Internal problem ID [10187]
Internal file name [OUTPUT/9134_Monday_June_06_2022_06_43_56_AM_1392925/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 8, system of first order odes
Problem number: 1865.
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 )&=a_{1} x \left (t \right )+b_{1} y \left (t \right )+c_{1}\\ y^{\prime }\left (t \right )&=a_{2} x \left (t \right )+b_{2} y \left (t \right )+c_{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 }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} c_{1} \\ c_{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}{cc} -\frac {\left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {b_{1} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ \frac {a_{2} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {\left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ] \end {align*}
Therefore the homogeneous solution is \begin {align*} \vec {x}_h(t) &= e^{A t} \vec {c} \\ &= \left [\begin {array}{cc} -\frac {\left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {b_{1} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ \frac {a_{2} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {\left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ] \left [\begin {array}{c} c_{1} \\ c_{2} \end {array}\right ] \\ &= \left [\begin {array}{c} -\frac {\left (\left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )\right ) c_{1}}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}+\frac {b_{1} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right ) c_{2}}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ \frac {a_{2} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right ) c_{1}}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}+\frac {\left (\left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )\right ) c_{2}}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ]\\ &= \left [\begin {array}{c} -\frac {\left (-c_{1} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+\left (a_{1} -b_{2} \right ) c_{1}+2 b_{1} c_{2}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (c_{1} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+\left (a_{1} -b_{2} \right ) c_{1}+2 b_{1} c_{2}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ \frac {\left (c_{2} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+\left (a_{1} -b_{2} \right ) c_{2}-2 a_{2} c_{1}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (-c_{2} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+\left (a_{1} -b_{2} \right ) c_{2}-2 a_{2} c_{1}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \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}{cc} -\frac {\left (\left (-a_{1} +b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )\right ) {\mathrm e}^{-\left (b_{2} +a_{1} \right ) t}}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {b_{1} {\mathrm e}^{-\left (b_{2} +a_{1} \right ) t} \left (-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ \frac {a_{2} {\mathrm e}^{-\left (b_{2} +a_{1} \right ) t} \left (-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {{\mathrm e}^{-\left (b_{2} +a_{1} \right ) t} \left (\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ] \end {align*}
Hence \begin {align*} \vec {x}_p (t) &= \left [\begin {array}{cc} -\frac {\left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {b_{1} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ \frac {a_{2} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {\left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ] \int { \left [\begin {array}{cc} -\frac {\left (\left (-a_{1} +b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )\right ) {\mathrm e}^{-\left (b_{2} +a_{1} \right ) t}}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {b_{1} {\mathrm e}^{-\left (b_{2} +a_{1} \right ) t} \left (-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ \frac {a_{2} {\mathrm e}^{-\left (b_{2} +a_{1} \right ) t} \left (-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {{\mathrm e}^{-\left (b_{2} +a_{1} \right ) t} \left (\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ] \left [\begin {array}{c} c_{1} \\ c_{2} \end {array}\right ]\,dt}\\ &= \left [\begin {array}{cc} -\frac {\left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {b_{1} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ \frac {a_{2} \left ({\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} & \frac {\left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-{\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ] \left [\begin {array}{c} \frac {2 \left (\left (b_{1} c_{2} -b_{2} c_{1} \right ) \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}-b_{2}^{2} c_{1} +\left (a_{1} c_{1} +b_{1} c_{2} \right ) b_{2} +\left (a_{1} c_{2} -2 a_{2} c_{1} \right ) b_{1} \right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-2 \,{\mathrm e}^{-\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (\left (-b_{1} c_{2} +b_{2} c_{1} \right ) \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}-b_{2}^{2} c_{1} +\left (a_{1} c_{1} +b_{1} c_{2} \right ) b_{2} +\left (a_{1} c_{2} -2 a_{2} c_{1} \right ) b_{1} \right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, \left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )} \\ \frac {2 \left (\left (-a_{1} c_{2} +a_{2} c_{1} \right ) \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}-a_{1}^{2} c_{2} +\left (a_{2} c_{1} +b_{2} c_{2} \right ) a_{1} +a_{2} \left (-2 b_{1} c_{2} +b_{2} c_{1} \right )\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-2 \,{\mathrm e}^{-\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (\left (a_{1} c_{2} -a_{2} c_{1} \right ) \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}-a_{1}^{2} c_{2} +\left (a_{2} c_{1} +b_{2} c_{2} \right ) a_{1} +a_{2} \left (-2 b_{1} c_{2} +b_{2} c_{1} \right )\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, \left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )} \end {array}\right ]\\ &= \left [\begin {array}{c} \frac {4 b_{1} c_{2} -4 b_{2} c_{1}}{\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )} \\ \frac {-4 a_{1} c_{2} +4 a_{2} c_{1}}{\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )} \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 {2 \left (\left (b_{2} a_{1} -a_{2} b_{1} \right ) \left (c_{1} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}-a_{1} c_{1}-2 b_{1} c_{2}+b_{2} c_{1}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+\left (b_{2} a_{1} -a_{2} b_{1} \right ) \left (c_{1} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+a_{1} c_{1}+2 b_{1} c_{2}-b_{2} c_{1}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-2 \left (-b_{1} c_{2} +b_{2} c_{1} \right ) \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, \left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (-b_{2} -a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )} \\ -\frac {2 \left (\left (b_{2} a_{1} -a_{2} b_{1} \right ) \left (c_{2} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+a_{1} c_{2}-2 a_{2} c_{1}-b_{2} c_{2}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+\left (b_{2} a_{1} -a_{2} b_{1} \right ) \left (c_{2} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}-a_{1} c_{2}+2 a_{2} c_{1}+b_{2} c_{2}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+2 \left (-a_{1} c_{2} +a_{2} c_{1} \right ) \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, \left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (-b_{2} -a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )} \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 }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} c_{1} \\ c_{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}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end {array}\right ]-\lambda \left [\begin {array}{cc} 1 & 0 \\ 0 & 1 \end {array}\right ]\right ) &= 0 \end {align*}
Therefore \begin {align*} \operatorname {det} \left (\left [\begin {array}{cc} a_{1} -\lambda & b_{1} \\ a_{2} & b_{2} -\lambda \end {array}\right ]\right ) &= 0 \end {align*}
Which gives the characteristic equation \begin {align*} \lambda ^{2}+\left (-b_{2} -a_{1} \right ) \lambda +b_{2} a_{1} -a_{2} b_{1}&=0 \end {align*}
The roots of the above are the eigenvalues. \begin {align*} \lambda _1 &= \frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\\ \lambda _2 &= \frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2} \end {align*}
This table summarises the above result
| eigenvalue | algebraic multiplicity | type of eigenvalue |
| \(\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\) | \(1\) | real eigenvalue |
| \(\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\) | \(1\) | real eigenvalue |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = \frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\)
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}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end {array}\right ] - \left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) \left [\begin {array}{cc} 1 & 0 \\ 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{cc} \frac {a_{1}}{2}-\frac {b_{2}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2} & b_{1} \\ a_{2} & \frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {a_{1}}{2}-\frac {b_{2}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}&b_{1}&0\\ a_{2}&\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {a_{2} R_{1}}{\frac {a_{1}}{2}-\frac {b_{2}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {a_{1}}{2}-\frac {b_{2}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}&b_{1}&0\\ 0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{cc} \frac {a_{1}}{2}-\frac {b_{2}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2} & b_{1} \\ 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{2}\}\) and the leading variables are \(\{v_{1}\}\). Let \(v_{2} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = -\frac {2 b_{1} t}{a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}\right \}\)
Hence the solution is \[ \left [\begin {array}{c} -\frac {2 b_{1} t}{a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {2 b_{1} t}{a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} -\frac {2 b_{1} t}{a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ t \end {array}\right ] = t \left [\begin {array}{c} -\frac {2 b_{1}}{a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} -\frac {2 b_{1} t}{a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {2 b_{1}}{a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{2} = \frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\)
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}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end {array}\right ] - \left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) \left [\begin {array}{cc} 1 & 0 \\ 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{cc} \frac {a_{1}}{2}-\frac {b_{2}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2} & b_{1} \\ a_{2} & \frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {a_{1}}{2}-\frac {b_{2}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}&b_{1}&0\\ a_{2}&\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {a_{2} R_{1}}{\frac {a_{1}}{2}-\frac {b_{2}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {a_{1}}{2}-\frac {b_{2}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}&b_{1}&0\\ 0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{cc} \frac {a_{1}}{2}-\frac {b_{2}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2} & b_{1} \\ 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{2}\}\) and the leading variables are \(\{v_{1}\}\). Let \(v_{2} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = \frac {2 b_{1} t}{-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}\right \}\)
Hence the solution is \[ \left [\begin {array}{c} \frac {2 b_{1} t}{-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ t \end {array}\right ] = \left [\begin {array}{c} \frac {2 b_{1} t}{-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} \frac {2 b_{1} t}{-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ t \end {array}\right ] = t \left [\begin {array}{c} \frac {2 b_{1}}{-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} \frac {2 b_{1} t}{-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ t \end {array}\right ] = \left [\begin {array}{c} \frac {2 b_{1}}{-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ 1 \end {array}\right ] \] 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 |
| \(\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ 1 \end {array}\right ]\) |
| \(\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ 1 \end {array}\right ]\) |
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 \(\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{1}(t) &= \vec {v}_{1} e^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t}\\ &= \left [\begin {array}{c} \frac {b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ 1 \end {array}\right ] e^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} \end {align*}
Since eigenvalue \(\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{2}(t) &= \vec {v}_{2} e^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t}\\ &= \left [\begin {array}{c} \frac {b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ 1 \end {array}\right ] e^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) 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) \end {align*}
Which is written as \begin {align*} \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] &= c_{1} \left [\begin {array}{c} \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} \end {array}\right ] + c_{2} \left [\begin {array}{c} \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) 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}{cc} \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} & \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} & {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) 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}{cc} \frac {\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}}{4 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, b_{1}} & \frac {\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ -\frac {\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}}{4 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, b_{1}} & \frac {\left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ] \end {align*}
Hence \begin {align*} \vec {x}_p(t) &= \left [\begin {array}{cc} \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} & \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} & {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} \end {array}\right ] \int { \left [\begin {array}{cc} \frac {\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}}{4 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, b_{1}} & \frac {\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \\ -\frac {\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}}{4 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, b_{1}} & \frac {\left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}}{2 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}} \end {array}\right ] \left [\begin {array}{c} c_{1} \\ c_{2} \end {array}\right ] \, dt}\\ &= \left [\begin {array}{cc} \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} & \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} & {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} \end {array}\right ] \int { \left [\begin {array}{c} -\frac {{\mathrm e}^{-\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (c_{1} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+\left (a_{1} -b_{2} \right ) c_{1} +2 b_{1} c_{2} \right )}{4 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, b_{1}} \\ \frac {\left (-c_{1} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+\left (a_{1} -b_{2} \right ) c_{1} +2 b_{1} c_{2} \right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right )}{4 \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, b_{1}} \end {array}\right ] \, dt}\\ &= \left [\begin {array}{cc} \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} & \frac {{\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} & {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} \end {array}\right ] \left [\begin {array}{c} \frac {\left (-b_{2} -a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (c_{1} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}+a_{1} c_{1} +2 b_{1} c_{2} -b_{2} c_{1} \right )}{8 \left (b_{2} a_{1} -a_{2} b_{1} \right ) \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, b_{1}} \\ \frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{-\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} \left (c_{1} \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}-a_{1} c_{1} -2 b_{1} c_{2} +b_{2} c_{1} \right )}{8 \left (b_{2} a_{1} -a_{2} b_{1} \right ) \sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\, b_{1}} \end {array}\right ] \\ &= \left [\begin {array}{c} \frac {b_{1} c_{2} -b_{2} c_{1}}{b_{2} a_{1} -a_{2} b_{1}} \\ \frac {-a_{1} c_{2} +a_{2} c_{1}}{b_{2} a_{1} -a_{2} b_{1}} \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 \left (t \right ) \end {array}\right ] &= \left [\begin {array}{c} \frac {c_{1} {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ c_{1} {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}+\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} \end {array}\right ] + \left [\begin {array}{c} \frac {c_{2} {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} b_{1}}{\frac {b_{2}}{2}-\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}} \\ c_{2} {\mathrm e}^{\left (\frac {b_{2}}{2}+\frac {a_{1}}{2}-\frac {\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} \end {array}\right ] + \left [\begin {array}{c} \frac {b_{1} c_{2} -b_{2} c_{1}}{b_{2} a_{1} -a_{2} b_{1}} \\ \frac {-a_{1} c_{2} +a_{2} c_{1}}{b_{2} a_{1} -a_{2} b_{1}} \end {array}\right ] \end {align*}
Which becomes \begin {align*} \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] = \left [\begin {array}{c} -\frac {2 \left (c_{2} \left (b_{2} a_{1} -a_{2} b_{1} \right ) \left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}-c_{1} \left (b_{2} a_{1} -a_{2} b_{1} \right ) \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+2 a_{2} \left (-b_{1} c_{2} +b_{2} c_{1} \right )\right ) b_{1}}{\left (-a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (a_{1} -b_{2} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) \left (b_{2} a_{1} -a_{2} b_{1} \right )} \\ \frac {c_{2} \left (b_{2} a_{1} -a_{2} b_{1} \right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} -\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+c_{1} \left (b_{2} a_{1} -a_{2} b_{1} \right ) {\mathrm e}^{\frac {\left (b_{2} +a_{1} +\sqrt {a_{1}^{2}-2 b_{2} a_{1} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}}+a_{2} c_{1} -a_{1} c_{2}}{b_{2} a_{1} -a_{2} b_{1}} \end {array}\right ] \end {align*}
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 334
dsolve({diff(x(t),t)=a__1*x(t)+b__1*y(t)+c__1,diff(y(t),t)=a__2*x(t)+b__2*y(t)+c__2},singsol=all)
\begin{align*} x \left (t \right ) &= {\mathrm e}^{\left (\frac {a_{1}}{2}+\frac {b_{2}}{2}+\frac {\sqrt {a_{1}^{2}-2 a_{1} b_{2} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} c_{4} +{\mathrm e}^{\left (\frac {a_{1}}{2}+\frac {b_{2}}{2}-\frac {\sqrt {a_{1}^{2}-2 a_{1} b_{2} +4 a_{2} b_{1} +b_{2}^{2}}}{2}\right ) t} c_{3} +\frac {b_{1} c_{2} -b_{2} c_{1}}{a_{1} b_{2} -a_{2} b_{1}} \\ y \left (t \right ) &= \frac {-\frac {a_{1} \left ({\mathrm e}^{\frac {\left (a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 a_{1} b_{2} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} c_{4} \left (a_{1} b_{2} -a_{2} b_{1} \right )+{\mathrm e}^{\frac {\left (a_{1} +b_{2} -\sqrt {a_{1}^{2}-2 a_{1} b_{2} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} c_{3} \left (a_{1} b_{2} -a_{2} b_{1} \right )-b_{2} c_{1} +b_{1} c_{2} \right ) \left (2 a_{1} b_{2} -2 a_{2} b_{1} \right )}{a_{1} b_{2} -a_{2} b_{1}}+\frac {\left (a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 a_{1} b_{2} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (a_{1} +b_{2} +\sqrt {a_{1}^{2}-2 a_{1} b_{2} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} c_{4} \left (2 a_{1} b_{2} -2 a_{2} b_{1} \right )}{2}+\frac {\left (a_{1} +b_{2} -\sqrt {a_{1}^{2}-2 a_{1} b_{2} +4 a_{2} b_{1} +b_{2}^{2}}\right ) {\mathrm e}^{\frac {\left (a_{1} +b_{2} -\sqrt {a_{1}^{2}-2 a_{1} b_{2} +4 a_{2} b_{1} +b_{2}^{2}}\right ) t}{2}} c_{3} \left (2 a_{1} b_{2} -2 a_{2} b_{1} \right )}{2}-c_{1} \left (2 a_{1} b_{2} -2 a_{2} b_{1} \right )}{\left (2 a_{1} b_{2} -2 a_{2} b_{1} \right ) b_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.359 (sec). Leaf size: 926
DSolve[{x'[t]==a1*x[t]+b1*y[t]+c1,y'[t]==a2*x[t]+b2*y[t]+c2},{x[t],y[t]},t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to \frac {2 e^{-\frac {1}{2} t \left (\sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}+\text {a1}+\text {b2}\right )} \left (2 \text {b2} \text {c1} \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2} e^{\frac {1}{2} t \left (\sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}+\text {a1}+\text {b2}\right )}+\text {b1} \left (-2 \text {c2} \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2} e^{\frac {1}{2} t \left (\sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}+\text {a1}+\text {b2}\right )}+\text {a2} e^{t (\text {a1}+\text {b2})} \left (\text {a1} c_1 \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )+c_1 \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2} \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}+1\right )+2 \text {b1} c_2 \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )\right )\right )+\text {a1} \text {b2}^2 c_1 e^{t (\text {a1}+\text {b2})} \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )-\text {b2} e^{t (\text {a1}+\text {b2})} \left (\text {a1}^2 c_1 \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )+\text {a1} c_1 \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2} \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}+1\right )+2 \text {a1} \text {b1} c_2 \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )+\text {a2} \text {b1} c_1 \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )\right )\right )}{(4 \text {a2} \text {b1}-4 \text {a1} \text {b2}) \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}} \\ y(t)\to \frac {e^{-\frac {1}{2} t \left (\sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}+\text {a1}+\text {b2}\right )} \left (4 \text {a2}^2 \text {b1} c_1 e^{t (\text {a1}+\text {b2})} \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )-4 \text {a2} \text {c1} \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2} e^{\frac {1}{2} t \left (\sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}+\text {a1}+\text {b2}\right )}+4 \text {a1} \text {c2} \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2} e^{\frac {1}{2} t \left (\sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}+\text {a1}+\text {b2}\right )}+2 \text {a2} e^{t (\text {a1}+\text {b2})} \left (\text {b1} c_2 \left (\text {b2} \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )+\sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2} \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}+1\right )\right )-\text {a1} (\text {b1} c_2+2 \text {b2} c_1) \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )\right )-2 \text {a1} \text {b2} c_2 e^{t (\text {a1}+\text {b2})} \left (\text {a1} \left (-e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}\right )+\text {b2} \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}-1\right )+\sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2} \left (e^{t \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}}+1\right )+\text {a1}\right )\right )}{(4 \text {a2} \text {b1}-4 \text {a1} \text {b2}) \sqrt {\text {a1}^2-2 \text {a1} \text {b2}+4 \text {a2} \text {b1}+\text {b2}^2}} \\ \end{align*}