Internal problem ID [1620]
Internal file name [OUTPUT/1621_Sunday_June_05_2022_02_24_48_AM_26617334/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 10 Linear system of Differential equations. Section 10.5, constant coefficient
homogeneous system II. Page 555
Problem number: section 10.5, problem 17.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "system of linear ODEs"
Solve \begin {align*} y_{1}^{\prime }\left (t \right )&=-7 y_{1} \left (t \right )+3 y_{2} \left (t \right )\\ y_{2}^{\prime }\left (t \right )&=-3 y_{1} \left (t \right )-y_{2} \left (t \right ) \end {align*}
With initial conditions \[ [y_{1} \left (0\right ) = 0, y_{2} \left (0\right ) = 2] \]
In this method, we will assume we have found the matrix exponential \(e^{A t}\) allready. There are different methods to determine this but will not be shown here. This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} y_{1}^{\prime }\left (t \right ) \\ y_{2}^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{cc} -7 & 3 \\ -3 & -1 \end {array}\right ]\, \left [\begin {array}{c} y_{1} \left (t \right ) \\ y_{2} \left (t \right ) \end {array}\right ] \end {align*}
For the above matrix \(A\), the matrix exponential can be found to be \begin {align*} e^{A t} &= \left [\begin {array}{cc} {\mathrm e}^{-4 t} \left (1-3 t \right ) & 3 t \,{\mathrm e}^{-4 t} \\ -3 t \,{\mathrm e}^{-4 t} & {\mathrm e}^{-4 t} \left (1+3 t \right ) \end {array}\right ] \end {align*}
Therefore the homogeneous solution is \begin {align*} \vec {x}_h(t) &= e^{A t} \vec {x}_0 \\ &= \left [\begin {array}{cc} {\mathrm e}^{-4 t} \left (1-3 t \right ) & 3 t \,{\mathrm e}^{-4 t} \\ -3 t \,{\mathrm e}^{-4 t} & {\mathrm e}^{-4 t} \left (1+3 t \right ) \end {array}\right ] \left [\begin {array}{c} 0 \\ 2 \end {array}\right ] \\ &= \left [\begin {array}{c} 6 t \,{\mathrm e}^{-4 t} \\ 2 \,{\mathrm e}^{-4 t} \left (1+3 t \right ) \end {array}\right ]\\ &= \left [\begin {array}{c} 6 t \,{\mathrm e}^{-4 t} \\ \left (2+6 t \right ) {\mathrm e}^{-4 t} \end {array}\right ] \end {align*}
Since no forcing function is given, then the final solution is \(\vec {x}_h(t)\) above.
This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} y_{1}^{\prime }\left (t \right ) \\ y_{2}^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{cc} -7 & 3 \\ -3 & -1 \end {array}\right ]\, \left [\begin {array}{c} y_{1} \left (t \right ) \\ y_{2} \left (t \right ) \end {array}\right ] \end {align*}
The first step is find the homogeneous solution. We start by finding the eigenvalues of \(A\). This is done by solving the following equation for the eigenvalues \(\lambda \) \begin {align*} \operatorname {det} \left ( A- \lambda I \right ) &= 0 \end {align*}
Expanding gives \begin {align*} \operatorname {det} \left (\left [\begin {array}{cc} -7 & 3 \\ -3 & -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} -7-\lambda & 3 \\ -3 & -1-\lambda \end {array}\right ]\right ) &= 0 \end {align*}
Which gives the characteristic equation \begin {align*} \lambda ^{2}+8 \lambda +16&=0 \end {align*}
The roots of the above are the eigenvalues. \begin {align*} \lambda _1 &= -4 \end {align*}
This table summarises the above result
| eigenvalue | algebraic multiplicity | type of eigenvalue |
| \(-4\) | \(1\) | real eigenvalue |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = -4\)
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} -7 & 3 \\ -3 & -1 \end {array}\right ] - \left (-4\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} -3 & 3 \\ -3 & 3 \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 \[ \left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -3&3&0\\ -3&3&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -3&3&0\\ 0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{cc} -3 & 3 \\ 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \] 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 \(\{v_{1} = t\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} t \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = t \left [\begin {array}{c} 1 \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} 1 \\ 1 \end {array}\right ] \] 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 |
| \(-4\) | \(2\) | \(1\) | Yes | \(\left [\begin {array}{c} 1 \\ 1 \end {array}\right ]\) |
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. eigenvalue \(-4\) is real and repated eigenvalue of multiplicity \(2\).There are two possible cases that can happen. This is illustrated in this diagram
This eigenvalue has algebraic multiplicity of \(2\), and geometric multiplicity \(1\), therefore this is defective eigenvalue. The defect is \(1\). This falls into case \(2\) shown above. We need to generate the missing additonal generalized eigevector \(\vec {v}_2\) by solving \[ \left ( A-\lambda I \right ) \vec {v}_2 = \vec {v}_1 \] Where \( \vec {v}_1\) is the normal (rank 1) eigenvector found above. Hence we need to solve \begin {align*} \left (\left [\begin {array}{cc} -7 & 3 \\ -3 & -1 \end {array}\right ]- \left (-4\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} 1 \\ 1 \end {array}\right ]\\ \left [\begin {array}{cc} -3 & 3 \\ -3 & 3 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ] &= \left [\begin {array}{c} 1 \\ 1 \end {array}\right ] \end {align*}
Solving for \(\vec {v}_2\) gives \[ \vec {v}_2 = \left [\begin {array}{c} \frac {2}{3} \\ 1 \end {array}\right ] \] We have found two generalized eigenvectors for eigenvalue \(-4\). Therefore the two basis solution associated with this eigenvalue are \begin {align*} \vec {x}_1(t) &= \vec {v}_1 e^{\lambda t}\\ &= \left [\begin {array}{c} 1 \\ 1 \end {array}\right ] {\mathrm e}^{-4 t}\\ &= \left [\begin {array}{c} {\mathrm e}^{-4 t} \\ {\mathrm e}^{-4 t} \end {array}\right ] \end {align*}
And \begin {align*} \vec {x}_2(t) &=\left ( \vec {v}_1 t + \vec {v}_2 \right ) e^{\lambda t} \\ &= \left (\left [\begin {array}{c} 1 \\ 1 \end {array}\right ] t + \left [\begin {array}{c} \frac {2}{3} \\ 1 \end {array}\right ]\right ) {\mathrm e}^{-4 t} \\ &=\left [\begin {array}{c} \frac {{\mathrm e}^{-4 t} \left (3 t +2\right )}{3} \\ {\mathrm e}^{-4 t} \left (t +1\right ) \end {array}\right ] \end {align*}
Therefore the final solution is \begin {align*} \vec {x}_h(t) &= c_{1} \vec {x}_{1}(t) + c_{2} \vec {x}_{2}(t) \end {align*}
Which is written as \begin {align*} \left [\begin {array}{c} y_{1} \left (t \right ) \\ y_{2} \left (t \right ) \end {array}\right ] &= c_{1} \left [\begin {array}{c} {\mathrm e}^{-4 t} \\ {\mathrm e}^{-4 t} \end {array}\right ] + c_{2} \left [\begin {array}{c} {\mathrm e}^{-4 t} \left (t +\frac {2}{3}\right ) \\ {\mathrm e}^{-4 t} \left (t +1\right ) \end {array}\right ] \end {align*}
Which becomes \begin {align*} \left [\begin {array}{c} y_{1} \left (t \right ) \\ y_{2} \left (t \right ) \end {array}\right ] = \left [\begin {array}{c} {\mathrm e}^{-4 t} \left (c_{1}+c_{2} t +\frac {2}{3} c_{2}\right ) \\ {\mathrm e}^{-4 t} \left (c_{2} t +c_{1}+c_{2}\right ) \end {array}\right ] \end {align*}
Since initial conditions are given, the solution above needs to be updated by solving for the constants of integrations using the given initial conditions \begin {align*} \left [\begin {array}{c} y_{1} \left (0\right )=0 \\ y_{2} \left (0\right )=2 \end {array}\right ]\tag {1} \end {align*}
Substituting initial conditions into the above solution at \(t=0\) gives \begin {align*} \left [\begin {array}{c} 0 \\ 2 \end {array}\right ] = \left [\begin {array}{c} c_{1}+\frac {2 c_{2}}{3} \\ c_{1}+c_{2} \end {array}\right ] \end {align*}
Solving for the constants of integrations gives \begin {align*} \left [\begin {array}{c} c_{1}=-4 \\ c_{2}=6 \end {array}\right ] \end {align*}
Substituting these constants back in original solution in Eq. (1) gives
\begin {align*} \left [\begin {array}{c} y_{1} \left (t \right ) \\ y_{2} \left (t \right ) \end {array}\right ] = \left [\begin {array}{c} 6 t \,{\mathrm e}^{-4 t} \\ \left (2+6 t \right ) {\mathrm e}^{-4 t} \end {array}\right ] \end {align*}
The following is the phase plot of the system.
The following are plots of each solution.
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 26
dsolve([diff(y__1(t),t) = -7*y__1(t)+3*y__2(t), diff(y__2(t),t) = -3*y__1(t)-y__2(t), y__1(0) = 0, y__2(0) = 2], singsol=all)
\begin{align*} y_{1} \left (t \right ) &= 6 \,{\mathrm e}^{-4 t} t \\ y_{2} \left (t \right ) &= \frac {{\mathrm e}^{-4 t} \left (18 t +6\right )}{3} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 27
DSolve[{y1'[t]==-7*y1[t]+3*y2[t],y2'[t]==-3*y1[t]-1*y2[t]},{y1[0]==0,y2[0]==2},{y1[t],y2[t]},t,IncludeSingularSolutions -> True]
\begin{align*} \text {y1}(t)\to 6 e^{-4 t} t \\ \text {y2}(t)\to e^{-4 t} (6 t+2) \\ \end{align*}