13.30 problem 47

Internal problem ID [14665]
Internal file name [OUTPUT/14345_Wednesday_April_03_2024_02_17_25_PM_76951881/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.5, page 175
Problem number: 47.
ODE order: 5.
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, _missing_x]]

\[ \boxed {8 y^{\left (5\right )}+4 y^{\prime \prime \prime \prime }+66 y^{\prime \prime \prime }-41 y^{\prime \prime }-37 y^{\prime }=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = 4, y^{\prime }\left (0\right ) = -14, y^{\prime \prime }\left (0\right ) = -14, y^{\prime \prime \prime }\left (0\right ) = 139, y^{\prime \prime \prime \prime }\left (0\right ) = -{\frac {29}{4}}\right ] \end {align*}

The characteristic equation is \[ 8 \lambda ^{5}+4 \lambda ^{4}+66 \lambda ^{3}-41 \lambda ^{2}-37 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 1\\ \lambda _3 &= -{\frac {1}{2}}\\ \lambda _4 &= -\frac {1}{2}-3 i\\ \lambda _5 &= -\frac {1}{2}+3 i \end {align*}

Therefore the homogeneous solution is \[ y_h(t)=c_{1} +c_{2} {\mathrm e}^{t}+{\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t} c_{3} +{\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t} c_{4} +{\mathrm e}^{-\frac {t}{2}} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= 1\\ y_2 &= {\mathrm e}^{t}\\ y_3 &= {\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t}\\ y_4 &= {\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t}\\ y_5 &= {\mathrm e}^{-\frac {t}{2}} \end {align*}

Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = c_{1} +c_{2} {\mathrm e}^{t}+{\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t} c_{3} +{\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t} c_{4} +{\mathrm e}^{-\frac {t}{2}} c_{5} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 4\) and \(t = 0\) in the above gives \begin {align*} 4 = c_{1} +c_{2} +c_{3} +c_{4} +c_{5}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{2} {\mathrm e}^{t}+\left (-\frac {1}{2}-3 i\right ) {\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t} c_{3} +\left (-\frac {1}{2}+3 i\right ) {\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t} c_{4} -\frac {{\mathrm e}^{-\frac {t}{2}} c_{5}}{2} \end {align*}

substituting \(y^{\prime } = -14\) and \(t = 0\) in the above gives \begin {align*} -14 = c_{2} +\left (-\frac {1}{2}-3 i\right ) c_{3} +\left (-\frac {1}{2}+3 i\right ) c_{4} -\frac {c_{5}}{2}\tag {2A} \end {align*}

Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{2} {\mathrm e}^{t}+\left (-\frac {35}{4}+3 i\right ) {\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t} c_{3} +\left (-\frac {35}{4}-3 i\right ) {\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t} c_{4} +\frac {{\mathrm e}^{-\frac {t}{2}} c_{5}}{4} \end {align*}

substituting \(y^{\prime \prime } = -14\) and \(t = 0\) in the above gives \begin {align*} -14 = c_{2} +\left (-\frac {35}{4}+3 i\right ) c_{3} +\left (-\frac {35}{4}-3 i\right ) c_{4} +\frac {c_{5}}{4}\tag {3A} \end {align*}

Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = c_{2} {\mathrm e}^{t}+\left (\frac {107}{8}+\frac {99 i}{4}\right ) {\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t} c_{3} +\left (\frac {107}{8}-\frac {99 i}{4}\right ) {\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t} c_{4} -\frac {{\mathrm e}^{-\frac {t}{2}} c_{5}}{8} \end {align*}

substituting \(y^{\prime \prime \prime } = 139\) and \(t = 0\) in the above gives \begin {align*} 139 = c_{2} +\left (\frac {107}{8}+\frac {99 i}{4}\right ) c_{3} +\left (\frac {107}{8}-\frac {99 i}{4}\right ) c_{4} -\frac {c_{5}}{8}\tag {4A} \end {align*}

Taking four derivatives of the solution gives \begin {align*} y^{\prime \prime \prime \prime } = c_{2} {\mathrm e}^{t}+\left (\frac {1081}{16}-\frac {105 i}{2}\right ) {\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t} c_{3} +\left (\frac {1081}{16}+\frac {105 i}{2}\right ) {\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t} c_{4} +\frac {{\mathrm e}^{-\frac {t}{2}} c_{5}}{16} \end {align*}

substituting \(y^{\prime \prime \prime \prime } = -{\frac {29}{4}}\) and \(t = 0\) in the above gives \begin {align*} -{\frac {29}{4}} = c_{2} +\left (\frac {1081}{16}-\frac {105 i}{2}\right ) c_{3} +\left (\frac {1081}{16}+\frac {105 i}{2}\right ) c_{4} +\frac {c_{5}}{16}\tag {5A} \end {align*}

Equations {1A,2A,3A,4A,5A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}, c_{5}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&=0\\ c_{3}&=\frac {3}{2}-2 i\\ c_{4}&=\frac {3}{2}+2 i\\ c_{5}&=1 \end {align*}

Substituting these values back in above solution results in \begin {align*} y = \frac {3 \,{\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t}}{2}-2 i {\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t}+\frac {3 \,{\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t}}{2}+2 i {\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t}+{\mathrm e}^{-\frac {t}{2}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\frac {3}{2}-2 i\right ) {\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t}+\left (\frac {3}{2}+2 i\right ) {\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t}+{\mathrm e}^{-\frac {t}{2}} \\ \end{align*}

Verification of solutions

\[ y = \left (\frac {3}{2}-2 i\right ) {\mathrm e}^{\left (-\frac {1}{2}-3 i\right ) t}+\left (\frac {3}{2}+2 i\right ) {\mathrm e}^{\left (-\frac {1}{2}+3 i\right ) t}+{\mathrm e}^{-\frac {t}{2}} \] Verified OK.

13.30.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [8 y^{\left (5\right )}+4 y^{\prime \prime \prime \prime }+66 y^{\prime \prime \prime }-41 y^{\prime \prime }-37 y^{\prime }=0, y \left (0\right )=4, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-14, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-14, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=139, y^{\prime \prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-\frac {29}{4}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 5 \\ {} & {} & y^{\left (5\right )} \\ \bullet & {} & \textrm {Isolate 5th derivative}\hspace {3pt} \\ {} & {} & y^{\left (5\right )}=-\frac {y^{\prime \prime \prime \prime }}{2}-\frac {33 y^{\prime \prime \prime }}{4}+\frac {41 y^{\prime \prime }}{8}+\frac {37 y^{\prime }}{8} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\left (5\right )}+\frac {y^{\prime \prime \prime \prime }}{2}+\frac {33 y^{\prime \prime \prime }}{4}-\frac {41 y^{\prime \prime }}{8}-\frac {37 y^{\prime }}{8}=0 \\ \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 {Define new variable}\hspace {3pt} y_{5}\left (t \right ) \\ {} & {} & y_{5}\left (t \right )=y^{\prime \prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{5}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{5}^{\prime }\left (t \right )=-\frac {y_{5}\left (t \right )}{2}-\frac {33 y_{4}\left (t \right )}{4}+\frac {41 y_{3}\left (t \right )}{8}+\frac {37 y_{2}\left (t \right )}{8} \\ & {} & \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_{5}\left (t \right )=y_{4}^{\prime }\left (t \right ), y_{5}^{\prime }\left (t \right )=-\frac {y_{5}\left (t \right )}{2}-\frac {33 y_{4}\left (t \right )}{4}+\frac {41 y_{3}\left (t \right )}{8}+\frac {37 y_{2}\left (t \right )}{8}\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 ) \\ y_{5}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & \frac {37}{8} & \frac {41}{8} & -\frac {33}{4} & -\frac {1}{2} \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & \frac {37}{8} & \frac {41}{8} & -\frac {33}{4} & -\frac {1}{2} \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 ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} 16 \\ -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}-3 \,\mathrm {I}, \left [\begin {array}{c} \frac {17296}{1874161}+\frac {13440 \,\mathrm {I}}{1874161} \\ \frac {856}{50653}-\frac {1584 \,\mathrm {I}}{50653} \\ -\frac {140}{1369}-\frac {48 \,\mathrm {I}}{1369} \\ -\frac {2}{37}+\frac {12 \,\mathrm {I}}{37} \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}+3 \,\mathrm {I}, \left [\begin {array}{c} \frac {17296}{1874161}-\frac {13440 \,\mathrm {I}}{1874161} \\ \frac {856}{50653}+\frac {1584 \,\mathrm {I}}{50653} \\ -\frac {140}{1369}+\frac {48 \,\mathrm {I}}{1369} \\ -\frac {2}{37}-\frac {12 \,\mathrm {I}}{37} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} 16 \\ -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} 16 \\ -8 \\ 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}-3 \,\mathrm {I}, \left [\begin {array}{c} \frac {17296}{1874161}+\frac {13440 \,\mathrm {I}}{1874161} \\ \frac {856}{50653}-\frac {1584 \,\mathrm {I}}{50653} \\ -\frac {140}{1369}-\frac {48 \,\mathrm {I}}{1369} \\ -\frac {2}{37}+\frac {12 \,\mathrm {I}}{37} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-\frac {1}{2}-3 \,\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} \frac {17296}{1874161}+\frac {13440 \,\mathrm {I}}{1874161} \\ \frac {856}{50653}-\frac {1584 \,\mathrm {I}}{50653} \\ -\frac {140}{1369}-\frac {48 \,\mathrm {I}}{1369} \\ -\frac {2}{37}+\frac {12 \,\mathrm {I}}{37} \\ 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}^{-\frac {t}{2}}\cdot \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right )\cdot \left [\begin {array}{c} \frac {17296}{1874161}+\frac {13440 \,\mathrm {I}}{1874161} \\ \frac {856}{50653}-\frac {1584 \,\mathrm {I}}{50653} \\ -\frac {140}{1369}-\frac {48 \,\mathrm {I}}{1369} \\ -\frac {2}{37}+\frac {12 \,\mathrm {I}}{37} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} \left (\frac {17296}{1874161}+\frac {13440 \,\mathrm {I}}{1874161}\right ) \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ \left (\frac {856}{50653}-\frac {1584 \,\mathrm {I}}{50653}\right ) \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ \left (-\frac {140}{1369}-\frac {48 \,\mathrm {I}}{1369}\right ) \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ \left (-\frac {2}{37}+\frac {12 \,\mathrm {I}}{37}\right ) \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ \cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{4}\left (t \right )={\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} \frac {17296 \cos \left (3 t \right )}{1874161}+\frac {13440 \sin \left (3 t \right )}{1874161} \\ \frac {856 \cos \left (3 t \right )}{50653}-\frac {1584 \sin \left (3 t \right )}{50653} \\ -\frac {140 \cos \left (3 t \right )}{1369}-\frac {48 \sin \left (3 t \right )}{1369} \\ -\frac {2 \cos \left (3 t \right )}{37}+\frac {12 \sin \left (3 t \right )}{37} \\ \cos \left (3 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{5}\left (t \right )={\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} -\frac {17296 \sin \left (3 t \right )}{1874161}+\frac {13440 \cos \left (3 t \right )}{1874161} \\ -\frac {856 \sin \left (3 t \right )}{50653}-\frac {1584 \cos \left (3 t \right )}{50653} \\ \frac {140 \sin \left (3 t \right )}{1369}-\frac {48 \cos \left (3 t \right )}{1369} \\ \frac {2 \sin \left (3 t \right )}{37}+\frac {12 \cos \left (3 t \right )}{37} \\ -\sin \left (3 t \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right )+c_{5} {\moverset {\rightarrow }{y}}_{5}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{2} {\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+{\mathrm e}^{-\frac {t}{2}} c_{3} \cdot \left [\begin {array}{c} 16 \\ -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]+c_{4} {\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} \frac {17296 \cos \left (3 t \right )}{1874161}+\frac {13440 \sin \left (3 t \right )}{1874161} \\ \frac {856 \cos \left (3 t \right )}{50653}-\frac {1584 \sin \left (3 t \right )}{50653} \\ -\frac {140 \cos \left (3 t \right )}{1369}-\frac {48 \sin \left (3 t \right )}{1369} \\ -\frac {2 \cos \left (3 t \right )}{37}+\frac {12 \sin \left (3 t \right )}{37} \\ \cos \left (3 t \right ) \end {array}\right ]+{\mathrm e}^{-\frac {t}{2}} c_{5} \cdot \left [\begin {array}{c} -\frac {17296 \sin \left (3 t \right )}{1874161}+\frac {13440 \cos \left (3 t \right )}{1874161} \\ -\frac {856 \sin \left (3 t \right )}{50653}-\frac {1584 \cos \left (3 t \right )}{50653} \\ \frac {140 \sin \left (3 t \right )}{1369}-\frac {48 \cos \left (3 t \right )}{1369} \\ \frac {2 \sin \left (3 t \right )}{37}+\frac {12 \cos \left (3 t \right )}{37} \\ -\sin \left (3 t \right ) \end {array}\right ]+\left [\begin {array}{c} c_{1} \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {16 \left (\left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )+\left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )+1874161 c_{3} \right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+c_{2} {\mathrm e}^{t}+c_{1} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=4 \\ {} & {} & 4=\frac {17296 c_{4}}{1874161}+\frac {13440 c_{5}}{1874161}+16 c_{3} +c_{2} +c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {16 \left (-3 \left (1081 c_{4} +840 c_{5} \right ) \sin \left (3 t \right )+3 \left (840 c_{4} -1081 c_{5} \right ) \cos \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}-\frac {8 \left (\left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )+\left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )+1874161 c_{3} \right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+c_{2} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-14 \\ {} & {} & -14=\frac {856 c_{4}}{50653}-\frac {1584 c_{5}}{50653}-8 c_{3} +c_{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {16 \left (-9 \left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )-9 \left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}-\frac {16 \left (-3 \left (1081 c_{4} +840 c_{5} \right ) \sin \left (3 t \right )+3 \left (840 c_{4} -1081 c_{5} \right ) \cos \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+\frac {4 \left (\left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )+\left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )+1874161 c_{3} \right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+c_{2} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-14 \\ {} & {} & -14=-\frac {140 c_{4}}{1369}-\frac {48 c_{5}}{1369}+4 c_{3} +c_{2} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=\frac {16 \left (27 \left (1081 c_{4} +840 c_{5} \right ) \sin \left (3 t \right )-27 \left (840 c_{4} -1081 c_{5} \right ) \cos \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}-\frac {24 \left (-9 \left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )-9 \left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+\frac {12 \left (-3 \left (1081 c_{4} +840 c_{5} \right ) \sin \left (3 t \right )+3 \left (840 c_{4} -1081 c_{5} \right ) \cos \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}-\frac {2 \left (\left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )+\left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )+1874161 c_{3} \right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+c_{2} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=139 \\ {} & {} & 139=-\frac {2 c_{4}}{37}+\frac {12 c_{5}}{37}-2 c_{3} +c_{2} \\ \bullet & {} & \textrm {Calculate the 4th derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=\frac {16 \left (81 \left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )+81 \left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}-\frac {32 \left (27 \left (1081 c_{4} +840 c_{5} \right ) \sin \left (3 t \right )-27 \left (840 c_{4} -1081 c_{5} \right ) \cos \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+\frac {24 \left (-9 \left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )-9 \left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}-\frac {8 \left (-3 \left (1081 c_{4} +840 c_{5} \right ) \sin \left (3 t \right )+3 \left (840 c_{4} -1081 c_{5} \right ) \cos \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+\frac {\left (\left (1081 c_{4} +840 c_{5} \right ) \cos \left (3 t \right )+\left (840 c_{4} -1081 c_{5} \right ) \sin \left (3 t \right )+1874161 c_{3} \right ) {\mathrm e}^{-\frac {t}{2}}}{1874161}+c_{2} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-\frac {29}{4} \\ {} & {} & -\frac {29}{4}=c_{4} +c_{3} +c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =0, c_{2} =0, c_{3} =\frac {1}{16}, c_{4} =-\frac {117}{16}, c_{5} =\frac {1711}{4}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (1+3 \cos \left (3 t \right )-4 \sin \left (3 t \right )\right ) {\mathrm e}^{-\frac {t}{2}} \end {array} \]

Maple trace

`Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

✓ Solution by Maple

Time used: 0.031 (sec). Leaf size: 23

dsolve([8*diff(y(t),t$5)+4*diff(y(t),t$4)+66*diff(y(t),t$3)-41*diff(y(t),t$2)-37*diff(y(t),t)=0,y(0) = 4, D(y)(0) = -14, (D@@2)(y)(0) = -14, (D@@3)(y)(0) = 139, (D@@4)(y)(0) = -29/4],y(t), singsol=all)
 

\[ y \left (t \right ) = {\mathrm e}^{-\frac {t}{2}} \left (1-4 \sin \left (3 t \right )+3 \cos \left (3 t \right )\right ) \]

✓ Solution by Mathematica

Time used: 0.326 (sec). Leaf size: 27

DSolve[{8*y'''''[t]+4*y''''[t]+66*y'''[t]-41*y''[t]-37*y'[t]==0,{y[0]==4,y'[0]==-14,y''[0]==-14,y'''[0]==139,y''''[0]==-29/4}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to e^{-t/2} (-4 \sin (3 t)+3 \cos (3 t)+1) \]