Internal problem ID [764]
Internal file name [OUTPUT/764_Sunday_June_05_2022_01_49_05_AM_58372908/index.tex]
Book: Elementary differential equations and boundary value problems, 10th ed., Boyce and
DiPrima
Section: Chapter 7.6, Complex Eigenvalues. page 417
Problem number: 24.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "system of linear ODEs"
Solve \begin {align*} x_{1}^{\prime }\left (t \right )&=-\frac {x_{1} \left (t \right )}{4}+x_{2} \left (t \right )\\ x_{2}^{\prime }\left (t \right )&=-x_{1} \left (t \right )-\frac {x_{2} \left (t \right )}{4}\\ x_{3}^{\prime }\left (t \right )&=\frac {x_{3} \left (t \right )}{10} \end {align*}
In this method, we will assume we have found the matrix exponential \(e^{A t}\) allready. There are different methods to determine this but will not be shown here. This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x_{1}^{\prime }\left (t \right ) \\ x_{2}^{\prime }\left (t \right ) \\ x_{3}^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ]\, \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \end {array}\right ] \end {align*}
For the above matrix \(A\), the matrix exponential can be found to be \begin {align*} e^{A t} &= \left [\begin {array}{ccc} {\mathrm e}^{-\frac {t}{4}} \cos \left (t \right ) & {\mathrm e}^{-\frac {t}{4}} \sin \left (t \right ) & 0 \\ -{\mathrm e}^{-\frac {t}{4}} \sin \left (t \right ) & {\mathrm e}^{-\frac {t}{4}} \cos \left (t \right ) & 0 \\ 0 & 0 & {\mathrm e}^{\frac {t}{10}} \end {array}\right ] \end {align*}
Therefore the homogeneous solution is \begin {align*} \vec {x}_h(t) &= e^{A t} \vec {c} \\ &= \left [\begin {array}{ccc} {\mathrm e}^{-\frac {t}{4}} \cos \left (t \right ) & {\mathrm e}^{-\frac {t}{4}} \sin \left (t \right ) & 0 \\ -{\mathrm e}^{-\frac {t}{4}} \sin \left (t \right ) & {\mathrm e}^{-\frac {t}{4}} \cos \left (t \right ) & 0 \\ 0 & 0 & {\mathrm e}^{\frac {t}{10}} \end {array}\right ] \left [\begin {array}{c} c_{1} \\ c_{2} \\ c_{3} \end {array}\right ] \\ &= \left [\begin {array}{c} {\mathrm e}^{-\frac {t}{4}} \cos \left (t \right ) c_{1}+{\mathrm e}^{-\frac {t}{4}} \sin \left (t \right ) c_{2} \\ -{\mathrm e}^{-\frac {t}{4}} \sin \left (t \right ) c_{1}+{\mathrm e}^{-\frac {t}{4}} \cos \left (t \right ) c_{2} \\ {\mathrm e}^{\frac {t}{10}} c_{3} \end {array}\right ]\\ &= \left [\begin {array}{c} {\mathrm e}^{-\frac {t}{4}} \left (\cos \left (t \right ) c_{1}+\sin \left (t \right ) c_{2}\right ) \\ {\mathrm e}^{-\frac {t}{4}} \left (-\sin \left (t \right ) c_{1}+\cos \left (t \right ) c_{2}\right ) \\ {\mathrm e}^{\frac {t}{10}} c_{3} \end {array}\right ] \end {align*}
Since no forcing function is given, then the final solution is \(\vec {x}_h(t)\) above.
This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x_{1}^{\prime }\left (t \right ) \\ x_{2}^{\prime }\left (t \right ) \\ x_{3}^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ]\, \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \end {array}\right ] \end {align*}
The first step is find the homogeneous solution. We start by finding the eigenvalues of \(A\). This is done by solving the following equation for the eigenvalues \(\lambda \) \begin {align*} \operatorname {det} \left ( A- \lambda I \right ) &= 0 \end {align*}
Expanding gives \begin {align*} \operatorname {det} \left (\left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ]-\lambda \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) &= 0 \end {align*}
Therefore \begin {align*} \operatorname {det} \left (\left [\begin {array}{ccc} -\frac {1}{4}-\lambda & 1 & 0 \\ -1 & -\frac {1}{4}-\lambda & 0 \\ 0 & 0 & \frac {1}{10}-\lambda \end {array}\right ]\right ) &= 0 \end {align*}
Which gives the characteristic equation \begin {align*} \lambda ^{3}+\frac {2}{5} \lambda ^{2}+\frac {81}{80} \lambda -\frac {17}{160}&=0 \end {align*}
The roots of the above are the eigenvalues. \begin {align*} \lambda _1 &= {\frac {1}{10}}\\ \lambda _2 &= -\frac {1}{4}+i\\ \lambda _3 &= -\frac {1}{4}-i \end {align*}
This table summarises the above result
| eigenvalue | algebraic multiplicity | type of eigenvalue |
| \(-\frac {1}{4}-i\) | \(1\) | complex eigenvalue |
| \(\frac {1}{10}\) | \(1\) | real eigenvalue |
| \(-\frac {1}{4}+i\) | \(1\) | complex eigenvalue |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = {\frac {1}{10}}\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes \begin {align*} \left (\left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ] - \left ({\frac {1}{10}}\right ) \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{ccc} -\frac {7}{20} & 1 & 0 \\ -1 & -\frac {7}{20} & 0 \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -{\frac {7}{20}}&1&0&0\\ -1&-{\frac {7}{20}}&0&0\\ 0&0&0&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {20 R_{1}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -{\frac {7}{20}}&1&0&0\\ 0&-{\frac {449}{140}}&0&0\\ 0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} -\frac {7}{20} & 1 & 0 \\ 0 & -\frac {449}{140} & 0 \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{3}\}\) and the leading variables are \(\{v_{1}, v_{2}\}\). Let \(v_{3} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = 0, v_{2} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ t \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ t \end {array}\right ] = t \left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ t \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{2} = -\frac {1}{4}-i\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes \begin {align*} \left (\left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ] - \left (-\frac {1}{4}-i\right ) \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{ccc} i & 1 & 0 \\ -1 & i & 0 \\ 0 & 0 & \frac {7}{20}+i \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} i&1&0&0\\ -1&i&0&0\\ 0&0&\frac {7}{20}+i&0 \end {array} \right ] \] \begin {align*} R_{2} = -i R_{1}+R_{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} i&1&0&0\\ 0&0&0&0\\ 0&0&\frac {7}{20}+i&0 \end {array} \right ] \end {align*}
Since the current pivot \(A(2,3)\) is zero, then the current pivot row is replaced with a row with a non-zero pivot. Swapping row \(2\) and row \(3\) gives \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} i&1&0&0\\ 0&0&\frac {7}{20}+i&0\\ 0&0&0&0 \end {array} \right ] \] Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} i & 1 & 0 \\ 0 & 0 & \frac {7}{20}+i \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{2}\}\) and the leading variables are \(\{v_{1}, v_{3}\}\). Let \(v_{2} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = i t, v_{3} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ t \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} i t \\ t \\ 0 \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ t \\ v_{3} \end {array}\right ] = t \left [\begin {array}{c} i \\ 1 \\ 0 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ t \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} i \\ 1 \\ 0 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{3} = -\frac {1}{4}+i\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes \begin {align*} \left (\left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ] - \left (-\frac {1}{4}+i\right ) \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{ccc} -i & 1 & 0 \\ -1 & -i & 0 \\ 0 & 0 & \frac {7}{20}-i \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -i&1&0&0\\ -1&-i&0&0\\ 0&0&\frac {7}{20}-i&0 \end {array} \right ] \] \begin {align*} R_{2} = i R_{1}+R_{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -i&1&0&0\\ 0&0&0&0\\ 0&0&\frac {7}{20}-i&0 \end {array} \right ] \end {align*}
Since the current pivot \(A(2,3)\) is zero, then the current pivot row is replaced with a row with a non-zero pivot. Swapping row \(2\) and row \(3\) gives \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -i&1&0&0\\ 0&0&\frac {7}{20}-i&0\\ 0&0&0&0 \end {array} \right ] \] Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} -i & 1 & 0 \\ 0 & 0 & \frac {7}{20}-i \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{2}\}\) and the leading variables are \(\{v_{1}, v_{3}\}\). Let \(v_{2} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = -i t, v_{3} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ t \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} -i t \\ t \\ 0 \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ t \\ v_{3} \end {array}\right ] = t \left [\begin {array}{c} -i \\ 1 \\ 0 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ t \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} -i \\ 1 \\ 0 \end {array}\right ] \] The following table gives a summary of this result. It shows for each eigenvalue the algebraic multiplicity \(m\), and its geometric multiplicity \(k\) and the eigenvectors associated with the eigenvalue. If \(m>k\) then the eigenvalue is defective which means the number of normal linearly independent eigenvectors associated with this eigenvalue (called the geometric multiplicity \(k\)) does not equal the algebraic multiplicity \(m\), and we need to determine an additional \(m-k\) generalized eigenvectors for this eigenvalue.
| multiplicity
| ||||
| eigenvalue | algebraic \(m\) | geometric \(k\) | defective? | eigenvectors |
| \(\frac {1}{10}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ]\) |
| \(-\frac {1}{4}+i\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} -i \\ 1 \\ 0 \end {array}\right ]\) |
| \(-\frac {1}{4}-i\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} i \\ 1 \\ 0 \end {array}\right ]\) |
Now that we found the eigenvalues and associated eigenvectors, we will go over each eigenvalue and generate the solution basis. The only problem we need to take care of is if the eigenvalue is defective. Since eigenvalue \(\frac {1}{10}\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{1}(t) &= \vec {v}_{1} e^{\frac {t}{10}}\\ &= \left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ] e^{\frac {t}{10}} \end {align*}
Therefore the final solution is \begin {align*} \vec {x}_h(t) &= c_{1} \vec {x}_{1}(t) + c_{2} \vec {x}_{2}(t) + c_{3} \vec {x}_{3}(t) \end {align*}
Which is written as \begin {align*} \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \end {array}\right ] &= c_{1} \left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{\frac {t}{10}} \end {array}\right ] + c_{2} \left [\begin {array}{c} -i {\mathrm e}^{\left (-\frac {1}{4}+i\right ) t} \\ {\mathrm e}^{\left (-\frac {1}{4}+i\right ) t} \\ 0 \end {array}\right ] + c_{3} \left [\begin {array}{c} i {\mathrm e}^{\left (-\frac {1}{4}-i\right ) t} \\ {\mathrm e}^{\left (-\frac {1}{4}-i\right ) t} \\ 0 \end {array}\right ] \end {align*}
Which becomes \begin {align*} \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \end {array}\right ] = \left [\begin {array}{c} i \left (c_{3} {\mathrm e}^{\left (-\frac {1}{4}-i\right ) t}-c_{2} {\mathrm e}^{\left (-\frac {1}{4}+i\right ) t}\right ) \\ c_{2} {\mathrm e}^{\left (-\frac {1}{4}+i\right ) t}+c_{3} {\mathrm e}^{\left (-\frac {1}{4}-i\right ) t} \\ c_{1} {\mathrm e}^{\frac {t}{10}} \end {array}\right ] \end {align*}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x_{1}^{\prime }\left (t \right )=-\frac {x_{1} \left (t \right )}{4}+x_{2} \left (t \right ), x_{2}^{\prime }\left (t \right )=-x_{1} \left (t \right )-\frac {x_{2} \left (t \right )}{4}, x_{3}^{\prime }\left (t \right )=\frac {x_{3} \left (t \right )}{10}\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{\textit {x\_\_}}}\left (t \right )=\left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Convert system into a vector equation}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{\textit {x\_\_}}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ]\cdot {\moverset {\rightarrow }{\textit {x\_\_}}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{\textit {x\_\_}}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ]\cdot {\moverset {\rightarrow }{\textit {x\_\_}}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} -\frac {1}{4} & 1 & 0 \\ -1 & -\frac {1}{4} & 0 \\ 0 & 0 & \frac {1}{10} \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{\textit {x\_\_}}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{\textit {x\_\_}}}\left (t \right ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [\frac {1}{10}, \left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{4}-\mathrm {I}, \left [\begin {array}{c} \mathrm {I} \\ 1 \\ 0 \end {array}\right ]\right ], \left [-\frac {1}{4}+\mathrm {I}, \left [\begin {array}{c} \mathrm {-I} \\ 1 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {1}{10}, \left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{\textit {x\_\_}}}_{1}={\mathrm e}^{\frac {t}{10}}\cdot \left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{4}-\mathrm {I}, \left [\begin {array}{c} \mathrm {I} \\ 1 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-\frac {1}{4}-\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} \mathrm {I} \\ 1 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & {\mathrm e}^{-\frac {t}{4}}\cdot \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right )\cdot \left [\begin {array}{c} \mathrm {I} \\ 1 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-\frac {t}{4}}\cdot \left [\begin {array}{c} \mathrm {I} \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \cos \left (t \right )-\mathrm {I} \sin \left (t \right ) \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{\textit {x\_\_}}}_{2}\left (t \right )={\mathrm e}^{-\frac {t}{4}}\cdot \left [\begin {array}{c} \sin \left (t \right ) \\ \cos \left (t \right ) \\ 0 \end {array}\right ], {\moverset {\rightarrow }{\textit {x\_\_}}}_{3}\left (t \right )={\mathrm e}^{-\frac {t}{4}}\cdot \left [\begin {array}{c} \cos \left (t \right ) \\ -\sin \left (t \right ) \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{\textit {x\_\_}}}=c_{1} {\moverset {\rightarrow }{\textit {x\_\_}}}_{1}+c_{2} {\moverset {\rightarrow }{\textit {x\_\_}}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{\textit {x\_\_}}}_{3}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{\textit {x\_\_}}}=c_{1} {\mathrm e}^{\frac {t}{10}}\cdot \left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-\frac {t}{4}}\cdot \left [\begin {array}{c} \sin \left (t \right ) \\ \cos \left (t \right ) \\ 0 \end {array}\right ]+c_{3} {\mathrm e}^{-\frac {t}{4}}\cdot \left [\begin {array}{c} \cos \left (t \right ) \\ -\sin \left (t \right ) \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Substitute in vector of dependent variables}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \end {array}\right ]=\left [\begin {array}{c} {\mathrm e}^{-\frac {t}{4}} \left (c_{2} \sin \left (t \right )+c_{3} \cos \left (t \right )\right ) \\ {\mathrm e}^{-\frac {t}{4}} \left (c_{2} \cos \left (t \right )-c_{3} \sin \left (t \right )\right ) \\ c_{1} {\mathrm e}^{\frac {t}{10}} \end {array}\right ] \\ \bullet & {} & \textrm {Solution to the system of ODEs}\hspace {3pt} \\ {} & {} & \left \{x_{1} \left (t \right )={\mathrm e}^{-\frac {t}{4}} \left (c_{2} \sin \left (t \right )+c_{3} \cos \left (t \right )\right ), x_{2} \left (t \right )={\mathrm e}^{-\frac {t}{4}} \left (c_{2} \cos \left (t \right )-c_{3} \sin \left (t \right )\right ), x_{3} \left (t \right )=c_{1} {\mathrm e}^{\frac {t}{10}}\right \} \end {array} \]
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 47
dsolve([diff(x__1(t),t)=-1/4*x__1(t)+1*x__2(t)+0*x__3(t),diff(x__2(t),t)=-1*x__1(t)-1/4*x__2(t)+0*x__3(t),diff(x__3(t),t)=0*x__1(t)-0*x__2(t)+1/10*x__3(t)],singsol=all)
\begin{align*} x_{1} \left (t \right ) &= {\mathrm e}^{-\frac {t}{4}} \left (c_{1} \sin \left (t \right )+c_{2} \cos \left (t \right )\right ) \\ x_{2} \left (t \right ) &= -{\mathrm e}^{-\frac {t}{4}} \left (c_{2} \sin \left (t \right )-c_{1} \cos \left (t \right )\right ) \\ x_{3} \left (t \right ) &= c_{3} {\mathrm e}^{\frac {t}{10}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.021 (sec). Leaf size: 110
DSolve[{x1'[t]==-1/4*x1[t]+1*x2[t]+0*x3[t],x2'[t]==-1*x1[t]-1/4*x2[t]+0*x3[t],x3'[t]==0*x1[t]-0*x2[t]+1/10*x3[t]},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
\begin{align*} \text {x1}(t)\to e^{-t/4} (c_1 \cos (t)+c_2 \sin (t)) \\ \text {x2}(t)\to e^{-t/4} (c_2 \cos (t)-c_1 \sin (t)) \\ \text {x3}(t)\to c_3 e^{t/10} \\ \text {x1}(t)\to e^{-t/4} (c_1 \cos (t)+c_2 \sin (t)) \\ \text {x2}(t)\to e^{-t/4} (c_2 \cos (t)-c_1 \sin (t)) \\ \text {x3}(t)\to 0 \\ \end{align*}