4.70 problem 67
Internal
problem
ID
[7939]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
67
Date
solved
:
Monday, October 21, 2024 at 04:35:23 PM
CAS
classification
:
system_of_ODEs
\begin{align*} \frac {d}{d t}x \left (t \right )&=x \left (t \right )+2 y \left (t \right )+2 t +1\\ \frac {d}{d t}y \left (t \right )&=5 x \left (t \right )+y \left (t \right )+3 t -1 \end{align*}
4.70.1 Solution using Matrix exponential method
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} \frac {d}{d t}x \left (t \right ) \\ \frac {d}{d t}y \left (t \right ) \end {array}\right ] &= \left [\begin {array}{cc} 1 & 2 \\ 5 & 1 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} 2 t +1 \\ 3 t -1 \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 {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2} & -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}}{10} \\ -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}}{4} & \frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{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 {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2} & -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}}{10} \\ -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}}{4} & \frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2} \end {array}\right ] \left [\begin {array}{c} c_{1} \\ c_{2} \end {array}\right ] \\ &= \left [\begin {array}{c} \left (\frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2}\right ) c_{1}-\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}\, c_{2}}{10} \\ -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}\, c_{1}}{4}+\left (\frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2}\right ) c_{2} \end {array}\right ]\\ &= \left [\begin {array}{c} \frac {\left (-c_{2} \sqrt {10}+5 c_{1}\right ) {\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{10}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t} \left (\frac {c_{2} \sqrt {10}}{5}+c_{1}\right )}{2} \\ \frac {\left (-c_{1} \sqrt {10}+2 c_{2}\right ) {\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{4}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t} \left (c_{1} \sqrt {10}+2 c_{2}\right )}{4} \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 {{\mathrm e}^{-2 t} \left ({\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}+{\mathrm e}^{\left (1+\sqrt {10}\right ) t}\right )}{2} & \frac {\sqrt {10}\, {\mathrm e}^{-2 t} \left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right )}{10} \\ \frac {\sqrt {10}\, {\mathrm e}^{-2 t} \left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right )}{4} & \frac {{\mathrm e}^{-2 t} \left ({\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}+{\mathrm e}^{\left (1+\sqrt {10}\right ) t}\right )}{2} \end {array}\right ] \end{align*}
Hence
\begin{align*} \vec {x}_p (t) &= \left [\begin {array}{cc} \frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2} & -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}}{10} \\ -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}}{4} & \frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2} \end {array}\right ] \int { \left [\begin {array}{cc} \frac {{\mathrm e}^{-2 t} \left ({\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}+{\mathrm e}^{\left (1+\sqrt {10}\right ) t}\right )}{2} & \frac {\sqrt {10}\, {\mathrm e}^{-2 t} \left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right )}{10} \\ \frac {\sqrt {10}\, {\mathrm e}^{-2 t} \left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right )}{4} & \frac {{\mathrm e}^{-2 t} \left ({\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}+{\mathrm e}^{\left (1+\sqrt {10}\right ) t}\right )}{2} \end {array}\right ] \left [\begin {array}{c} 2 t +1 \\ 3 t -1 \end {array}\right ]\,dt}\\ &= \left [\begin {array}{cc} \frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2} & -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}}{10} \\ -\frac {\left (-{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}\right ) \sqrt {10}}{4} & \frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{2}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{2} \end {array}\right ] \left [\begin {array}{c} \frac {\left (\left (-63 t -67\right ) \sqrt {10}-180 t +85\right ) {\mathrm e}^{-\left (1+\sqrt {10}\right ) t}}{810}+\frac {7 \left (\left (t +\frac {67}{63}\right ) \sqrt {10}-\frac {20 t}{7}+\frac {85}{63}\right ) {\mathrm e}^{\left (-1+\sqrt {10}\right ) t}}{90} \\ \frac {\left (\left (-36 t +17\right ) \sqrt {10}-126 t -134\right ) {\mathrm e}^{-\left (1+\sqrt {10}\right ) t}}{324}+\frac {\left (\left (t -\frac {17}{36}\right ) \sqrt {10}-\frac {7 t}{2}-\frac {67}{18}\right ) {\mathrm e}^{\left (-1+\sqrt {10}\right ) t}}{9} \end {array}\right ]\\ &= \left [\begin {array}{c} -\frac {4 t}{9}+\frac {17}{81} \\ -\frac {7 t}{9}-\frac {67}{81} \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 {17}{81}+\frac {\left (-c_{2} \sqrt {10}+5 c_{1}\right ) {\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{10}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t} \left (c_{2} \sqrt {10}+5 c_{1}\right )}{10}-\frac {4 t}{9} \\ \frac {\left (-c_{1} \sqrt {10}+2 c_{2}\right ) {\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}}{4}+\frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t} \left (c_{1} \sqrt {10}+2 c_{2}\right )}{4}-\frac {7 t}{9}-\frac {67}{81} \end {array}\right ] \end{align*}
4.70.2 Solution using explicit Eigenvalue and Eigenvector method
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} \frac {d}{d t}x \left (t \right ) \\ \frac {d}{d t}y \left (t \right ) \end {array}\right ] &= \left [\begin {array}{cc} 1 & 2 \\ 5 & 1 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} 2 t +1 \\ 3 t -1 \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} 1 & 2 \\ 5 & 1 \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} 1-\lambda & 2 \\ 5 & 1-\lambda \end {array}\right ]\right ) &= 0 \end{align*}
Which gives the characteristic equation
\begin{align*} \lambda ^{2}-2 \lambda -9&=0 \end{align*}
The roots of the above are the eigenvalues.
\begin{align*} \lambda _1 &= 1+\sqrt {10}\\ \lambda _2 &= 1-\sqrt {10} \end{align*}
This table summarises the above result
| | |
| eigenvalue |
algebraic multiplicity |
type of eigenvalue |
| | |
| \(1-\sqrt {10}\) | \(1\) | real eigenvalue |
| | |
| \(1+\sqrt {10}\) | \(1\) | real eigenvalue |
| | |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = 1-\sqrt {10}\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes
\begin{align*} \left (\left [\begin {array}{cc} 1 & 2 \\ 5 & 1 \end {array}\right ] - \left (1-\sqrt {10}\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} \sqrt {10} & 2 \\ 5 & \sqrt {10} \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
\[ \left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \sqrt {10}&2&0\\ 5&\sqrt {10}&0 \end {array} \right ] \]
\begin{align*} R_{2} = R_{2}-\frac {\sqrt {10}\, R_{1}}{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \sqrt {10}&2&0\\ 0&0&0 \end {array} \right ] \end{align*}
Therefore the system in Echelon form is
\[ \left [\begin {array}{cc} \sqrt {10} & 2 \\ 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \]
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 \(\left \{v_{1} = -\frac {t \sqrt {10}}{5}\right \}\)
Hence the solution is
\[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {t \sqrt {10}}{5} \\ t \end {array}\right ] \]
Since there is one free Variable, we have found one eigenvector
associated with this eigenvalue. The above can be written as
\[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = t \left [\begin {array}{c} -\frac {\sqrt {10}}{5} \\ 1 \end {array}\right ] \]
Let \(t = 1\) the eigenvector becomes
\[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {\sqrt {10}}{5} \\ 1 \end {array}\right ] \]
Which is normalized to
\[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} -\sqrt {10} \\ 5 \end {array}\right ] \]
Considering the eigenvalue \(\lambda _{2} = 1+\sqrt {10}\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes
\begin{align*} \left (\left [\begin {array}{cc} 1 & 2 \\ 5 & 1 \end {array}\right ] - \left (1+\sqrt {10}\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} -\sqrt {10} & 2 \\ 5 & -\sqrt {10} \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
\[ \left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -\sqrt {10}&2&0\\ 5&-\sqrt {10}&0 \end {array} \right ] \]
\begin{align*} R_{2} = R_{2}+\frac {\sqrt {10}\, R_{1}}{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -\sqrt {10}&2&0\\ 0&0&0 \end {array} \right ] \end{align*}
Therefore the system in Echelon form is
\[ \left [\begin {array}{cc} -\sqrt {10} & 2 \\ 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \]
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 \(\left \{v_{1} = \frac {t \sqrt {10}}{5}\right \}\)
Hence the solution is
\[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} \frac {t \sqrt {10}}{5} \\ t \end {array}\right ] \]
Since there is one free Variable, we have found one eigenvector
associated with this eigenvalue. The above can be written as
\[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = t \left [\begin {array}{c} \frac {\sqrt {10}}{5} \\ 1 \end {array}\right ] \]
Let \(t = 1\) the eigenvector becomes
\[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} \frac {\sqrt {10}}{5} \\ 1 \end {array}\right ] \]
Which is normalized to
\[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} \sqrt {10} \\ 5 \end {array}\right ] \]
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 |
| | | | |
| \(1+\sqrt {10}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {\sqrt {10}}{5} \\ 1 \end {array}\right ]\) |
| | | | |
| \(1-\sqrt {10}\) |
\(1\) |
\(1\) |
No |
\(\left [\begin {array}{c} -\frac {\sqrt {10}}{5} \\ 1 \end {array}\right ]\) |
| | | | |
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 \(1+\sqrt {10}\) is real and distinct then the corresponding
eigenvector solution is
\begin{align*} \vec {x}_{1}(t) &= \vec {v}_{1} e^{\left (1+\sqrt {10}\right ) t}\\ &= \left [\begin {array}{c} \frac {\sqrt {10}}{5} \\ 1 \end {array}\right ] e^{\left (1+\sqrt {10}\right ) t} \end{align*}
Since eigenvalue \(1-\sqrt {10}\) is real and distinct then the corresponding eigenvector solution is
\begin{align*} \vec {x}_{2}(t) &= \vec {v}_{2} e^{\left (1-\sqrt {10}\right ) t}\\ &= \left [\begin {array}{c} -\frac {\sqrt {10}}{5} \\ 1 \end {array}\right ] e^{\left (1-\sqrt {10}\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 {\sqrt {10}\, {\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{5} \\ {\mathrm e}^{\left (1+\sqrt {10}\right ) t} \end {array}\right ] + c_{2} \left [\begin {array}{c} -\frac {{\mathrm e}^{\left (1-\sqrt {10}\right ) t} \sqrt {10}}{5} \\ {\mathrm e}^{\left (1-\sqrt {10}\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 {\sqrt {10}\, {\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{5} & -\frac {{\mathrm e}^{\left (1-\sqrt {10}\right ) t} \sqrt {10}}{5} \\ {\mathrm e}^{\left (1+\sqrt {10}\right ) t} & {\mathrm e}^{\left (1-\sqrt {10}\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 {\sqrt {10}\, {\mathrm e}^{-\left (1+\sqrt {10}\right ) t}}{4} & \frac {{\mathrm e}^{-\left (1+\sqrt {10}\right ) t}}{2} \\ -\frac {\sqrt {10}\, {\mathrm e}^{\left (-1+\sqrt {10}\right ) t}}{4} & \frac {{\mathrm e}^{\left (-1+\sqrt {10}\right ) t}}{2} \end {array}\right ] \end{align*}
Hence
\begin{align*} \vec {x}_p(t) &= \left [\begin {array}{cc} \frac {\sqrt {10}\, {\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{5} & -\frac {{\mathrm e}^{\left (1-\sqrt {10}\right ) t} \sqrt {10}}{5} \\ {\mathrm e}^{\left (1+\sqrt {10}\right ) t} & {\mathrm e}^{\left (1-\sqrt {10}\right ) t} \end {array}\right ] \int { \left [\begin {array}{cc} \frac {\sqrt {10}\, {\mathrm e}^{-\left (1+\sqrt {10}\right ) t}}{4} & \frac {{\mathrm e}^{-\left (1+\sqrt {10}\right ) t}}{2} \\ -\frac {\sqrt {10}\, {\mathrm e}^{\left (-1+\sqrt {10}\right ) t}}{4} & \frac {{\mathrm e}^{\left (-1+\sqrt {10}\right ) t}}{2} \end {array}\right ] \left [\begin {array}{c} 2 t +1 \\ 3 t -1 \end {array}\right ] \, dt}\\ &= \left [\begin {array}{cc} \frac {\sqrt {10}\, {\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{5} & -\frac {{\mathrm e}^{\left (1-\sqrt {10}\right ) t} \sqrt {10}}{5} \\ {\mathrm e}^{\left (1+\sqrt {10}\right ) t} & {\mathrm e}^{\left (1-\sqrt {10}\right ) t} \end {array}\right ] \int { \left [\begin {array}{c} \frac {{\mathrm e}^{-\left (1+\sqrt {10}\right ) t} \left (2 t \sqrt {10}+\sqrt {10}+6 t -2\right )}{4} \\ -\frac {{\mathrm e}^{\left (-1+\sqrt {10}\right ) t} \left (2 t \sqrt {10}+\sqrt {10}-6 t +2\right )}{4} \end {array}\right ] \, dt}\\ &= \left [\begin {array}{cc} \frac {\sqrt {10}\, {\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{5} & -\frac {{\mathrm e}^{\left (1-\sqrt {10}\right ) t} \sqrt {10}}{5} \\ {\mathrm e}^{\left (1+\sqrt {10}\right ) t} & {\mathrm e}^{\left (1-\sqrt {10}\right ) t} \end {array}\right ] \left [\begin {array}{c} \frac {\left (18 t \sqrt {10}+185 \sqrt {10}-18 t -572\right ) {\mathrm e}^{-\left (1+\sqrt {10}\right ) t} \left (2 t \sqrt {10}+\sqrt {10}+6 t -2\right )}{-648 t -5184+1620 \sqrt {10}} \\ -\frac {\left (18 t \sqrt {10}+185 \sqrt {10}+18 t +572\right ) {\mathrm e}^{\left (-1+\sqrt {10}\right ) t} \left (2 t \sqrt {10}+\sqrt {10}-6 t +2\right )}{324 \left (2 t +16+5 \sqrt {10}\right )} \end {array}\right ] \\ &= \left [\begin {array}{c} \frac {-72 t^{3}-1118 t^{2}+436 t +51}{162 t^{2}+2592 t +243} \\ \frac {-126 t^{3}-2150 t^{2}-2333 t -201}{162 t^{2}+2592 t +243} \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 \sqrt {10}\, {\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{5} \\ c_1 \,{\mathrm e}^{\left (1+\sqrt {10}\right ) t} \end {array}\right ] + \left [\begin {array}{c} -\frac {c_2 \,{\mathrm e}^{\left (1-\sqrt {10}\right ) t} \sqrt {10}}{5} \\ c_2 \,{\mathrm e}^{\left (1-\sqrt {10}\right ) t} \end {array}\right ] + \left [\begin {array}{c} \frac {-72 t^{3}-1118 t^{2}+436 t +51}{162 t^{2}+2592 t +243} \\ \frac {-126 t^{3}-2150 t^{2}-2333 t -201}{162 t^{2}+2592 t +243} \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 {c_1 \sqrt {10}\, {\mathrm e}^{\left (1+\sqrt {10}\right ) t}}{5}-\frac {c_2 \,{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t} \sqrt {10}}{5}-\frac {4 t}{9}+\frac {17}{81} \\ c_1 \,{\mathrm e}^{\left (1+\sqrt {10}\right ) t}+c_2 \,{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t}-\frac {7 t}{9}-\frac {67}{81} \end {array}\right ] \end{align*}
4.70.3 Maple step by step solution
4.70.4 Maple dsolve solution
Solving time : 0.054 (sec)
Leaf size : 67
dsolve([diff(x(t),t) = x(t)+2*y(t)+2*t+1, diff(y(t),t) = 5*x(t)+y(t)+3*t-1]
,{op([x(t), y(t)])})
\begin{align*}
x \left (t \right ) &= {\mathrm e}^{\left (1+\sqrt {10}\right ) t} c_2 +{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t} c_1 -\frac {4 t}{9}+\frac {17}{81} \\
y \left (t \right ) &= \frac {{\mathrm e}^{\left (1+\sqrt {10}\right ) t} c_2 \sqrt {10}}{2}-\frac {{\mathrm e}^{-\left (-1+\sqrt {10}\right ) t} c_1 \sqrt {10}}{2}-\frac {7 t}{9}-\frac {67}{81} \\
\end{align*}
4.70.5 Mathematica DSolve solution
Solving time : 11.175 (sec)
Leaf size : 158
DSolve[{{D[x[t],t]==x[t]+2*y[t]+2*t+1,D[y[t],t]==5*x[t]+y[t]+3*t-1},{}},
{x[t],y[t]},t,IncludeSingularSolutions->True]
\begin{align*}
x(t)\to \frac {1}{810} e^{t-\sqrt {10} t} \left (e^{\left (\sqrt {10}-1\right ) t} (170-360 t)+81 \left (5 c_1+\sqrt {10} c_2\right ) e^{2 \sqrt {10} t}+81 \left (5 c_1-\sqrt {10} c_2\right )\right ) \\
y(t)\to \frac {1}{324} e^{t-\sqrt {10} t} \left (-4 e^{\left (\sqrt {10}-1\right ) t} (63 t+67)+81 \left (\sqrt {10} c_1+2 c_2\right ) e^{2 \sqrt {10} t}-81 \left (\sqrt {10} c_1-2 c_2\right )\right ) \\
\end{align*}