Internal problem ID [13074]
Internal file name [OUTPUT/11730_Sunday_December_03_2023_07_16_09_PM_26869848/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 3. Linear Systems. Exercises section 3.1. page 258
Problem number: 34.
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 )&=1\\ y^{\prime }\left (t \right )&=x \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) + \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} 0 & 0 \\ 1 & 0 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} 1 \\ 0 \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} 1 & 0 \\ t & 1 \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} 1 & 0 \\ t & 1 \end {array}\right ] \left [\begin {array}{c} c_{1} \\ c_{2} \end {array}\right ] \\ &= \left [\begin {array}{c} c_{1} \\ t c_{1}+c_{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} 1 & 0 \\ -t & 1 \end {array}\right ] \end {align*}
Hence \begin {align*} \vec {x}_p (t) &= \left [\begin {array}{cc} 1 & 0 \\ t & 1 \end {array}\right ] \int { \left [\begin {array}{cc} 1 & 0 \\ -t & 1 \end {array}\right ] \left [\begin {array}{c} 1 \\ 0 \end {array}\right ]\,dt}\\ &= \left [\begin {array}{cc} 1 & 0 \\ t & 1 \end {array}\right ] \left [\begin {array}{c} t \\ -\frac {t^{2}}{2} \end {array}\right ]\\ &= \left [\begin {array}{c} t \\ \frac {t^{2}}{2} \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} c_{1}+t \\ t c_{1}+c_{2}+\frac {1}{2} t^{2} \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} 0 & 0 \\ 1 & 0 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} 1 \\ 0 \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} 0 & 0 \\ 1 & 0 \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} -\lambda & 0 \\ 1 & -\lambda \end {array}\right ]\right ) &= 0 \end {align*}
Since the matrix \(A\) is triangular matrix, then the determinant is the product of the elements along the diagonal. Therefore the above becomes \begin {align*} (-\lambda )(-\lambda )&=0 \end {align*}
The roots of the above are the eigenvalues.
\begin {align*} \lambda _1 &= 0 \end {align*}
This table summarises the above result
| eigenvalue | algebraic multiplicity | type of eigenvalue |
| \(0\) | \(1\) | real eigenvalue |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = 0\)
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} 0 & 0 \\ 1 & 0 \end {array}\right ] - \left (0\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} 0 & 0 \\ 1 & 0 \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 \[ \left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 0&0&0\\ 1&0&0 \end {array} \right ] \] Since the current pivot \(A(1,1)\) is zero, then the current pivot row is replaced with a row with a non-zero pivot. Swapping row \(1\) and row \(2\) gives \[ \left [\begin {array}{@{}cc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1&0&0\\ 0&0&0 \end {array} \right ] \] Therefore the system in Echelon form is \[ \left [\begin {array}{cc} 1 & 0 \\ 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \end {array}\right ] \] 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 \(\{v_{1} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} 0 \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = t \left [\begin {array}{c} 0 \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ t \end {array}\right ] = \left [\begin {array}{c} 0 \\ 1 \end {array}\right ] \] 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 |
| \(0\) | \(2\) | \(1\) | Yes | \(\left [\begin {array}{c} 0 \\ 1 \end {array}\right ]\) |
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 \(0\) 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} 0 & 0 \\ 1 & 0 \end {array}\right ]- \left (0\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 \\ 1 \end {array}\right ]\\ \left [\begin {array}{cc} 0 & 0 \\ 1 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 1 \end {array}\right ] \end {align*}
Solving for \(\vec {v}_2\) gives \[ \vec {v}_2 = \left [\begin {array}{c} 1 \\ 1 \end {array}\right ] \] We have found two generalized eigenvectors for eigenvalue \(0\). 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} 0 \\ 1 \end {array}\right ] 1\\ &= \left [\begin {array}{c} 0 \\ 1 \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} 0 \\ 1 \end {array}\right ] t + \left [\begin {array}{c} 1 \\ 1 \end {array}\right ]\right ) 1 \\ &=\left [\begin {array}{c} 1 \\ 1+t \end {array}\right ] \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} 0 \\ 1 \end {array}\right ] + c_{2} \left [\begin {array}{c} 1 \\ 1+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} 0 & 1 \\ 1 & 1+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} -1-t & 1 \\ 1 & 0 \end {array}\right ] \end {align*}
Hence \begin {align*} \vec {x}_p(t) &= \left [\begin {array}{cc} 0 & 1 \\ 1 & 1+t \end {array}\right ] \int { \left [\begin {array}{cc} -1-t & 1 \\ 1 & 0 \end {array}\right ] \left [\begin {array}{c} 1 \\ 0 \end {array}\right ] \, dt}\\ &= \left [\begin {array}{cc} 0 & 1 \\ 1 & 1+t \end {array}\right ] \int { \left [\begin {array}{c} -1-t \\ 1 \end {array}\right ] \, dt}\\ &= \left [\begin {array}{cc} 0 & 1 \\ 1 & 1+t \end {array}\right ] \left [\begin {array}{c} -t -\frac {1}{2} t^{2} \\ t \end {array}\right ] \\ &= \left [\begin {array}{c} t \\ \frac {t^{2}}{2} \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} 0 \\ c_{1} \end {array}\right ] + \left [\begin {array}{c} c_{2} \\ c_{2} \left (1+t \right ) \end {array}\right ] + \left [\begin {array}{c} t \\ \frac {t^{2}}{2} \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} c_{2}+t \\ c_{1}+c_{2} t +c_{2}+\frac {1}{2} t^{2} \end {array}\right ] \end {align*}
The following is the phase plot of the system.
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 21
dsolve([diff(x(t),t)=1,diff(y(t),t)=x(t)],singsol=all)
\begin{align*} x \left (t \right ) &= c_{2} +t \\ y \left (t \right ) &= c_{2} t +\frac {1}{2} t^{2}+c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.009 (sec). Leaf size: 26
DSolve[{x'[t]==1,y'[t]==x[t]},{x[t],y[t]},t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to t+c_1 \\ y(t)\to \frac {t^2}{2}+c_1 t+c_2 \\ \end{align*}