Question

In Problems \({\bf{33}}\) and \({\bf{34}}\) use (14) to solve the given initial-value problem.

\(34.{{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{l}}1&{ - 1}\\1&{ - 1}\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{l}}{1/t}\\{1/t}\end{array}} \right),{\bf{X}}(1) = \left( {\begin{array}{*{20}{r}}2\\{ - 1}\end{array}} \right)\)

Short answer

The general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{l}}1&{ - 1}\\1&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}{1/t}\\{1/t}\end{array}} \right)\) is \(X = \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right)t - \left( {\begin{array}{*{20}{l}}1\\4\end{array}} \right) + \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\ln (t)\)

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

Given

\({X^\prime } = \left( {\begin{array}{*{20}{l}}1&{ - 1}\\1&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\frac{1}{t}}\\{\frac{1}{t}}\end{array}} \right)\)

Where

\(A = \left( {\begin{array}{*{20}{l}}1&{ - 1}\\1&{ - 1}\end{array}} \right)\), and

\(F(t) = \left( {\begin{array}{*{20}{c}}{\frac{1}{t}}\\{\frac{1}{t}}\end{array}} \right)\)

Obtaining the eigenvalues of the coefficient matrix,

\(\det (A - \lambda I) = 0\; \to \;\;\left| {\begin{array}{*{20}{c}}{1 - \lambda }&{ - 1}\\1&{ - 1 - \lambda }\end{array}} \right| = 0\)

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

\( - 1 - \lambda + \lambda + {\lambda ^2} + 1 = 0\)

\({\lambda ^2} = 0\)

We have\(\lambda = 0\), which is an eigenvalue of multiplicity two.

Determine the eigenvector and a corresponding solution vector

Solving,

\((A - \lambda I)K = 0 \to \left( {\begin{array}{*{20}{c}}{1 - 0}&{ - 1}\\1&{ - 1 - 0}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{k_1}}\\{{k_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}1&{ - 1}\\1&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{k_1}}\\{{k_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right)\)

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

\(\left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{k_1}}\\{{k_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right)\)

Here we have

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

Choosing v yields\({k_1} = 1\).

This gives an eigenvector and a corresponding solution vector

\(K = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right),\;\;\;{X_{\bf{1}}} = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\)

\((A - \lambda I)P = K\;\;\; \to \left( {\begin{array}{*{20}{c}}{1 - 0}&{ - 1}\\1&{ - 1 - 0}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{p_1}}\\{{p_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}1&{ - 1}\\1&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{p_1}}\\{{p_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\)

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

\(\left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{p_1}}\\{{p_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\)

Here we have

\({k_1} - {k_2} = 1\)

Choosing \({k_2} = 0\) yields\({k_1} = 1\).

This gives an eigenvector and a corresponding solution vector

\(P = \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right),\;\;\;{X_2} = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)t + \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\)

Determine the general solution of the system

The entries in \({X_1}\)form the first column of\(\Phi (t)\), and the entries in \({X_2}\)form the second column of\(\Phi (t)\).

Therefore,

\(\Phi (t) = \left( {\begin{array}{*{20}{c}}1&{1 + t}\\1&t\end{array}} \right)\)

We want to make sure that \(\Phi (t)\)is an invertible matrix by checking the determinant, where

\(|\Phi (t)| = \left| {\begin{array}{*{20}{c}}1&{1 + t}\\1&t\end{array}} \right|\)

\( = t - (1 + t)\)

\( = - 1 \ne 0\)

Since the determinant does not equal zero, the matrix is in fact, an invertible matrix.

So now,

\({\Phi ^{ - 1}}(t) = - \left( {\begin{array}{*{20}{c}}t&{ - 1 - t}\\{ - 1}&1\end{array}} \right)\)

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

Using

\(X = \Phi (t){\Phi ^{ - 1}}(1)X(1) + \Phi (t)\int_0^t {{\Phi ^{ - 1}}} (s)F(s)ds\)

Where

\(X(1) = \left( {\begin{array}{*{20}{r}}2\\{ - 1}\end{array}} \right)\)

First, we have

\(\Phi (t){\Phi ^{ - 1}}(1)X(1){\rm{ }} = \left( {\begin{array}{*{20}{c}}1&{1 + t}\\1&t\end{array}} \right)\left( {\begin{array}{*{20}{r}}{ - 1}&2\\1&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{r}}2\\{ - 1}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}1&{1 + t}\\1&t\end{array}} \right)\left( {\begin{array}{*{20}{r}}{ - 4}\\3\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - 1 + 3t}\\{ - 4 + 3t}\end{array}} \right)\)

Substituting into (1),

\(X = \left( {\begin{array}{*{20}{c}}{ - 1 + 3t}\\{ - 4 + 3t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&{1 + t}\\1&t\end{array}} \right)\int_0^t {\left( {\begin{array}{*{20}{c}}{ - s}&{1 + s}\\1&{ - 1}\end{array}} \right)} \left( {\begin{array}{*{20}{c}}{\frac{1}{s}}\\{\frac{1}{s}}\end{array}} \right)ds\)

\( = \left( {\begin{array}{*{20}{c}}{ - 1 + 3t}\\{ - 4 + 3t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&{1 + t}\\1&t\end{array}} \right)\int_0^t {\left( {\begin{array}{*{20}{c}}{\frac{1}{s}}\\0\end{array}} \right)} ds\)

\( = \left( {\begin{array}{*{20}{c}}{ - 1 + 3t}\\{ - 4 + 3t}\end{array}} \right) + \left. {\left( {\begin{array}{*{20}{c}}1&{1 + t}\\1&t\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\ln (s)}\\0\end{array}} \right)} \right|_0^t\)

\( = \left( {\begin{array}{*{20}{c}}{ - 1 + 3t}\\{ - 4 + 3t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&{1 + t}\\1&t\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\ln (t)}\\0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{l}}{ - 1 + 3t}\\{ - 4 + 3t}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}{\ln (t)}\\{\ln (t)}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right)t - \left( {\begin{array}{*{20}{l}}1\\4\end{array}} \right) + \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\ln (t)\)

Therefore, the general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{l}}1&{ - 1}\\1&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}{1/t}\\{1/t}\end{array}} \right)\) is \(X = \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right)t - \left( {\begin{array}{*{20}{l}}1\\4\end{array}} \right) + \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\ln (t)\)