) = - R{I_2} + \frac{1}{C}{q_3}\)

Applying Kirchhoff's current law at the point \(A\)one will get \(0 = - {I_1} + {I_2} + {I_3}\) and the point \(B\)gives us that\(0 = {I_1} - {I_2} - {I_3}\), so in both cases, one has that\({I_1} = {I_2} + {I_3}\)

So, one has to solve the following system:

\(\begin{aligned}{c}0 = L\frac{{d{I_1}}}{{dt}} + R{I_2}\\E\left( t \right) = - R{I_2} + \frac{1}{C}{q_3}\\{I_1} = {I_2} + {I_3}\end{aligned}\)

Expressing \({I_1}\) in terms of \({I_2}\) and \({I_3}\) in the first two equations of the system one will get

\(\begin{aligned}{c}0 = L\frac{{d{I_2}}}{{dt}} + L\frac{{d{I_3}}}{{dt}} + R{I_2}\\E\left( t \right) = - R{I_2} + \frac{1}{C}{q_3}\end{aligned}\)

Also, one can express\({I_i} = \frac{{d{q_i}}}{{dt}},i = \overline {1,3} \), so one has

\(\begin{aligned}{c}0 = L\frac{{{d^2}{q_2}}}{{d{t^2}}} + L\frac{{{d^2}{q_3}}}{{d{t^2}}} + R\frac{{d{q_2}}}{{dt}}\\E\left( t \right) = - R\frac{{d{q_2}}}{{dt}} + \frac{1}{C}{q_3}\end{aligned}\)

Substituting the values for\(E\),\(L\),\(R\), \(C\) our system becomes

\(\begin{aligned}{c}0.5\frac{{{d^2}{q_2}}}{{d{t^2}}} + 0.5\frac{{{d^2}{q_3}}}{{d{t^2}}} + \frac{{d{q_2}}}{{dt}} = 0\\ - \frac{{d{q_2}}}{{dt}} + 2{q_3} = \cos 3t\end{aligned}\)

Multiplying the first equation by \(2\)one will get

\(\begin{aligned}{c}\frac{{{d^2}{q_2}}}{{d{t^2}}} + \frac{{{d^2}{q_3}}}{{d{t^2}}} + 2\frac{{d{q_2}}}{{dt}} = 0\\ - \frac{{d{q_2}}}{{dt}} + 2{q_3} = \cos 3t\end{aligned}\)

One will solve this by the elimination method. First, it needs to rewrite the system in operator form:

\(\begin{aligned}{c}\left( {{D^2} + 2D} \right)\left( {{q_2}} \right) + {D^2}\left( {{q_3}} \right) = 0\\ - D\left( {{q_2}} \right) + 2\left( {{q_3}} \right) = \cos 3t\end{aligned}\)

Multiplying the second equation by \( - {D^2}\) and the first by \(2\) and then adding those two equations together one will get:

\(\begin{aligned}{c}2\left( {{D^2} + 2D} \right)\left( {{q_2}} \right) + 2{D^2}\left( {{q_3}} \right) = 0\\{D^3}\left( {{q_2}} \right) - 2{D^2}\left( {{q_3}} \right) = - {D^2}\left( {\cos 3t} \right)\\\left( {{D^3} + 2{D^2} + 4D} \right)\left( {{q_2}} \right) = - {D^2}\left( {\cos 3t} \right)\\\left( {{D^3} + 2{D^2} + 4D} \right)\left( {{q_2}} \right) = 9\cos 3t\end{aligned}\)

First, one can find a homogeneous solution. The auxiliary equation is \({r^3} + 2{r^2} + 4r = r\left( {{r^2} + 2r + 4} \right) = 0\) and its roots are\({r_{1,2}} = \frac{{ - 2 \pm \sqrt {4 - 16} }}{2} = - 1 \pm i\sqrt 3 ,\;\;\;{r_3} = 0\)

So, the homogeneous solution for \({q_2}\) is\({q_{{2_h}}}\left( t \right) = {c_1}{e^{ - t}}\cos \sqrt 3 t + {c_2}{e^{ - t}}\sin \sqrt 3 t + {c_3}\)

One can find a particular solution by the method of undetermined coefficients. Assume that \({q_{{2_p}}}\left( t \right) = A\cos 3t + B\sin 3t\).

Then

\(\begin{aligned}{c}{q_{{2_p}}}'\left( t \right) = - 3A\sin 3t + 3B\cos 3t\\{q_{{2_p}}}''\left( t \right) = - 9A\cos 3t - 9B\sin 3t\\{q_{{2_p}}}'''\left( t \right) = 27A\sin 3t - 27B\cos 3t\end{aligned}\)

Substituting that in the differential equation one has that

\(\begin{aligned}{c}\left( {{D^3} + 2{D^2} + 4D} \right)\left( {{q_{{2_p}}}} \right) = {q_{{2_p}}}'''\left( t \right) + 2{q_{{2_p}}}''\left( t \right) + 4{q_{{2_p}}}'\left( t \right)\\ = 27A\sin 3t - 27B\cos 3t - 18A\cos 3t - 18B\sin 3t - 12A\sin 3t + 12B\cos 3t\\ = \left( {15A - 18B} \right)\sin 3t + \left( { - 15B - 18A} \right)\cos 3t\\ = 9\cos 3t\end{aligned}\)

\( \Rightarrow 15A - 18B = 0,\;\;\; - 15B - 18A = 9\)

The first equation gives us that \(A = \frac{{6B}}{5}\), so substituting this into the second equation one will get\(A = - \frac{{18}}{{61}}\), and \(B = - \frac{{15}}{{61}}\)

So now one has that the particular solution for \({q_2}\) is\({q_{{2_p}}}\left( t \right) = - \frac{{18}}{{61}}\cos 3t - \frac{{15}}{{61}}\sin 3t\)and the general solution for \({q_2}\) is\({q_2}(t) = {q_{{2_h}}}(t) + {q_{{2_p}}} = {c_1}{e^{ - t}}\cos \sqrt 3 t + {c_2}{e^{ - t}}\sin \sqrt 3 t + {c_3} - \frac{{18}}{{61}}\cos 3t - \frac{{15}}{{61}}\sin 3t\).

But one needs to find an equation for the current, so one will differentiate \({q_2}\left( t \right)\) to obtain \({I_2}\)

\(\begin{aligned}{c}\frac{{dq}}{{dt}} = - {c_1}{e^{ - t}}\cos \sqrt 3 t - \sqrt 3 {c_1}{e^{ - t}}\sin \sqrt 3 t - {c_2}{e^{ - t}}\sin \sqrt 3 t + \sqrt 3 {c_2}{e^{ - t}}\cos \sqrt 3 t + \frac{{54}}{{61}}\sin 3t - \frac{{45}}{{61}}\cos 3t\\{I_2}\left( t \right) = \left( { - {c_1} + \sqrt 3 {c_2}} \right){e^{ - t}}\cos \sqrt 3 t - \left( {\sqrt 3 {c_1} + {c_2}} \right){e^{ - t}}\sin \sqrt 3 t + \frac{{54}}{{61}}\sin 3t - \frac{{45}}{{61}}\cos 3t\end{aligned}\)

One will find the charge \({q_3}\) from \( - D\left( {{q_2}} \right) + 2\left( {{q_3}} \right) = \cos 3t\)i.e., from \(2{q_3} = \cos 3t + {I_2}\):

\(\begin{aligned}{c}2{q_3}\left( t \right) = \cos 3t + \left( { - {c_1} + \sqrt 3 {c_2}} \right){e^{ - t}}\cos \sqrt 3 t - \left( {\sqrt 3 {c_1} + {c_2}} \right){e^{ - t}}\sin \sqrt 3 t + \frac{{54}}{{61}}\sin 3t - \frac{{45}}{{61}}\cos 3t\\ = \left( { - {c_1} + \sqrt 3 {c_2}} \right){e^{ - t}}\cos \sqrt 3 t - \left( {\sqrt 3 {c_1} + {c_2}} \right){e^{ - t}}\sin \sqrt 3 t + \frac{{54}}{{61}}\sin 3t + \frac{{16}}{{61}}\cos 3t\\{q_3}\left( t \right) = \frac{{ - {c_1} + \sqrt 3 {c_2}}}{2}{e^{ - t}}\cos \sqrt 3 t - \frac{{\sqrt 3 {c_1} + {c_2}}}{2}{e^{ - t}}\sin \sqrt 3 t + \frac{{27}}{{61}}\sin 3t + \frac{8}{{61}}\cos 3t\end{aligned}\)

As before, one will differentiate \({q_3}\) to get the current \({I_3}\)

\(\begin{aligned}{c}\frac{{d{q_3}}}{{dt}} = \frac{{{c_1} - \sqrt 3 {c_2}}}{2}{e^{ - t}}\cos \sqrt 3 t + \frac{{\sqrt 3 {c_1} - 3{c_2}}}{2}{e^{ - t}}\sin \sqrt 3 t + \frac{{\sqrt 3 {c_1} + {c_2}}}{2}{e^{ - t}}\sin \sqrt 3 t\\ - \frac{{3{c_1} + \sqrt 3 {c_2}}}{2}{e^{ - t}}\cos \sqrt 3 t + \frac{{81}}{{61}}\cos 3t - \frac{{24}}{{61}}\sin 3t\\{I_3}\left( t \right) = - \left( {{c_1} + \sqrt 3 {c_2}} \right){e^{ - t}}\cos \sqrt 3 t + \left( {\sqrt 3 {c_1} - {c_2}} \right){e^{ - t}}\sin \sqrt 3 t + \frac{{81}}{{61}}\cos 3t - \frac{{24}}{{61}}\sin 3t\end{aligned}\)

Finally, one will find \({I_1}\left( t \right)\) from \({I_1}\left( t \right) = {I_2}\left( t \right) + {I_3}\left( t \right)\)

\(\begin{aligned}{c}{I_1}\left( t \right) = \left( { - {c_1} + \sqrt 3 {c_2}} \right){e^{ - t}}\cos \sqrt 3 t - \left( {\sqrt 3 {c_1} + {c_2}} \right){e^{ - t}}\sin \sqrt 3 t + \frac{{54}}{{61}}\sin 3t - \frac{{45}}{{61}}\cos 3t\\ = - \left( {{c_1} + \sqrt 3 {c_2}} \right){e^{ - t}}\cos \sqrt 3 t + \left( {\sqrt 3 {c_1} - {c_2}} \right){e^{ - t}}\sin \sqrt 3 t + \frac{{81}}{{61}}\cos 3t - \frac{{24}}{{61}}\sin 3t\\{I_1}\left( t \right) = - 2{c_1}{e^{ - t}}\cos \sqrt 3 t - 2{c_2}{e^{ - t}}\sin \sqrt 3 t + \frac{{36}}{{61}}\cos 3t + \frac{{30}}{{61}}\sin 3t\end{aligned}\)

One will find the constant\({c_1}\), \({c_2}\) from the initial conditions which are\({I_1}\left( 0 \right) = {I_2}\left( 0 \right) = {I_3}\left( 0 \right) = 0\), so one has a system

\(\begin{aligned}{c}{I_1}\left( 0 \right) = - 2{c_1} + \frac{{36}}{{61}}\\ = 0\end{aligned}\)

\(\begin{aligned}{c}{I_2}\left( 0 \right) = - {c_1} + \sqrt 3 {c_2} - \frac{{45}}{{61}}\\ = 0\end{aligned}\)