Internal problem ID [6715]
Internal file name [OUTPUT/5963_Sunday_June_05_2022_04_04_56_PM_20748891/index.tex]
Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL,
WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th
edition.
Section: CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS.
EXERCISES 8.1. Page 332
Problem number: 5.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "system of linear ODEs"
Solve \begin {align*} x^{\prime }\left (t \right )&=x \left (t \right )-y+z \left (t \right )+t -1\\ y^{\prime }&=2 x \left (t \right )+y-z \left (t \right )-3 t^{2}\\ z^{\prime }\left (t \right )&=x \left (t \right )+y+z \left (t \right )+t^{2}-t +2 \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) + \vec {G}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x^{\prime }\left (t \right ) \\ y^{\prime } \\ z^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} t -1 \\ -3 t^{2} \\ t^{2}-t +2 \end {array}\right ] \end {align*}
Since the system is nonhomogeneous, then the solution is given by \begin {align*} \vec {x}(t) &= \vec {x}_h(t) + \vec {x}_p(t) \end {align*}
Where \(\vec {x}_h(t)\) is the homogeneous solution to \(\vec {x}'(t) = A\, \vec {x}(t)\) and \(\vec {x}_p(t)\) is a particular solution to \(\vec {x}'(t) = A\, \vec {x}(t) + \vec {G}(t)\). The particular solution will be found using variation of parameters method applied to the fundamental matrix. For the above matrix \(A\), the matrix exponential can be found to be \begin {align*} e^{A t} &= \left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} & -\frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & -\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} \\ -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} & \frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11}+{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right ) & -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {13 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} \\ -\frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} & \frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & \frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} \end {array}\right ]\\ &= \left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} & -\frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & -\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} \\ -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} & \frac {{\mathrm e}^{\frac {t}{2}} \left (\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \cos \left (\frac {\sqrt {11}\, t}{2}\right )\right )}{11} & -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {13 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} \\ -\frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} & \frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & \frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} \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} \frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} & -\frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & -\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} \\ -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} & \frac {{\mathrm e}^{\frac {t}{2}} \left (\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \cos \left (\frac {\sqrt {11}\, t}{2}\right )\right )}{11} & -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {13 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} \\ -\frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} & \frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & \frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} \end {array}\right ] \left [\begin {array}{c} c_{1} \\ c_{2} \\ c_{3} \end {array}\right ] \\ &= \left [\begin {array}{c} \left (\frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5}\right ) c_{1}-\frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) c_{2}}{11}+\left (-\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5}\right ) c_{3} \\ \left (-\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5}\right ) c_{1}+\frac {{\mathrm e}^{\frac {t}{2}} \left (\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \cos \left (\frac {\sqrt {11}\, t}{2}\right )\right ) c_{2}}{11}+\left (-\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {13 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5}\right ) c_{3} \\ \left (-\frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5}\right ) c_{1}+\frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) c_{2}}{11}+\left (\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5}\right ) c_{3} \end {array}\right ]\\ &= \left [\begin {array}{c} -\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \left (c_{1}+10 c_{2}-4 c_{3}\right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \left (c_{1}-\frac {2 c_{3}}{3}\right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {2 \,{\mathrm e}^{2 t} \left (c_{1}+c_{3}\right )}{5} \\ -\frac {{\mathrm e}^{\frac {t}{2}} \left (c_{1}-5 c_{2}+c_{3}\right ) \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \left (c_{1}+\frac {5 c_{2}}{17}-\frac {13 c_{3}}{17}\right ) {\mathrm e}^{\frac {t}{2}} \sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t} \left (c_{1}+c_{3}\right )}{5} \\ \frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \left (c_{1}+10 c_{2}-4 c_{3}\right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}-\frac {3 \left (c_{1}-\frac {2 c_{3}}{3}\right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {3 \,{\mathrm e}^{2 t} \left (c_{1}+c_{3}\right )}{5} \end {array}\right ] \end {align*}
The particular solution given by \begin {align*} \vec {x}_p (t) &= e^{A t} \int { e^{-A t} \vec {G}(t) \,dt} \end {align*}
But \begin {align*} e^{-A t} &= (e^{A t})^{-1} \\ &= \left [\begin {array}{ccc} \frac {{\mathrm e}^{-2 t} \left (\sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+33 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )+22\right )}{55} & \frac {2 \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \sqrt {11}\, {\mathrm e}^{-\frac {t}{2}}}{11} & -\frac {2 \,{\mathrm e}^{-2 t} \left (2 \sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )-11\right )}{55} \\ -\frac {{\mathrm e}^{-2 t} \left (17 \sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )-11\right )}{55} & -\frac {\left (\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )-11 \cos \left (\frac {\sqrt {11}\, t}{2}\right )\right ) {\mathrm e}^{-\frac {t}{2}}}{11} & \frac {{\mathrm e}^{-2 t} \left (13 \sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )-11 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )+11\right )}{55} \\ -\frac {{\mathrm e}^{-2 t} \left (\sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+33 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )-33\right )}{55} & -\frac {2 \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \sqrt {11}\, {\mathrm e}^{-\frac {t}{2}}}{11} & \frac {{\mathrm e}^{-2 t} \left (4 \sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+22 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )+33\right )}{55} \end {array}\right ] \end {align*}
Hence \begin {align*} \vec {x}_p (t) &= \left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} & -\frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & -\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} \\ -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} & \frac {{\mathrm e}^{\frac {t}{2}} \left (\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \cos \left (\frac {\sqrt {11}\, t}{2}\right )\right )}{11} & -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {13 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} \\ -\frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} & \frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & \frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} \end {array}\right ] \int { \left [\begin {array}{ccc} \frac {{\mathrm e}^{-2 t} \left (\sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+33 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )+22\right )}{55} & \frac {2 \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \sqrt {11}\, {\mathrm e}^{-\frac {t}{2}}}{11} & -\frac {2 \,{\mathrm e}^{-2 t} \left (2 \sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )-11\right )}{55} \\ -\frac {{\mathrm e}^{-2 t} \left (17 \sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )-11\right )}{55} & -\frac {\left (\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )-11 \cos \left (\frac {\sqrt {11}\, t}{2}\right )\right ) {\mathrm e}^{-\frac {t}{2}}}{11} & \frac {{\mathrm e}^{-2 t} \left (13 \sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )-11 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )+11\right )}{55} \\ -\frac {{\mathrm e}^{-2 t} \left (\sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+33 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )-33\right )}{55} & -\frac {2 \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \sqrt {11}\, {\mathrm e}^{-\frac {t}{2}}}{11} & \frac {{\mathrm e}^{-2 t} \left (4 \sqrt {11}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+22 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )+33\right )}{55} \end {array}\right ] \left [\begin {array}{c} t -1 \\ -3 t^{2} \\ t^{2}-t +2 \end {array}\right ]\,dt}\\ &= \left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} & -\frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & -\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} \\ -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} & \frac {{\mathrm e}^{\frac {t}{2}} \left (\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )+11 \cos \left (\frac {\sqrt {11}\, t}{2}\right )\right )}{11} & -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {13 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} \\ -\frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} & \frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & \frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} \end {array}\right ] \left [\begin {array}{c} \frac {6 \,{\mathrm e}^{-\frac {t}{2}} \left (t^{2}+\frac {1}{6} t +\frac {1}{9}\right ) \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {2 \,{\mathrm e}^{-\frac {t}{2}} \sqrt {11}\, \left (t^{2}-\frac {13}{2} t -\frac {23}{3}\right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}-\frac {{\mathrm e}^{-2 t} \left (t^{2}+t +\frac {3}{2}\right )}{5} \\ -\frac {2 \,{\mathrm e}^{-\frac {t}{2}} \left (t^{2}+\frac {7}{2} t +4\right ) \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {34 \,{\mathrm e}^{-\frac {t}{2}} \sqrt {11}\, \left (t^{2}-\frac {1}{34} t -\frac {2}{17}\right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}-\frac {{\mathrm e}^{-2 t} \left (t^{2}+t +\frac {3}{2}\right )}{10} \\ -\frac {6 \,{\mathrm e}^{-\frac {t}{2}} \left (t^{2}+\frac {1}{6} t +\frac {1}{9}\right ) \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {2 \,{\mathrm e}^{-\frac {t}{2}} \sqrt {11}\, \left (t^{2}-\frac {13}{2} t -\frac {23}{3}\right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}-\frac {3 \,{\mathrm e}^{-2 t} \left (t^{2}+t +\frac {3}{2}\right )}{10} \end {array}\right ]\\ &= \left [\begin {array}{c} t^{2}-\frac {1}{6} \\ -\frac {1}{2} t^{2}-\frac {3}{2} t -\frac {7}{4} \\ -\frac {3}{2} t^{2}-\frac {1}{2} t -\frac {7}{12} \end {array}\right ] \end {align*}
Hence the complete solution is \begin {align*} \vec {x}(t) &= \vec {x}_h(t) + \vec {x}_p (t) \\ &= \left [\begin {array}{c} -\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \left (c_{1}+10 c_{2}-4 c_{3}\right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \left (c_{1}-\frac {2 c_{3}}{3}\right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {2 \,{\mathrm e}^{2 t} \left (c_{1}+c_{3}\right )}{5}+t^{2}-\frac {1}{6} \\ -\frac {{\mathrm e}^{\frac {t}{2}} \left (c_{1}-5 c_{2}+c_{3}\right ) \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \left (c_{1}+\frac {5 c_{2}}{17}-\frac {13 c_{3}}{17}\right ) {\mathrm e}^{\frac {t}{2}} \sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t} \left (c_{1}+c_{3}\right )}{5}-\frac {t^{2}}{2}-\frac {3 t}{2}-\frac {7}{4} \\ \frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \left (c_{1}+10 c_{2}-4 c_{3}\right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}-\frac {3 \left (c_{1}-\frac {2 c_{3}}{3}\right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {3 \,{\mathrm e}^{2 t} \left (c_{1}+c_{3}\right )}{5}-\frac {3 t^{2}}{2}-\frac {t}{2}-\frac {7}{12} \end {array}\right ] \end {align*}
This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) + \vec {G}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x^{\prime }\left (t \right ) \\ y^{\prime } \\ z^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} t -1 \\ -3 t^{2} \\ t^{2}-t +2 \end {array}\right ] \end {align*}
Since the system is nonhomogeneous, then the solution is given by \begin {align*} \vec {x}(t) &= \vec {x}_h(t) + \vec {x}_p(t) \end {align*}
Where \(\vec {x}_h(t)\) is the homogeneous solution to \(\vec {x}'(t) = A\, \vec {x}(t)\) and \(\vec {x}_p(t)\) is a particular solution to \(\vec {x}'(t) = A\, \vec {x}(t) + \vec {G}(t)\). The particular solution will be found using variation of parameters method applied to the fundamental matrix.
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} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \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} 1-\lambda & -1 & 1 \\ 2 & 1-\lambda & -1 \\ 1 & 1 & 1-\lambda \end {array}\right ]\right ) &= 0 \end {align*}
Which gives the characteristic equation \begin {align*} \lambda ^{3}-3 \lambda ^{2}+5 \lambda -6&=0 \end {align*}
The roots of the above are the eigenvalues. \begin {align*} \lambda _1 &= \frac {1}{2}+\frac {i \sqrt {11}}{2}\\ \lambda _2 &= \frac {1}{2}-\frac {i \sqrt {11}}{2}\\ \lambda _3 &= 2 \end {align*}
This table summarises the above result
| eigenvalue | algebraic multiplicity | type of eigenvalue |
| \(2\) | \(1\) | real eigenvalue |
| \(\frac {1}{2}-\frac {i \sqrt {11}}{2}\) | \(1\) | complex eigenvalue |
| \(\frac {1}{2}+\frac {i \sqrt {11}}{2}\) | \(1\) | complex eigenvalue |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = 2\)
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} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end {array}\right ] - \left (2\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} -1 & -1 & 1 \\ 2 & -1 & -1 \\ 1 & 1 & -1 \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 \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -1&-1&1&0\\ 2&-1&-1&0\\ 1&1&-1&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}+2 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -1&-1&1&0\\ 0&-3&1&0\\ 1&1&-1&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}+R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -1&-1&1&0\\ 0&-3&1&0\\ 0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} -1 & -1 & 1 \\ 0 & -3 & 1 \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] 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 \(\left \{v_{1} = \frac {2 t}{3}, v_{2} = \frac {t}{3}\right \}\)
Hence the solution is \[ \left [\begin {array}{c} \frac {2 t}{3} \\ \frac {t}{3} \\ t \end {array}\right ] = \left [\begin {array}{c} \frac {2 t}{3} \\ \frac {t}{3} \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} \frac {2 t}{3} \\ \frac {t}{3} \\ t \end {array}\right ] = t \left [\begin {array}{c} \frac {2}{3} \\ \frac {1}{3} \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} \frac {2 t}{3} \\ \frac {t}{3} \\ t \end {array}\right ] = \left [\begin {array}{c} \frac {2}{3} \\ \frac {1}{3} \\ 1 \end {array}\right ] \] Which is normalized to \[ \left [\begin {array}{c} \frac {2 t}{3} \\ \frac {t}{3} \\ t \end {array}\right ] = \left [\begin {array}{c} 2 \\ 1 \\ 3 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{2} = \frac {1}{2}-\frac {i \sqrt {11}}{2}\)
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} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end {array}\right ] - \left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\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 {1}{2}+\frac {i \sqrt {11}}{2} & -1 & 1 \\ 2 & \frac {1}{2}+\frac {i \sqrt {11}}{2} & -1 \\ 1 & 1 & \frac {1}{2}+\frac {i \sqrt {11}}{2} \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 \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {1}{2}+\frac {i \sqrt {11}}{2}&-1&1&0\\ 2&\frac {1}{2}+\frac {i \sqrt {11}}{2}&-1&0\\ 1&1&\frac {1}{2}+\frac {i \sqrt {11}}{2}&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {2 R_{1}}{\frac {1}{2}+\frac {i \sqrt {11}}{2}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {1}{2}+\frac {i \sqrt {11}}{2}&-1&1&0\\ 0&\frac {-\sqrt {11}-i}{i-\sqrt {11}}&\frac {\sqrt {11}-5 i}{i-\sqrt {11}}&0\\ 1&1&\frac {1}{2}+\frac {i \sqrt {11}}{2}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {R_{1}}{\frac {1}{2}+\frac {i \sqrt {11}}{2}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {1}{2}+\frac {i \sqrt {11}}{2}&-1&1&0\\ 0&\frac {-\sqrt {11}-i}{i-\sqrt {11}}&\frac {\sqrt {11}-5 i}{i-\sqrt {11}}&0\\ 0&\frac {i \sqrt {11}+3}{1+i \sqrt {11}}&\frac {-\sqrt {11}-7 i}{i-\sqrt {11}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {\left (i \sqrt {11}+3\right ) \left (i-\sqrt {11}\right ) R_{2}}{\left (1+i \sqrt {11}\right ) \left (-\sqrt {11}-i\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {1}{2}+\frac {i \sqrt {11}}{2}&-1&1&0\\ 0&\frac {-\sqrt {11}-i}{i-\sqrt {11}}&\frac {\sqrt {11}-5 i}{i-\sqrt {11}}&0\\ 0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} \frac {1}{2}+\frac {i \sqrt {11}}{2} & -1 & 1 \\ 0 & \frac {-\sqrt {11}-i}{i-\sqrt {11}} & \frac {\sqrt {11}-5 i}{i-\sqrt {11}} \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] 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 \(\left \{v_{1} = -t, v_{2} = \frac {t \left (i \sqrt {11}+5\right )}{i \sqrt {11}-1}\right \}\)
Hence the solution is \[ \left [\begin {array}{c} -t \\ \frac {t \left (\operatorname {I} \sqrt {11}+5\right )}{\operatorname {I} \sqrt {11}-1} \\ t \end {array}\right ] = \left [\begin {array}{c} -t \\ \frac {t \left (i \sqrt {11}+5\right )}{i \sqrt {11}-1} \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} -t \\ \frac {t \left (\operatorname {I} \sqrt {11}+5\right )}{\operatorname {I} \sqrt {11}-1} \\ t \end {array}\right ] = t \left [\begin {array}{c} -1 \\ \frac {i \sqrt {11}+5}{i \sqrt {11}-1} \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} -t \\ \frac {t \left (\operatorname {I} \sqrt {11}+5\right )}{\operatorname {I} \sqrt {11}-1} \\ t \end {array}\right ] = \left [\begin {array}{c} -1 \\ \frac {i \sqrt {11}+5}{i \sqrt {11}-1} \\ 1 \end {array}\right ] \] Which is normalized to \[ \left [\begin {array}{c} -t \\ \frac {t \left (\operatorname {I} \sqrt {11}+5\right )}{\operatorname {I} \sqrt {11}-1} \\ t \end {array}\right ] = \left [\begin {array}{c} -1 \\ \frac {i \sqrt {11}+5}{i \sqrt {11}-1} \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{3} = \frac {1}{2}+\frac {i \sqrt {11}}{2}\)
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} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end {array}\right ] - \left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\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 {1}{2}-\frac {i \sqrt {11}}{2} & -1 & 1 \\ 2 & \frac {1}{2}-\frac {i \sqrt {11}}{2} & -1 \\ 1 & 1 & \frac {1}{2}-\frac {i \sqrt {11}}{2} \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 \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {1}{2}-\frac {i \sqrt {11}}{2}&-1&1&0\\ 2&\frac {1}{2}-\frac {i \sqrt {11}}{2}&-1&0\\ 1&1&\frac {1}{2}-\frac {i \sqrt {11}}{2}&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {2 R_{1}}{\frac {1}{2}-\frac {i \sqrt {11}}{2}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {1}{2}-\frac {i \sqrt {11}}{2}&-1&1&0\\ 0&\frac {-i+\sqrt {11}}{\sqrt {11}+i}&\frac {-\sqrt {11}-5 i}{\sqrt {11}+i}&0\\ 1&1&\frac {1}{2}-\frac {i \sqrt {11}}{2}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {R_{1}}{\frac {1}{2}-\frac {i \sqrt {11}}{2}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {1}{2}-\frac {i \sqrt {11}}{2}&-1&1&0\\ 0&\frac {-i+\sqrt {11}}{\sqrt {11}+i}&\frac {-\sqrt {11}-5 i}{\sqrt {11}+i}&0\\ 0&\frac {3 i+\sqrt {11}}{\sqrt {11}+i}&\frac {\sqrt {11}-7 i}{\sqrt {11}+i}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {\left (3 i+\sqrt {11}\right ) R_{2}}{-i+\sqrt {11}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {1}{2}-\frac {i \sqrt {11}}{2}&-1&1&0\\ 0&\frac {-i+\sqrt {11}}{\sqrt {11}+i}&\frac {-\sqrt {11}-5 i}{\sqrt {11}+i}&0\\ 0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} \frac {1}{2}-\frac {i \sqrt {11}}{2} & -1 & 1 \\ 0 & \frac {-i+\sqrt {11}}{\sqrt {11}+i} & \frac {-\sqrt {11}-5 i}{\sqrt {11}+i} \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{3}\}\) 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 \(\left \{v_{1} = -t, v_{2} = -\frac {\left (\sqrt {11}+5 i\right ) t}{i-\sqrt {11}}\right \}\)
Hence the solution is \[ \left [\begin {array}{c} -t \\ -\frac {\left (\sqrt {11}+5 \,\operatorname {I}\right ) t}{\operatorname {I}-\sqrt {11}} \\ t \end {array}\right ] = \left [\begin {array}{c} -t \\ -\frac {\left (\sqrt {11}+5 i\right ) t}{i-\sqrt {11}} \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} -t \\ -\frac {\left (\sqrt {11}+5 \,\operatorname {I}\right ) t}{\operatorname {I}-\sqrt {11}} \\ t \end {array}\right ] = t \left [\begin {array}{c} -1 \\ -\frac {\sqrt {11}+5 i}{i-\sqrt {11}} \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} -t \\ -\frac {\left (\sqrt {11}+5 \,\operatorname {I}\right ) t}{\operatorname {I}-\sqrt {11}} \\ t \end {array}\right ] = \left [\begin {array}{c} -1 \\ -\frac {\sqrt {11}+5 i}{i-\sqrt {11}} \\ 1 \end {array}\right ] \] Which is normalized to \[ \left [\begin {array}{c} -t \\ -\frac {\left (\sqrt {11}+5 \,\operatorname {I}\right ) t}{\operatorname {I}-\sqrt {11}} \\ t \end {array}\right ] = \left [\begin {array}{c} -1 \\ -\frac {\sqrt {11}+5 i}{i-\sqrt {11}} \\ 1 \end {array}\right ] \] 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 |
| \(\frac {1}{2}+\frac {i \sqrt {11}}{2}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {\frac {3}{2}+\frac {3 i \sqrt {11}}{2}}{\left (-\frac {5}{2}+\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )} \\ -\frac {4+i \sqrt {11}}{-\frac {5}{2}+\frac {i \sqrt {11}}{2}} \\ 1 \end {array}\right ]\) |
| \(\frac {1}{2}-\frac {i \sqrt {11}}{2}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {\frac {3}{2}-\frac {3 i \sqrt {11}}{2}}{\left (-\frac {5}{2}-\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )} \\ -\frac {4-i \sqrt {11}}{-\frac {5}{2}-\frac {i \sqrt {11}}{2}} \\ 1 \end {array}\right ]\) |
| \(2\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {2}{3} \\ \frac {1}{3} \\ 1 \end {array}\right ]\) |
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 \(2\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{1}(t) &= \vec {v}_{1} e^{2 t}\\ &= \left [\begin {array}{c} \frac {2}{3} \\ \frac {1}{3} \\ 1 \end {array}\right ] e^{2 t} \end {align*}
Therefore the homogeneous 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 \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] &= c_{1} \left [\begin {array}{c} \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}+\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )} \\ -\frac {{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (4+i \sqrt {11}\right )}{-\frac {5}{2}+\frac {i \sqrt {11}}{2}} \\ {\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \end {array}\right ] + c_{2} \left [\begin {array}{c} \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )} \\ -\frac {{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (4-i \sqrt {11}\right )}{-\frac {5}{2}-\frac {i \sqrt {11}}{2}} \\ {\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \end {array}\right ] + c_{3} \left [\begin {array}{c} \frac {2 \,{\mathrm e}^{2 t}}{3} \\ \frac {{\mathrm e}^{2 t}}{3} \\ {\mathrm e}^{2 t} \end {array}\right ] \end {align*}
Now that we found homogeneous solution above, we need to find a particular solution \(\vec {x}_p(t)\). We will use Variation of parameters. The fundamental matrix is \[ \Phi =\begin {bmatrix} \vec {x}_{1} & \vec {x}_{2} & \cdots \end {bmatrix} \] Where \(\vec {x}_i\) are the solution basis found above. Therefore the fundamental matrix is \begin {align*} \Phi (t)&= \left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}+\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )} & \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )} & \frac {2 \,{\mathrm e}^{2 t}}{3} \\ -\frac {{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (4+i \sqrt {11}\right )}{-\frac {5}{2}+\frac {i \sqrt {11}}{2}} & -\frac {{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (4-i \sqrt {11}\right )}{-\frac {5}{2}-\frac {i \sqrt {11}}{2}} & \frac {{\mathrm e}^{2 t}}{3} \\ {\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} & {\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} & {\mathrm e}^{2 t} \end {array}\right ] \end {align*}
The particular solution is then given by \begin {align*} \vec {x}_p(t) &= \Phi \int { \Phi ^{-1} \vec {G}(t) \, dt}\\ \end {align*}
But \begin {align*} \Phi ^{-1} &= \left [\begin {array}{ccc} -\frac {\left (i \sqrt {11}+33\right ) {\mathrm e}^{-\frac {\left (1+i \sqrt {11}\right ) t}{2}}}{110} & -\frac {i \sqrt {11}\, {\mathrm e}^{-\frac {\left (1+i \sqrt {11}\right ) t}{2}}}{11} & -\frac {\left (i \sqrt {11}-1\right ) \sqrt {11}\, {\mathrm e}^{-\frac {\left (1+i \sqrt {11}\right ) t}{2}} \left (i \sqrt {11}+3\right ) \left (\sqrt {11}+5 i\right )}{1320} \\ \frac {7 \sqrt {11}\, \left (1+i \sqrt {11}\right ) \left (i \sqrt {11}+5\right ) {\mathrm e}^{\frac {\left (i \sqrt {11}-1\right ) t}{2}} \left (\sqrt {11}+i\right ) \left (i \sqrt {11}+\frac {19}{7}\right )}{23760} & \frac {i \sqrt {11}\, {\mathrm e}^{\frac {\left (i \sqrt {11}-1\right ) t}{2}}}{11} & -\frac {\left (1+i \sqrt {11}\right ) \sqrt {11}\, {\mathrm e}^{\frac {\left (i \sqrt {11}-1\right ) t}{2}} \left (i \sqrt {11}+5\right ) \left (3 i+\sqrt {11}\right )}{1320} \\ \frac {3 \,{\mathrm e}^{-2 t}}{5} & 0 & \frac {3 \,{\mathrm e}^{-2 t}}{5} \end {array}\right ] \end {align*}
Hence \begin {align*} \vec {x}_p(t) &= \left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}+\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )} & \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )} & \frac {2 \,{\mathrm e}^{2 t}}{3} \\ -\frac {{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (4+i \sqrt {11}\right )}{-\frac {5}{2}+\frac {i \sqrt {11}}{2}} & -\frac {{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (4-i \sqrt {11}\right )}{-\frac {5}{2}-\frac {i \sqrt {11}}{2}} & \frac {{\mathrm e}^{2 t}}{3} \\ {\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} & {\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} & {\mathrm e}^{2 t} \end {array}\right ] \int { \left [\begin {array}{ccc} -\frac {\left (i \sqrt {11}+33\right ) {\mathrm e}^{-\frac {\left (1+i \sqrt {11}\right ) t}{2}}}{110} & -\frac {i \sqrt {11}\, {\mathrm e}^{-\frac {\left (1+i \sqrt {11}\right ) t}{2}}}{11} & -\frac {\left (i \sqrt {11}-1\right ) \sqrt {11}\, {\mathrm e}^{-\frac {\left (1+i \sqrt {11}\right ) t}{2}} \left (i \sqrt {11}+3\right ) \left (\sqrt {11}+5 i\right )}{1320} \\ \frac {7 \sqrt {11}\, \left (1+i \sqrt {11}\right ) \left (i \sqrt {11}+5\right ) {\mathrm e}^{\frac {\left (i \sqrt {11}-1\right ) t}{2}} \left (\sqrt {11}+i\right ) \left (i \sqrt {11}+\frac {19}{7}\right )}{23760} & \frac {i \sqrt {11}\, {\mathrm e}^{\frac {\left (i \sqrt {11}-1\right ) t}{2}}}{11} & -\frac {\left (1+i \sqrt {11}\right ) \sqrt {11}\, {\mathrm e}^{\frac {\left (i \sqrt {11}-1\right ) t}{2}} \left (i \sqrt {11}+5\right ) \left (3 i+\sqrt {11}\right )}{1320} \\ \frac {3 \,{\mathrm e}^{-2 t}}{5} & 0 & \frac {3 \,{\mathrm e}^{-2 t}}{5} \end {array}\right ] \left [\begin {array}{c} t -1 \\ -3 t^{2} \\ t^{2}-t +2 \end {array}\right ] \, dt}\\ &= \left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}+\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )} & \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )} & \frac {2 \,{\mathrm e}^{2 t}}{3} \\ -\frac {{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (4+i \sqrt {11}\right )}{-\frac {5}{2}+\frac {i \sqrt {11}}{2}} & -\frac {{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (4-i \sqrt {11}\right )}{-\frac {5}{2}-\frac {i \sqrt {11}}{2}} & \frac {{\mathrm e}^{2 t}}{3} \\ {\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} & {\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} & {\mathrm e}^{2 t} \end {array}\right ] \int { \left [\begin {array}{c} \frac {{\mathrm e}^{-\frac {\left (1+i \sqrt {11}\right ) t}{2}} \left (77+i \left (34 t^{2}-5 t +9\right ) \sqrt {11}+22 t^{2}-55 t \right )}{110} \\ \frac {{\mathrm e}^{\frac {\left (i \sqrt {11}-1\right ) t}{2}} \left (77+i \left (-34 t^{2}+5 t -9\right ) \sqrt {11}+22 t^{2}-55 t \right )}{110} \\ \frac {3 \,{\mathrm e}^{-2 t} \left (t^{2}+1\right )}{5} \end {array}\right ] \, dt}\\ &= \left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}+\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )} & \frac {3 \,{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )} & \frac {2 \,{\mathrm e}^{2 t}}{3} \\ -\frac {{\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (4+i \sqrt {11}\right )}{-\frac {5}{2}+\frac {i \sqrt {11}}{2}} & -\frac {{\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (4-i \sqrt {11}\right )}{-\frac {5}{2}-\frac {i \sqrt {11}}{2}} & \frac {{\mathrm e}^{2 t}}{3} \\ {\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} & {\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} & {\mathrm e}^{2 t} \end {array}\right ] \left [\begin {array}{c} \frac {{\mathrm e}^{-\frac {\left (1+i \sqrt {11}\right ) t}{2}} \left (34 i \sqrt {11}\, t^{2}-5 i \sqrt {11}\, t +9 i \sqrt {11}+22 t^{2}-55 t +77\right ) \left (30 i \sqrt {11}\, t^{2}+9 i \sqrt {11}\, t +8 i \sqrt {11}-30 t^{2}+63 t +76\right )}{7590+330 i \left (8 t -11\right ) \sqrt {11}+19800 t^{2}-4620 t} \\ \frac {\sqrt {11}\, {\mathrm e}^{\frac {\left (i \sqrt {11}-1\right ) t}{2}} \left (30 \sqrt {11}\, t^{2}+9 \sqrt {11}\, t +8 \sqrt {11}-30 i t^{2}+63 i t +76 i\right ) \left (34 t^{2}+2 i \sqrt {11}\, t^{2}-5 t -5 i \sqrt {11}\, t +9+7 i \sqrt {11}\right )}{-7590+330 i \left (8 t -11\right ) \sqrt {11}-19800 t^{2}+4620 t} \\ -\frac {3 \,{\mathrm e}^{-2 t} \left (2 t^{2}+2 t +3\right )}{20} \end {array}\right ] \\ &= \left [\begin {array}{c} \frac {360 t^{6}-168 t^{5}+306 t^{4}-230 t^{3}+125 t^{2}+43 t -31}{360 t^{4}-168 t^{3}+366 t^{2}-258 t +186} \\ \frac {-120 t^{6}-304 t^{5}-374 t^{4}-84 t^{3}-231 t^{2}+115 t -217}{240 t^{4}-112 t^{3}+244 t^{2}-172 t +124} \\ \frac {-1080 t^{6}+144 t^{5}-1350 t^{4}+604 t^{3}-727 t^{2}+115 t -217}{720 t^{4}-336 t^{3}+732 t^{2}-516 t +372} \end {array}\right ] \end {align*}
Now that we found particular solution, the final solution is \begin {align*} \vec {x}(t) &= \vec {x}_h(t) + \vec {x}_p(t)\\ \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] &= \left [\begin {array}{c} \frac {3 c_{1} {\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}+\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}+\frac {i \sqrt {11}}{2}\right )} \\ -\frac {c_{1} {\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \left (4+i \sqrt {11}\right )}{-\frac {5}{2}+\frac {i \sqrt {11}}{2}} \\ c_{1} {\mathrm e}^{\left (\frac {1}{2}+\frac {i \sqrt {11}}{2}\right ) t} \end {array}\right ] + \left [\begin {array}{c} \frac {3 c_{2} {\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {i \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {11}}{2}\right )} \\ -\frac {c_{2} {\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \left (4-i \sqrt {11}\right )}{-\frac {5}{2}-\frac {i \sqrt {11}}{2}} \\ c_{2} {\mathrm e}^{\left (\frac {1}{2}-\frac {i \sqrt {11}}{2}\right ) t} \end {array}\right ] + \left [\begin {array}{c} \frac {2 c_{3} {\mathrm e}^{2 t}}{3} \\ \frac {c_{3} {\mathrm e}^{2 t}}{3} \\ c_{3} {\mathrm e}^{2 t} \end {array}\right ] + \left [\begin {array}{c} \frac {360 t^{6}-168 t^{5}+306 t^{4}-230 t^{3}+125 t^{2}+43 t -31}{360 t^{4}-168 t^{3}+366 t^{2}-258 t +186} \\ \frac {-120 t^{6}-304 t^{5}-374 t^{4}-84 t^{3}-231 t^{2}+115 t -217}{240 t^{4}-112 t^{3}+244 t^{2}-172 t +124} \\ \frac {-1080 t^{6}+144 t^{5}-1350 t^{4}+604 t^{3}-727 t^{2}+115 t -217}{720 t^{4}-336 t^{3}+732 t^{2}-516 t +372} \end {array}\right ] \end {align*}
Which becomes \begin {align*} \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] = \left [\begin {array}{c} -{\mathrm e}^{\frac {\left (1+i \sqrt {11}\right ) t}{2}} c_{1}-{\mathrm e}^{-\frac {\left (i \sqrt {11}-1\right ) t}{2}} c_{2}+\frac {2 c_{3} {\mathrm e}^{2 t}}{3}+t^{2}-\frac {1}{6} \\ -\frac {7}{4}+\frac {\left (1+i \sqrt {11}\right ) c_{1} {\mathrm e}^{\frac {\left (1+i \sqrt {11}\right ) t}{2}}}{2}+\frac {c_{2} \left (1-i \sqrt {11}\right ) {\mathrm e}^{-\frac {\left (i \sqrt {11}-1\right ) t}{2}}}{2}-\frac {t^{2}}{2}+\frac {c_{3} {\mathrm e}^{2 t}}{3}-\frac {3 t}{2} \\ {\mathrm e}^{\frac {\left (1+i \sqrt {11}\right ) t}{2}} c_{1}+{\mathrm e}^{-\frac {\left (i \sqrt {11}-1\right ) t}{2}} c_{2}+c_{3} {\mathrm e}^{2 t}-\frac {3 t^{2}}{2}-\frac {t}{2}-\frac {7}{12} \end {array}\right ] \end {align*}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }\left (t \right )=x \left (t \right )-y+z \left (t \right )+t -1, y^{\prime }=2 x \left (t \right )+y-z \left (t \right )-3 t^{2}, z^{\prime }\left (t \right )=x \left (t \right )+y+z \left (t \right )+t^{2}-t +2\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=\left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Convert system into a vector equation}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{x}}\left (t \right )+\left [\begin {array}{c} t -1 \\ -3 t^{2} \\ t^{2}-t +2 \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{x}}\left (t \right )+\left [\begin {array}{c} t -1 \\ -3 t^{2} \\ t^{2}-t +2 \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} t -1 \\ -3 t^{2} \\ t^{2}-t +2 \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{x}}\left (t \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [2, \left [\begin {array}{c} \frac {2}{3} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}, \left [\begin {array}{c} \frac {3 \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )} \\ -\frac {4-\mathrm {I} \sqrt {11}}{-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}} \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}+\frac {\mathrm {I} \sqrt {11}}{2}, \left [\begin {array}{c} \frac {3 \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}+\frac {\mathrm {I} \sqrt {11}}{2}\right ) \left (-\frac {1}{2}+\frac {\mathrm {I} \sqrt {11}}{2}\right )} \\ -\frac {4+\mathrm {I} \sqrt {11}}{-\frac {5}{2}+\frac {\mathrm {I} \sqrt {11}}{2}} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {2}{3} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{1}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {2}{3} \\ \frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}, \left [\begin {array}{c} \frac {3 \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )} \\ -\frac {4-\mathrm {I} \sqrt {11}}{-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right ) t}\cdot \left [\begin {array}{c} \frac {3 \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )} \\ -\frac {4-\mathrm {I} \sqrt {11}}{-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}} \\ 1 \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}{2}}\cdot \left (\cos \left (\frac {\sqrt {11}\, t}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {11}\, t}{2}\right )\right )\cdot \left [\begin {array}{c} \frac {3 \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )} \\ -\frac {4-\mathrm {I} \sqrt {11}}{-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\frac {t}{2}}\cdot \left [\begin {array}{c} \frac {3 \left (\cos \left (\frac {\sqrt {11}\, t}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {11}\, t}{2}\right )\right ) \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )}{\left (-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right ) \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {11}}{2}\right )} \\ -\frac {\left (\cos \left (\frac {\sqrt {11}\, t}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {11}\, t}{2}\right )\right ) \left (4-\mathrm {I} \sqrt {11}\right )}{-\frac {5}{2}-\frac {\mathrm {I} \sqrt {11}}{2}} \\ \cos \left (\frac {\sqrt {11}\, t}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{x}}_{2}\left (t \right )={\mathrm e}^{\frac {t}{2}}\cdot \left [\begin {array}{c} -\cos \left (\frac {\sqrt {11}\, t}{2}\right ) \\ \frac {\cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2} \\ \cos \left (\frac {\sqrt {11}\, t}{2}\right ) \end {array}\right ], {\moverset {\rightarrow }{x}}_{3}\left (t \right )={\mathrm e}^{\frac {t}{2}}\cdot \left [\begin {array}{c} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \\ -\frac {\sqrt {11}\, \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {\sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2} \\ -\sin \left (\frac {\sqrt {11}\, t}{2}\right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{x}}_{p}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=c_{1} {\moverset {\rightarrow }{x}}_{1}+c_{2} {\moverset {\rightarrow }{x}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{x}}_{3}\left (t \right )+{\moverset {\rightarrow }{x}}_{p}\left (t \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (t \right )=\left [\begin {array}{ccc} \frac {2 \,{\mathrm e}^{2 t}}{3} & -{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right ) & {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \\ \frac {{\mathrm e}^{2 t}}{3} & {\mathrm e}^{\frac {t}{2}} \left (\frac {\cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2}\right ) & {\mathrm e}^{\frac {t}{2}} \left (-\frac {\sqrt {11}\, \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {\sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2}\right ) \\ {\mathrm e}^{2 t} & {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right ) & -{\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (t \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\phi \left (t \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} \frac {2 \,{\mathrm e}^{2 t}}{3} & -{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right ) & {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \\ \frac {{\mathrm e}^{2 t}}{3} & {\mathrm e}^{\frac {t}{2}} \left (\frac {\cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {\sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2}\right ) & {\mathrm e}^{\frac {t}{2}} \left (-\frac {\sqrt {11}\, \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {\sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2}\right ) \\ {\mathrm e}^{2 t} & {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right ) & -{\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {2}{3} & -1 & 0 \\ \frac {1}{3} & \frac {1}{2} & -\frac {\sqrt {11}}{2} \\ 1 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} & -\frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & -\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {2 \,{\mathrm e}^{2 t}}{5} \\ -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {17 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} & \frac {\left (\sqrt {11}\, \cos \left (\frac {\sqrt {11}\, t}{2}\right )+\sin \left (\frac {\sqrt {11}\, t}{2}\right )\right ) \sqrt {11}\, {\mathrm e}^{\frac {t}{2}}}{11} & -\frac {{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {13 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {{\mathrm e}^{2 t}}{5} \\ -\frac {3 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}+\frac {\sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} & \frac {2 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{11} & \frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {3 \,{\mathrm e}^{2 t}}{5} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{x}}_{p}\left (t \right )=\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{p}^{\prime }\left (t \right )=\Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )={\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=\frac {1}{\Phi \left (t \right )}\cdot {\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (t \right )=\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{p}\left (t \right )=\Phi \left (t \right )\cdot \left (\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{p}\left (t \right )=\left [\begin {array}{c} \frac {3 \,{\mathrm e}^{2 t}}{10}-\frac {46 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{165}-\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{15}+t^{2}-\frac {1}{6} \\ \frac {3 \,{\mathrm e}^{2 t}}{20}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {8 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {t^{2}}{2}-\frac {3 t}{2}-\frac {7}{4} \\ \frac {9 \,{\mathrm e}^{2 t}}{20}+\frac {46 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{165}+\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{15}-\frac {3 t^{2}}{2}-\frac {t}{2}-\frac {7}{12} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=c_{1} {\moverset {\rightarrow }{x}}_{1}+c_{2} {\moverset {\rightarrow }{x}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{x}}_{3}\left (t \right )+\left [\begin {array}{c} \frac {3 \,{\mathrm e}^{2 t}}{10}-\frac {46 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{165}-\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{15}+t^{2}-\frac {1}{6} \\ \frac {3 \,{\mathrm e}^{2 t}}{20}+\frac {4 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{55}+\frac {8 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{5}-\frac {t^{2}}{2}-\frac {3 t}{2}-\frac {7}{4} \\ \frac {9 \,{\mathrm e}^{2 t}}{20}+\frac {46 \sqrt {11}\, {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{165}+\frac {2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{15}-\frac {3 t^{2}}{2}-\frac {t}{2}-\frac {7}{12} \end {array}\right ] \\ \bullet & {} & \textrm {Substitute in vector of dependent variables}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ]=\left [\begin {array}{c} -\frac {1}{6}+\frac {\left (-2-15 c_{2} \right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{15}+\left (c_{3} -\frac {46 \sqrt {11}}{165}\right ) {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+\frac {\left (9+20 c_{1} \right ) {\mathrm e}^{2 t}}{30}+t^{2} \\ -\frac {7}{4}+\frac {\left (-\sqrt {11}\, c_{3} +c_{2} +\frac {16}{5}\right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {{\mathrm e}^{\frac {t}{2}} \left (\left (c_{2} -\frac {8}{55}\right ) \sqrt {11}+c_{3} \right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2}+\frac {\left (9+20 c_{1} \right ) {\mathrm e}^{2 t}}{60}-\frac {t^{2}}{2}-\frac {3 t}{2} \\ -\frac {7}{12}+\frac {\left (15 c_{2} +2\right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{15}-\left (c_{3} -\frac {46 \sqrt {11}}{165}\right ) {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+\frac {\left (9+20 c_{1} \right ) {\mathrm e}^{2 t}}{20}-\frac {3 t^{2}}{2}-\frac {t}{2} \end {array}\right ] \\ \bullet & {} & \textrm {Solution to the system of ODEs}\hspace {3pt} \\ {} & {} & \left \{x \left (t \right )=-\frac {1}{6}+\frac {\left (-2-15 c_{2} \right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{15}+\left (c_{3} -\frac {46 \sqrt {11}}{165}\right ) {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+\frac {\left (9+20 c_{1} \right ) {\mathrm e}^{2 t}}{30}+t^{2}, y=-\frac {7}{4}+\frac {\left (-\sqrt {11}\, c_{3} +c_{2} +\frac {16}{5}\right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {{\mathrm e}^{\frac {t}{2}} \left (\left (c_{2} -\frac {8}{55}\right ) \sqrt {11}+c_{3} \right ) \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2}+\frac {\left (9+20 c_{1} \right ) {\mathrm e}^{2 t}}{60}-\frac {t^{2}}{2}-\frac {3 t}{2}, z \left (t \right )=-\frac {7}{12}+\frac {\left (15 c_{2} +2\right ) {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{15}-\left (c_{3} -\frac {46 \sqrt {11}}{165}\right ) {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )+\frac {\left (9+20 c_{1} \right ) {\mathrm e}^{2 t}}{20}-\frac {3 t^{2}}{2}-\frac {t}{2}\right \} \end {array} \]
✓ Solution by Maple
Time used: 0.203 (sec). Leaf size: 172
dsolve([diff(x(t),t)=x(t)-y(t)+z(t)+t-1,diff(y(t),t)=2*x(t)+y(t)-z(t)-3*t^2,diff(z(t),t)=x(t)+y(t)+z(t)+t^2-t+2],singsol=all)
\begin{align*} x \left (t \right ) &= t^{2}-\frac {1}{6}+c_{1} {\mathrm e}^{2 t}+c_{2} {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )+c_{3} {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right ) \\ y \left (t \right ) &= -\frac {t^{2}}{2}-\frac {7}{4}+\frac {c_{1} {\mathrm e}^{2 t}}{2}-\frac {c_{2} {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}+\frac {c_{2} {\mathrm e}^{\frac {t}{2}} \sqrt {11}\, \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {c_{3} {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {c_{3} {\mathrm e}^{\frac {t}{2}} \sqrt {11}\, \cos \left (\frac {\sqrt {11}\, t}{2}\right )}{2}-\frac {3 t}{2} \\ z \left (t \right ) &= -\frac {t}{2}+\frac {3 c_{1} {\mathrm e}^{2 t}}{2}-c_{2} {\mathrm e}^{\frac {t}{2}} \cos \left (\frac {\sqrt {11}\, t}{2}\right )-c_{3} {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {11}\, t}{2}\right )-\frac {3 t^{2}}{2}-\frac {7}{12} \\ \end{align*}
✓ Solution by Mathematica
Time used: 15.906 (sec). Leaf size: 304
DSolve[{x'[t]==x[t]-y[t]+z[t]+t-1,y'[t]==2*x[t]+y[t]-z[t]-3*t^2,z'[t]==x[t]+y[t]+z[t]+t^2-t+2},{x[t],y[t],z[t]},t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to t^2+\frac {2}{5} c_1 e^{2 t}+\frac {2}{5} c_3 e^{2 t}+\frac {1}{5} (3 c_1-2 c_3) e^{t/2} \cos \left (\frac {\sqrt {11} t}{2}\right )-\frac {(c_1+10 c_2-4 c_3) e^{t/2} \sin \left (\frac {\sqrt {11} t}{2}\right )}{5 \sqrt {11}}-\frac {1}{6} \\ y(t)\to \frac {1}{220} \left (-11 \left (10 t^2+30 t-4 (c_1+c_3) e^{2 t}+35\right )-44 (c_1-5 c_2+c_3) e^{t/2} \cos \left (\frac {\sqrt {11} t}{2}\right )+4 \sqrt {11} (17 c_1+5 c_2-13 c_3) e^{t/2} \sin \left (\frac {\sqrt {11} t}{2}\right )\right ) \\ z(t)\to -\frac {3 t^2}{2}-\frac {t}{2}+\frac {3}{5} (c_1+c_3) e^{2 t}-\frac {1}{5} (3 c_1-2 c_3) e^{t/2} \cos \left (\frac {\sqrt {11} t}{2}\right )+\frac {(c_1+10 c_2-4 c_3) e^{t/2} \sin \left (\frac {\sqrt {11} t}{2}\right )}{5 \sqrt {11}}-\frac {7}{12} \\ \end{align*}