Question

\({\bf{12}}\)

(a) The system of differential equations for the currents \({i_2}(t)\)and \({i_3}(t)\) in the electrical network shown in Figure \({\bf{8}}.{\bf{3}}.{\bf{2}}\) is

\(\frac{d}{{dt}}\left( {\begin{array}{*{20}{l}}{{i_2}}\\{{i_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - {R_1}/{L_1}}&{ - {R_1}/{L_1}}\\{ - {R_1}/{L_2}}&{ - \left( {{R_1} + {R_2}} \right)/{L_2}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{i_2}}\\{{i_3}}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}{E/{L_1}}\\{E/{L_2}}\end{array}} \right)\)

Use the method of undetermined coefficients to solve the system if\({R_1} = 2\Omega ,{R_2} = 3\Omega ,{L_1} = 1\;{\rm{h}},{L_2} = 1\;{\rm{h}}\),\(E = 60\;{\rm{V}},{i_2}(0) = 0\) and\({i_3}(0) = 0\).

(b) Determine the current\({i_1}(t)\).

Short answer

(a) The general solution of the system is \(X(t) = 12\left( {\begin{array}{*{20}{r}}{ - 2}\\1\end{array}} \right){e^{ - t}} - 6\left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right){e^{ - 6t}} + \left( {\begin{array}{*{20}{c}}{30}\\0\end{array}} \right)\).

(b) The current is \({i_1}(t) = - 12{e^{ - t}} - 18{e^{ - 6t}} + 30\).

Step-by-step solution

Determine the complex Eigen values

The polynomial of degree \(n\) of the variable \(\lambda \) is the characteristic polynomial of the \(n \times n\) matrix\(A\).

\(p(\lambda ) = \det (\lambda I - A)\)

Its root is the eigenvalue of\(A\).

Determine the Eigen values

Substituting the gives values into the given system, we get

\(\frac{d}{{dt}}\left( {\begin{array}{*{20}{l}}{{i_2}}\\{{i_3}}\end{array}} \right){\rm{ }} = \left( {\begin{array}{*{20}{l}}{ - 2}&{ - 2}\\{ - 2}&{ - 5}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{i_2}}\\{{i_3}}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}{60}\\{60}\end{array}} \right)\)

\({X^\prime } = \left( {\begin{array}{*{20}{l}}{ - 2}&{ - 2}\\{ - 2}&{ - 5}\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}{60}\\{60}\end{array}} \right)\)

Where

\(X = \left( {\begin{array}{*{20}{l}}{{i_2}}\\{{i_3}}\end{array}} \right)\)

\(A = \left( {\begin{array}{*{20}{l}}{ - 2}&{ - 2}\\{ - 2}&{ - 5}\end{array}} \right)\)

And

\(F(t) = \left( {\begin{array}{*{20}{l}}{60}\\{60}\end{array}} \right)\)

Obtaining the eigenvalues of the coefficient matrix,

\(\det (A - \lambda I) = 0\;\;\; \to \left| {\begin{array}{*{20}{c}}{ - 2 - \lambda }&{ - 2}\\{ - 2}&{ - 5 - \lambda }\end{array}} \right| = 0\)

\(( - 2 - \lambda )( - 5 - \lambda ) - 4 = 0\)

\(10 + 2\lambda + 5\lambda + {\lambda ^2} - 4 = 0\)

\({\lambda ^2} + 7\lambda + 6 = 0\)

\((\lambda + 1)(\lambda + 6) = 0\)

So our eigenvalues are \({\lambda _1} = - 1\)and\({\lambda _2} = - 6\).

Determine the Eigen vector and a corresponding solution vector

For \({\lambda _1} = - 1:\)

\((A - I\mid 0) = \left( {\begin{array}{*{20}{c}}{ - 2 - ( - 1)}&{ - 2}&0\\{ - 2}&{ - 5 - ( - 1)}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - 2 - ( - 1)}&{ - 2}&0\\{ - 2}&{ - 5 - ( - 1)}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 2}&0\\{ - 2}&{ - 4}&0\end{array}} \right)\)

Apply row operation\({R_2} - 2{R_1} \to {R_2}\):

\( = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 2}&0\\0&0&0\end{array}} \right)\)

So here we have

\( - {k_1} - 2{k_2} = 0\;\;\; \to \;\;\;{k_1} = - 2{k_2}\)

Choosing \({k_2} = 1\)yields\({k_1} = - 2\).

This gives an eigenvector and a corresponding solution vector

\({K_{\bf{1}}} = \left( {\begin{array}{*{20}{r}}{ - 2}\\1\end{array}} \right),\;\;\;{X_{\bf{1}}} = \left( {\begin{array}{*{20}{r}}{ - 2}\\1\end{array}} \right){e^{ - t}}\)

For \({\lambda _1} = - 6:\)

\((A - 6I\mid 0) = \left( {\begin{array}{*{20}{c}}{ - 2 - ( - 6)}&{ - 2}&0\\{ - 2}&{ - 5 - ( - 6)}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}4&{ - 2}&0\\{ - 2}&1&0\end{array}} \right)\)

Apply row operation\(2{R_2} + {R_1} \to {R_2}\):

\( = \left( {\begin{array}{*{20}{c}}4&{ - 2}&0\\0&0&0\end{array}} \right)\)

So here we have

\(4{k_1} - 2{k_2} = 0\;\;\; \to \;\;\;{k_1} = \frac{1}{2}{k_2}\)

Choosing \({k_2} = 2\) yields\({k_1} = 1\).

This gives an eigenvector and a corresponding solution vector

\({K_{\bf{2}}} = \left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right),\;\;\;{X_2} = \left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right){e^{ - 6t}}\)

Hence, the complementary solution is\({X_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{r}}{ - 2}\\1\end{array}} \right){e^{ - t}} + {c_2}\left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right){e^{ - 6t}}\)

Determine the general solution of the system

Since\(F(t) = \left( {\begin{array}{*{20}{l}}{60}\\{60}\end{array}} \right)\), we shall try to find a particular solution of the system that possesses the same form:

\({X_{\bf{p}}} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\end{array}} \right)\)

Differentiating,

\(X_{\rm{p}}^\prime = \left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right)\)

Substituting into\(\left( 1 \right)\),

\(\left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right) = \left( {\begin{array}{*{20}{l}}{ - 2}&{ - 2}\\{ - 2}&{ - 5}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}{60}\\{60}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{l}}{ - 2{a_1} - 2{a_2}}\\{ - 2{a_1} - 5{a_2}}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}{60}\\{60}\end{array}} \right)\)

So here we have

\( - 2{a_1} - 2{a_2} = - 60.....(1)\)

And,

\( - 2{a_1} - 5{a_2} = - 60........(2)\)

Multiplying \(\left( 2 \right)\) from \(\left( 1 \right)\) yields

\(3{a_2} = 0\)

\({a_2} = 0\)

To solve for \({a_1}\) we substitute the value we obtained for \({a_1}\) into one of the equations above,

\( - 2{a_1} - 2(0) = - 60\;\;\; \to \;\;\;{a_1} = 30\)

Therefore, the particular solution is

\({X_{\bf{p}}} = \left( {\begin{array}{*{20}{c}}{30}\\0\end{array}} \right)\)

The general solution of the system is

\(X = {X_{\bf{c}}} + {X_{\bf{p}}}\)

\( = {c_1}\left( {\begin{array}{*{20}{r}}{ - 2}\\1\end{array}} \right){e^{ - t}} + {c_2}\left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right){e^{ - 6t}} + \left( {\begin{array}{*{20}{c}}{30}\\0\end{array}} \right)\)

Applying the initial value condition

\(X(0) = \left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{r}}{ - 2}\\1\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{30}\\0\end{array}} \right)\)

Where,

\( - 2{c_1} + {c_2} = - 30\)

\({c_1} + 2{c_2} = 0\)

Rearranging equation \(\left( 1 \right).\)

\({c_2} = 2{c_1} - 30\)

Substituting into \(\left( 2 \right),\)

\({c_1} + 2\left( {2{c_1} - 30} \right) = 0\)

\({c_1} + 4{c_1} - 60 = 0\)

\(5{c_1} = 60\)

\({c_1} = 12\)

To solve for \({c_2}\)we substitute the value we obtained for \({c_1}\) into one of the equations above,

\(12 + 2{c_2} = 0\;\;\; \to \;\;\;{c_2} = - 6\)

Therefore, the solution is

\(X(t) = 12\left( {\begin{array}{*{20}{r}}{ - 2}\\1\end{array}} \right){e^{ - t}} - 6\left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right){e^{ - 6t}} + \left( {\begin{array}{*{20}{c}}{30}\\0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - 24}\\{12}\end{array}} \right){e^{ - t}} + \left( {\begin{array}{*{20}{c}}{ - 6}\\{ - 12}\end{array}} \right){e^{ - 6t}} + \left( {\begin{array}{*{20}{c}}{30}\\0\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}{{i_2}}\\{{i_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 24{e^{ - t}} - 6{e^{ - 6t}} + 30}\\{12{e^{ - t}} - 12{e^{ - 6t}}}\end{array}} \right)\)

Therefore, the general solution of the system is\(X(t) = 12\left( {\begin{array}{*{20}{r}}{ - 2}\\1\end{array}} \right){e^{ - t}} - 6\left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right){e^{ - 6t}} + \left( {\begin{array}{*{20}{c}}{30}\\0\end{array}} \right)\).

Determine the current

From the network, we see that \({i_1}(t)\) is going into the node while \({i_2}(t)\) and \({i_3}(t)\)is going out of the node, therefore,

\({i_1}(t) = {i_2}(t) + {i_3}(t)\)

\( = - 24{e^{ - t}} - 6{e^{ - 6t}} + 30 + 12{e^{ - t}} - 12{e^{ - 6t}}\)

\( = - 12{e^{ - t}} - 18{e^{ - 6t}} + 30\)

Therefore, the current is \({i_1}(t) = - 12{e^{ - t}} - 18{e^{ - 6t}} + 30\).