2.1.72 Internal
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[8210]Book
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1.0 Problem
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72Date
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Sunday, November 10, 2024 at 03:22:03 AMCAS
classification
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system_of_ODEs
\begin{align*} \frac {d}{d t}x \left (t \right )&=x \left (t \right )-3 y \left (t \right )\\ \frac {d}{d t}y \left (t \right )&=3 x \left (t \right )+7 y \left (t \right ) \end{align*}
In this method, we will assume we have found the matrix exponential \(e^{A t}\) allready. There are
different methods to determine this but will not be shown here. This is a system of linear
ODE’s given as
\begin{align*} \vec {x}'(t) &= A\, \vec {x}(t) \end{align*}
Or
\begin{align*} \left [\begin {array}{c} \frac {d}{d t}x \left (t \right ) \\ \frac {d}{d t}y \left (t \right ) \end {array}\right ] &= \left [\begin {array}{cc} 1 & -3 \\ 3 & 7 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \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 {c} \\ &= \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} c_{1} \\ c_{2} \end {array}\right ] \\ &= \left [\begin {array}{c} {\mathrm e}^{4 t} \left (1-3 t \right ) c_{1}-3 t \,{\mathrm e}^{4 t} c_{2} \\ 3 t \,{\mathrm e}^{4 t} c_{1}+{\mathrm e}^{4 t} \left (1+3 t \right ) c_{2} \end {array}\right ]\\ &= \left [\begin {array}{c} \left (c_{1} \left (1-3 t \right )-3 c_{2} t \right ) {\mathrm e}^{4 t} \\ {\mathrm e}^{4 t} \left (3 t c_{1}+3 c_{2} t +c_{2}\right ) \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} \frac {d}{d t}x \left (t \right ) \\ \frac {d}{d t}y \left (t \right ) \end {array}\right ] &= \left [\begin {array}{cc} 1 & -3 \\ 3 & 7 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \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} 1 & -3 \\ 3 & 7 \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 & -3 \\ 3 & 7-\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} 1 & -3 \\ 3 & 7 \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} 1 & -3 \\ 3 & 7 \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} x \left (t \right ) \\ y \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} x \left (t \right ) \\ y \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*}
Figure 2.155: Phase plot \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}x \left (t \right )=x \left (t \right )-3 y \left (t \right ), \frac {d}{d t}y \left (t \right )=3 x \left (t \right )+7 y \left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=\left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Convert system into a vector equation}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{x}}\left (t \right )=\left [\begin {array}{cc} 1 & -3 \\ 3 & 7 \end {array}\right ]\cdot {\moverset {\rightarrow }{x}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{x}}\left (t \right )=\left [\begin {array}{cc} 1 & -3 \\ 3 & 7 \end {array}\right ]\cdot {\moverset {\rightarrow }{x}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cc} 1 & -3 \\ 3 & 7 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{x}}\left (t \right )=A \cdot {\moverset {\rightarrow }{x}}\left (t \right ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [4, \left [\begin {array}{c} -1 \\ 1 \end {array}\right ]\right ], \left [4, \left [\begin {array}{c} 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [4, \left [\begin {array}{c} -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} 4 \\ {} & {} & {\moverset {\rightarrow }{x}}_{1}\left (t \right )={\mathrm e}^{4 t}\cdot \left [\begin {array}{c} -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =4\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{2}\left (t \right )={\mathrm e}^{\lambda t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} t \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =4 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{x}}_{2}\left (t \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda t} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda t} A \right )\cdot \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda t} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda t} \left (\lambda t {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{x}}_{2}\left (t \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} 4 \\ {} & {} & \left (\left [\begin {array}{cc} 1 & -3 \\ 3 & 7 \end {array}\right ]-4\cdot \left [\begin {array}{cc} 1 & 0 \\ 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{3} \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} 4 \\ {} & {} & {\moverset {\rightarrow }{x}}_{2}\left (t \right )={\mathrm e}^{4 t}\cdot \left (t \cdot \left [\begin {array}{c} -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} \frac {1}{3} \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}=\mathit {C1} {\moverset {\rightarrow }{x}}_{1}\left (t \right )+\mathit {C2} {\moverset {\rightarrow }{x}}_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}=\mathit {C1} \,{\mathrm e}^{4 t}\cdot \left [\begin {array}{c} -1 \\ 1 \end {array}\right ]+\mathit {C2} \,{\mathrm e}^{4 t}\cdot \left (t \cdot \left [\begin {array}{c} -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} \frac {1}{3} \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Substitute in vector of dependent variables}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ]=\left [\begin {array}{c} {\mathrm e}^{4 t} \left (-\mathit {C1} -\mathit {C2} t +\frac {1}{3} \mathit {C2} \right ) \\ {\mathrm e}^{4 t} \left (\mathit {C2} t +\mathit {C1} \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Solution to the system of ODEs}\hspace {3pt} \\ {} & {} & \left \{x \left (t \right )={\mathrm e}^{4 t} \left (-\mathit {C1} -\mathit {C2} t +\frac {1}{3} \mathit {C2} \right ), y \left (t \right )={\mathrm e}^{4 t} \left (\mathit {C2} t +\mathit {C1} \right )\right \} \end {array} \]
Solving time : 0.034
dsolve ([ diff ( x ( t ), t ) = x(t)-3*y(t), diff ( y ( t ), t ) = 3*x(t)+7*y(t)]
,{ op ([ x ( t ), y(t)])})
\begin{align*}
x \left (t \right ) &= {\mathrm e}^{4 t} \left (c_{2} t +c_{1} \right ) \\
y \left (t \right ) &= -\frac {{\mathrm e}^{4 t} \left (3 c_{2} t +3 c_{1} +c_{2} \right )}{3} \\
\end{align*}
Solving time : 0.003
DSolve [{{ D [ x [ t ], t ]== x [ t ]-3* y [ t ], D [ y [ t ], t ]==3* x [ t ]+7* y [ t ]},{}}, {x[t],y[t]},t,IncludeSingularSolutions-> True ]
\begin{align*}
x(t)\to -e^{4 t} (c_1 (3 t-1)+3 c_2 t) \\
y(t)\to e^{4 t} (3 (c_1+c_2) t+c_2) \\
\end{align*}