Internal problem ID [14684]
Internal file name [OUTPUT/14364_Wednesday_April_03_2024_02_17_40_PM_81751103/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.6, page 187
Problem number: 10.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_high_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime \prime \prime }-10 y^{\prime \prime \prime }+38 y^{\prime \prime }-64 y^{\prime }+40 y=153 \,{\mathrm e}^{-t}} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime \prime }-10 y^{\prime \prime \prime }+38 y^{\prime \prime }-64 y^{\prime }+40 y = 0 \] The characteristic equation is \[ \lambda ^{4}-10 \lambda ^{3}+38 \lambda ^{2}-64 \lambda +40 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 3-i\\ \lambda _2 &= 3+i\\ \lambda _3 &= 2\\ \lambda _4 &= 2 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{2 t}+c_{2} {\mathrm e}^{2 t} t +{\mathrm e}^{\left (3-i\right ) t} c_{3} +{\mathrm e}^{\left (3+i\right ) t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{2 t} \\ y_2 &= t \,{\mathrm e}^{2 t} \\ y_3 &= {\mathrm e}^{\left (3-i\right ) t} \\ y_4 &= {\mathrm e}^{\left (3+i\right ) t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-10 y^{\prime \prime \prime }+38 y^{\prime \prime }-64 y^{\prime }+40 y = 153 \,{\mathrm e}^{-t} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 153 \,{\mathrm e}^{-t} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-t}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{t \,{\mathrm e}^{2 t}, {\mathrm e}^{\left (3-i\right ) t}, {\mathrm e}^{\left (3+i\right ) t}, {\mathrm e}^{2 t}\} \] Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set. \[ y_p = A_{1} {\mathrm e}^{-t} \] The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives \[ 153 A_{1} {\mathrm e}^{-t} = 153 \,{\mathrm e}^{-t} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = {\mathrm e}^{-t} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{2 t}+c_{2} {\mathrm e}^{2 t} t +{\mathrm e}^{\left (3-i\right ) t} c_{3} +{\mathrm e}^{\left (3+i\right ) t} c_{4}\right ) + \left ({\mathrm e}^{-t}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{\left (3-i\right ) t} c_{3} +{\mathrm e}^{\left (3+i\right ) t} c_{4} +{\mathrm e}^{2 t} \left (c_{2} t +c_{1} \right )+{\mathrm e}^{-t} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\left (3-i\right ) t} c_{3} +{\mathrm e}^{\left (3+i\right ) t} c_{4} +{\mathrm e}^{2 t} \left (c_{2} t +c_{1} \right )+{\mathrm e}^{-t} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{\left (3-i\right ) t} c_{3} +{\mathrm e}^{\left (3+i\right ) t} c_{4} +{\mathrm e}^{2 t} \left (c_{2} t +c_{1} \right )+{\mathrm e}^{-t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }-10 y^{\prime \prime \prime }+38 y^{\prime \prime }-64 y^{\prime }+40 y=153 \,{\mathrm e}^{-t} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (t \right ) \\ {} & {} & y_{4}\left (t \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (t \right )=153 \,{\mathrm e}^{-t}+10 y_{4}\left (t \right )-38 y_{3}\left (t \right )+64 y_{2}\left (t \right )-40 y_{1}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=y_{1}^{\prime }\left (t \right ), y_{3}\left (t \right )=y_{2}^{\prime }\left (t \right ), y_{4}\left (t \right )=y_{3}^{\prime }\left (t \right ), y_{4}^{\prime }\left (t \right )=153 \,{\mathrm e}^{-t}+10 y_{4}\left (t \right )-38 y_{3}\left (t \right )+64 y_{2}\left (t \right )-40 y_{1}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \\ y_{4}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -40 & 64 & -38 & 10 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 153 \,{\mathrm e}^{-t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 153 \,{\mathrm e}^{-t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -40 & 64 & -38 & 10 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\left (t \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [3-\mathrm {I}, \left [\begin {array}{c} \frac {9}{500}+\frac {13 \,\mathrm {I}}{500} \\ \frac {2}{25}+\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}+\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ]\right ], \left [3+\mathrm {I}, \left [\begin {array}{c} \frac {9}{500}-\frac {13 \,\mathrm {I}}{500} \\ \frac {2}{25}-\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}-\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} 2 \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (t \right )={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 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 =2\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{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 =2 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{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 }{y}}_{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} 2 \\ {} & {} & \left (\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -40 & 64 & -38 & 10 \end {array}\right ]-2\cdot \left [\begin {array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -\frac {1}{16} \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} 2 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (t \right )={\mathrm e}^{2 t}\cdot \left (t \cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {1}{16} \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [3-\mathrm {I}, \left [\begin {array}{c} \frac {9}{500}+\frac {13 \,\mathrm {I}}{500} \\ \frac {2}{25}+\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}+\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (3-\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} \frac {9}{500}+\frac {13 \,\mathrm {I}}{500} \\ \frac {2}{25}+\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}+\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & {\mathrm e}^{3 t}\cdot \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right )\cdot \left [\begin {array}{c} \frac {9}{500}+\frac {13 \,\mathrm {I}}{500} \\ \frac {2}{25}+\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}+\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \left (\frac {9}{500}+\frac {13 \,\mathrm {I}}{500}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \left (\frac {2}{25}+\frac {3 \,\mathrm {I}}{50}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \left (\frac {3}{10}+\frac {\mathrm {I}}{10}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \cos \left (t \right )-\mathrm {I} \sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \\ \frac {2 \cos \left (t \right )}{25}+\frac {3 \sin \left (t \right )}{50} \\ \frac {3 \cos \left (t \right )}{10}+\frac {\sin \left (t \right )}{10} \\ \cos \left (t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (t \right )={\mathrm e}^{3 t}\cdot \left [\begin {array}{c} -\frac {9 \sin \left (t \right )}{500}+\frac {13 \cos \left (t \right )}{500} \\ -\frac {2 \sin \left (t \right )}{25}+\frac {3 \cos \left (t \right )}{50} \\ -\frac {3 \sin \left (t \right )}{10}+\frac {\cos \left (t \right )}{10} \\ -\sin \left (t \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}\left (t \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right )+{\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (t \right )=\left [\begin {array}{cccc} \frac {{\mathrm e}^{2 t}}{8} & {\mathrm e}^{2 t} \left (\frac {t}{8}-\frac {1}{16}\right ) & {\mathrm e}^{3 t} \left (\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500}\right ) & {\mathrm e}^{3 t} \left (-\frac {9 \sin \left (t \right )}{500}+\frac {13 \cos \left (t \right )}{500}\right ) \\ \frac {{\mathrm e}^{2 t}}{4} & \frac {t \,{\mathrm e}^{2 t}}{4} & {\mathrm e}^{3 t} \left (\frac {2 \cos \left (t \right )}{25}+\frac {3 \sin \left (t \right )}{50}\right ) & {\mathrm e}^{3 t} \left (-\frac {2 \sin \left (t \right )}{25}+\frac {3 \cos \left (t \right )}{50}\right ) \\ \frac {{\mathrm e}^{2 t}}{2} & \frac {t \,{\mathrm e}^{2 t}}{2} & {\mathrm e}^{3 t} \left (\frac {3 \cos \left (t \right )}{10}+\frac {\sin \left (t \right )}{10}\right ) & {\mathrm e}^{3 t} \left (-\frac {3 \sin \left (t \right )}{10}+\frac {\cos \left (t \right )}{10}\right ) \\ {\mathrm e}^{2 t} & t \,{\mathrm e}^{2 t} & {\mathrm e}^{3 t} \cos \left (t \right ) & -{\mathrm e}^{3 t} \sin \left (t \right ) \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (t \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\phi \left (t \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{cccc} \frac {{\mathrm e}^{2 t}}{8} & {\mathrm e}^{2 t} \left (\frac {t}{8}-\frac {1}{16}\right ) & {\mathrm e}^{3 t} \left (\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500}\right ) & {\mathrm e}^{3 t} \left (-\frac {9 \sin \left (t \right )}{500}+\frac {13 \cos \left (t \right )}{500}\right ) \\ \frac {{\mathrm e}^{2 t}}{4} & \frac {t \,{\mathrm e}^{2 t}}{4} & {\mathrm e}^{3 t} \left (\frac {2 \cos \left (t \right )}{25}+\frac {3 \sin \left (t \right )}{50}\right ) & {\mathrm e}^{3 t} \left (-\frac {2 \sin \left (t \right )}{25}+\frac {3 \cos \left (t \right )}{50}\right ) \\ \frac {{\mathrm e}^{2 t}}{2} & \frac {t \,{\mathrm e}^{2 t}}{2} & {\mathrm e}^{3 t} \left (\frac {3 \cos \left (t \right )}{10}+\frac {\sin \left (t \right )}{10}\right ) & {\mathrm e}^{3 t} \left (-\frac {3 \sin \left (t \right )}{10}+\frac {\cos \left (t \right )}{10}\right ) \\ {\mathrm e}^{2 t} & t \,{\mathrm e}^{2 t} & {\mathrm e}^{3 t} \cos \left (t \right ) & -{\mathrm e}^{3 t} \sin \left (t \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} \frac {1}{8} & -\frac {1}{16} & \frac {9}{500} & \frac {13}{500} \\ \frac {1}{4} & 0 & \frac {2}{25} & \frac {3}{50} \\ \frac {1}{2} & 0 & \frac {3}{10} & \frac {1}{10} \\ 1 & 0 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{cccc} \left (1-2 t \right ) {\mathrm e}^{2 t} & \frac {\left (11 t +7\right ) {\mathrm e}^{2 t}}{5}-\frac {7 \,{\mathrm e}^{3 t} \left (\cos \left (t \right )-\frac {\sin \left (t \right )}{7}\right )}{5} & \frac {\left (-8 t -11\right ) {\mathrm e}^{2 t}}{10}+\frac {11 \,{\mathrm e}^{3 t} \left (\cos \left (t \right )-\frac {3 \sin \left (t \right )}{11}\right )}{10} & \frac {{\mathrm e}^{2 t} \left (2+t \right )}{10}-\frac {\left (\cos \left (t \right )-\frac {\sin \left (t \right )}{2}\right ) {\mathrm e}^{3 t}}{5} \\ -4 t \,{\mathrm e}^{2 t} & \frac {\left (22 t +25\right ) {\mathrm e}^{2 t}}{5}-4 \left (\cos \left (t \right )-\frac {\sin \left (t \right )}{2}\right ) {\mathrm e}^{3 t} & \frac {\left (-8 t -15\right ) {\mathrm e}^{2 t}}{5}+3 \,{\mathrm e}^{3 t} \left (\cos \left (t \right )-\frac {2 \sin \left (t \right )}{3}\right ) & \frac {\left (2 t +5\right ) {\mathrm e}^{2 t}}{10}-\frac {{\mathrm e}^{3 t} \left (-\sin \left (t \right )+\cos \left (t \right )\right )}{2} \\ -8 t \,{\mathrm e}^{2 t} & \frac {\left (44 t +50\right ) {\mathrm e}^{2 t}}{5}-10 \,{\mathrm e}^{3 t} \left (-\sin \left (t \right )+\cos \left (t \right )\right ) & \frac {2 \left (-8 t -15\right ) {\mathrm e}^{2 t}}{5}+7 \left (\cos \left (t \right )-\frac {9 \sin \left (t \right )}{7}\right ) {\mathrm e}^{3 t} & \frac {\left (2 t +5\right ) {\mathrm e}^{2 t}}{5}-{\mathrm e}^{3 t} \left (\cos \left (t \right )-2 \sin \left (t \right )\right ) \\ -16 t \,{\mathrm e}^{2 t} & \frac {4 \left (22 t +25\right ) {\mathrm e}^{2 t}}{5}-20 \,{\mathrm e}^{3 t} \left (\cos \left (t \right )-2 \sin \left (t \right )\right ) & \frac {4 \left (-8 t -15\right ) {\mathrm e}^{2 t}}{5}+12 \,{\mathrm e}^{3 t} \left (\cos \left (t \right )-\frac {17 \sin \left (t \right )}{6}\right ) & \frac {2 \left (2 t +5\right ) {\mathrm e}^{2 t}}{5}-{\mathrm e}^{3 t} \left (\cos \left (t \right )-7 \sin \left (t \right )\right ) \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (t \right )=\Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )={\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=\frac {1}{\Phi \left (t \right )}\cdot {\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (t \right )=\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot \left (\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\left [\begin {array}{c} \frac {{\mathrm e}^{-t} \left (-81 \cos \left (t \right ) {\mathrm e}^{4 t}+18 \,{\mathrm e}^{4 t} \sin \left (t \right )+51 \,{\mathrm e}^{3 t} t +85 \,{\mathrm e}^{3 t}-4\right )}{10} \\ \frac {{\mathrm e}^{-t} \left (-225 \cos \left (t \right ) {\mathrm e}^{4 t}+135 \,{\mathrm e}^{4 t} \sin \left (t \right )+102 \,{\mathrm e}^{3 t} t +221 \,{\mathrm e}^{3 t}+4\right )}{10} \\ \frac {17 \left (6 t +13\right ) {\mathrm e}^{2 t}}{5}+9 \left (-6 \cos \left (t \right )+7 \sin \left (t \right )\right ) {\mathrm e}^{3 t}+\frac {49 \,{\mathrm e}^{-t}}{5} \\ \frac {34 \left (6 t +13\right ) {\mathrm e}^{2 t}}{5}+9 \left (-11 \cos \left (t \right )+27 \sin \left (t \right )\right ) {\mathrm e}^{3 t}+\frac {53 \,{\mathrm e}^{-t}}{5} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}\left (t \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right )+\left [\begin {array}{c} \frac {{\mathrm e}^{-t} \left (-81 \cos \left (t \right ) {\mathrm e}^{4 t}+18 \,{\mathrm e}^{4 t} \sin \left (t \right )+51 \,{\mathrm e}^{3 t} t +85 \,{\mathrm e}^{3 t}-4\right )}{10} \\ \frac {{\mathrm e}^{-t} \left (-225 \cos \left (t \right ) {\mathrm e}^{4 t}+135 \,{\mathrm e}^{4 t} \sin \left (t \right )+102 \,{\mathrm e}^{3 t} t +221 \,{\mathrm e}^{3 t}+4\right )}{10} \\ \frac {17 \left (6 t +13\right ) {\mathrm e}^{2 t}}{5}+9 \left (-6 \cos \left (t \right )+7 \sin \left (t \right )\right ) {\mathrm e}^{3 t}+\frac {49 \,{\mathrm e}^{-t}}{5} \\ \frac {34 \left (6 t +13\right ) {\mathrm e}^{2 t}}{5}+9 \left (-11 \cos \left (t \right )+27 \sin \left (t \right )\right ) {\mathrm e}^{3 t}+\frac {53 \,{\mathrm e}^{-t}}{5} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (36 c_{3} +52 c_{4} -16200\right ) \cos \left (t \right )+52 \sin \left (t \right ) \left (c_{3} -\frac {9 c_{4}}{13}+\frac {900}{13}\right )\right ) {\mathrm e}^{3 t}}{2000}+\frac {\left (\left (250 t -125\right ) c_{2} +10200 t +250 c_{1} +17000\right ) {\mathrm e}^{2 t}}{2000}-\frac {2 \,{\mathrm e}^{-t}}{5} \end {array} \]
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 4; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 33
dsolve(diff(y(t),t$4)-10*diff(y(t),t$3)+38*diff(y(t),t$2)-64*diff(y(t),t)+40*y(t)=153*exp(-t),y(t), singsol=all)
\[ y \left (t \right ) = \left (t c_{2} +c_{1} \right ) {\mathrm e}^{2 t}+\left (\cos \left (t \right ) c_{3} +\sin \left (t \right ) c_{4} \right ) {\mathrm e}^{3 t}+{\mathrm e}^{-t} \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 48
DSolve[y''''[t]-10*y'''[t]+38*y''[t]-64*y'[t]+40*y[t]==153*Exp[-t],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{-t}+c_3 e^{2 t}+c_4 e^{2 t} t+c_2 e^{3 t} \cos (t)+c_1 e^{3 t} \sin (t) \]