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 \({L_1} = 10H,{L_2} = 30H,{R_1} = 10\Omega ,{R_2} = 40\Omega \) and\(E\left( t \right) = 20\;V\). Applying Kirchhoff's voltage law on the first loop one will get\(E\left( t \right) = {L_1}\frac{{d{I_1}}}{{dt}} + {R_1}{I_3}\)and from the second loop, one has that\(0 = {L_2}\frac{{d{I_2}}}{{dt}} + {R_2}{I_2} - {R_1}{I_3}\).

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

\({I_1} = {I_2} + {I_3}\)

One has to solve the following system:

\(\begin{aligned}{c}E\left( t \right) &= {L_1}\frac{{d{I_1}}}{{dt}} + {R_1}{I_3}\\0 &= {L_2}\frac{{d{I_2}}}{{dt}} + {R_2}{I_2} - {R_1}{I_3}\\{I_1} &= {I_2} + {I_3}\end{aligned}\)

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

\(\begin{aligned}{c}E\left( t \right) &= {L_1}\frac{{d{I_2}}}{{dt}} + {L_1}\frac{{d{I_3}}}{{dt}} + {R_1}{I_3},\\0 &= {L_2}\frac{{d{I_2}}}{{dt}} + {R_2}{I_2} - {R_1}{I_3}\end{aligned}\)

Substituting the values for\(E,\,{L_1},\,{L_2},\,{R_1}\,{\rm{and}}\,{R_2}\) our system becomes

\(\begin{aligned}{c}20 = 10\frac{{d{I_2}}}{{dt}} + 10\frac{{d{I_3}}}{{dt}} + 10{I_3}\\0 &= 30\frac{{d{I_2}}}{{dt}} + 40{I_2} - 10{I_3}\end{aligned}\)

Dividing both equations by \(10\)one will get

\(\begin{aligned}{c}2 &= \frac{{d{I_2}}}{{dt}} + \frac{{d{I_3}}}{{dt}} + {I_3}\\0 &= 3\frac{{d{I_2}}}{{dt}} + 4{I_2} - {I_3}\end{aligned}\)

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

\(\begin{aligned}{c}D\left( {{I_2}} \right) + \left( {D + 1} \right)\left( {{I_3}} \right) &= 20\\\left( {3D + 4} \right)\left( {{I_2}} \right) - {I_3} &= 0\end{aligned}\)

From the second equation one has that\({I_3} = \left( {3D + 4} \right)\left( {{I_2}} \right)\), so substituting this into the first equation one has

\(\begin{aligned}{c}D\left( {{I_2}} \right) + \left( {D + 1} \right)\left( {3D + 4} \right)\left( {{I_2}} \right) &= 20\\\left( {3{D^2} + 8D + 4} \right)\left( {{I_2}} \right) &= 20\end{aligned}\)

First, one can find a homogeneous solution. The auxiliary equation is \(3{r^2} + 8r + 4 = 0\) and its roots are:

\(\begin{aligned}{c}{r_{1,2}} &= \frac{{ - 8 \pm \sqrt {64 - 48} }}{6}\\ &= \frac{{ - 8 \pm 4}}{6}\\{r_1} &= - 2\\{r_2} = - \frac{2}{3}\end{aligned}\)

So, the homogeneous solution for \({I_2}\) is\({I_{{2_h}}}\left( t \right) = {c_1}{e^{ - 2t}} + {c_2}{e^{\frac{{ - 2t}}{3}}}\)

One can find a particular solution by the method of undetermined coefficients. Assume that\({I_{{2_p}}}\left( t \right) = A\). Then

\(\begin{aligned}{c}{I_{{2_p}}}'\left( t \right) &= 0,{I_{{2_p}}}\\''\left( t \right) &= 0\end{aligned}\).

Substituting that in the differential equation one has that

\(\begin{aligned}{c}\left( {3{D^2} + 8D + 4} \right)\left( {{I_{{2_p}}}} \right) &= 0 + 0 + 4\\A = 4A &= 20\\A = 5\end{aligned}\)

So, the particular solution is \({I_{{2_p}}}\left( t \right) = 5\) and the general solution for \({I_2}\) is\({I_2}\left( t \right) = {I_{{2_h}}}\left( t \right) + {I_{{2_p}}}\left( t \right) = {c_1}{e^{ - 2t}} + {c_2}{e^{\frac{{ - 2t}}{3}}} + 5\)

The current \({I_3}\) can be derived from\({I_3} = \left( {3D + 4} \right)\left( {{I_2}} \right)\), so, one will first find the first derivative of \({I_2}\)and then use it to find\({I_3}\).

\(\begin{aligned}{c}D\left( {{I_2}} \right) &= - 2{c_1}{e^{ - 2t}} - \frac{2}{3}{c_2}{e^{\frac{{ - 2t}}{3}}}\\{I_3}\left( t \right) &= 3D\left( {{I_2}\left( t \right)} \right) + 4\left( {{I_2}\left( t \right)} \right)\\ &= 3\left( { - 2{c_1}{e^{ - 2t}} - \frac{2}{3}{c_2}{e^{\frac{{ - 2t}}{3}}}} \right) + 4\left( {{c_1}{e^{ - 2t}} + {c_2}{e^{\frac{{ - 2t}}{3}}} + 5} \right)\\ &= - 6{c_1}{e^{ - 2t}} - 2{c_2}{e^{\frac{{ - 2t}}{3}}} + 4{c_1}{e^{ - 2t}} + 4{c_2}{e^{\frac{{ - 2t}}{3}}} + 20\end{aligned}\)

\( \Rightarrow {I_3}\left( t \right) = - 2{c_1}{e^{ - 2t}} + 2{c_2}{e^{\frac{{ - 2t}}{3}}} + 20\)

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

\(\begin{aligned}{c}{I_1}\left( t \right) &= {c_1}{e^{ - 2t}} + {c_2}{e^{\frac{{ - 2t}}{3}}} + 5 - 2{c_1}{e^{ - 2t}} + 2{c_2}{e^{\frac{{ - 2t}}{3}}} + 20\\{I_1}\left( t \right) = - {c_1}{e^{ - 2t}} + 3{c_2}{e^{\frac{{ - 2t}}{3}}} + 25\end{aligned}\)

One will find the constant \({c_1}\) and \({c_2}\) 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

\(\begin{aligned}{c}{I_1}\left( 0 \right) &= - {c_1} + 3{c_2} + 25\\ &= 0\end{aligned}\)

\(\begin{aligned}{c}{I_2}\left( 0 \right) &= {c_1} + {c_2} + 5\\ &= 0\end{aligned}\)

\(\begin{aligned}{c}{I_3}\left( 0 \right) &= - 2{c_1} + 2{c_2} + 20\\ &= 0\end{aligned}\)

Adding the first and the second equation of the system one gets that\(4{c_2} + 30 = 0\), so\({c_2} = - \frac{{15}}{2}\). Substituting the value for \({c_2}\) in any of the equations of the system one will get that\({c_1} = \frac{5}{2}\), so the currents in the given network are:

\(\begin{aligned}{c}{{\bf{I}}_{\bf{1}}}\left( {\bf{t}} \right){\bf{ = - }}\frac{{\bf{5}}}{{\bf{2}}}{{\bf{e}}^{{\bf{ - 2t}}}}{\bf{ - }}\frac{{{\bf{45}}}}{{\bf{2}}}{{\bf{e}}^{\frac{{{\bf{ - 2t}}}}{{\bf{3}}}}}{\bf{ + 25}}\\{{\bf{I}}_{\bf{2}}}\left( {\bf{t}} \right){\bf{ = }}\frac{{\bf{5}}}{{\bf{2}}}{{\bf{e}}^{{\bf{ - 2t}}}}{\bf{ - }}\frac{{{\bf{15}}}}{{\bf{2}}}{{\bf{e}}^{\frac{{{\bf{ - 2t}}}}{{\bf{3}}}}}{\bf{ + 5}}\\{{\bf{I}}_{\bf{3}}}\left( {\bf{t}} \right){\bf{ = - 5}}{{\bf{e}}^{{\bf{ - 2t}}}}{\bf{ - 15}}{{\bf{e}}^{\frac{{{\bf{ - 2t}}}}{{\bf{3}}}}}{\bf{ + 20}}\end{aligned}\)