Internal problem ID [14666]
Internal file name [OUTPUT/14346_Wednesday_April_03_2024_02_17_25_PM_83505057/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: 48.
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]]
\[ \boxed {2 y^{\left (5\right )}+7 y^{\prime \prime \prime \prime }+17 y^{\prime \prime \prime }+17 y^{\prime \prime }+5 y^{\prime }=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = -3, y^{\prime }\left (0\right ) = {\frac {15}{2}}, y^{\prime \prime }\left (0\right ) = {\frac {17}{4}}, y^{\prime \prime \prime }\left (0\right ) = -{\frac {385}{8}}, y^{\prime \prime \prime \prime }\left (0\right ) = {\frac {1217}{16}}\right ] \end {align*}
The characteristic equation is \[ 2 \lambda ^{5}+7 \lambda ^{4}+17 \lambda ^{3}+17 \lambda ^{2}+5 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= -{\frac {1}{2}}\\ \lambda _3 &= -1\\ \lambda _4 &= -1-2 i\\ \lambda _5 &= -1+2 i \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-t}+c_{2} +{\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +c_{4} {\mathrm e}^{-\frac {t}{2}}+{\mathrm e}^{\left (-1+2 i\right ) t} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-t}\\ y_2 &= 1\\ y_3 &= {\mathrm e}^{\left (-1-2 i\right ) t}\\ y_4 &= {\mathrm e}^{-\frac {t}{2}}\\ y_5 &= {\mathrm e}^{\left (-1+2 i\right ) t} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{-t}+c_{2} +{\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +c_{4} {\mathrm e}^{-\frac {t}{2}}+{\mathrm e}^{\left (-1+2 i\right ) t} 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 \(y = -3\) and \(t = 0\) in the above gives \begin {align*} -3 = 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_{1} {\mathrm e}^{-t}+\left (-1-2 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} c_{3} -\frac {c_{4} {\mathrm e}^{-\frac {t}{2}}}{2}+\left (-1+2 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} c_{5} \end {align*}
substituting \(y^{\prime } = {\frac {15}{2}}\) and \(t = 0\) in the above gives \begin {align*} {\frac {15}{2}} = -c_{1} +\left (-1-2 i\right ) c_{3} -\frac {c_{4}}{2}+\left (-1+2 i\right ) c_{5}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{-t}+\left (-3+4 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +\frac {c_{4} {\mathrm e}^{-\frac {t}{2}}}{4}+\left (-3-4 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} c_{5} \end {align*}
substituting \(y^{\prime \prime } = {\frac {17}{4}}\) and \(t = 0\) in the above gives \begin {align*} {\frac {17}{4}} = c_{1} +\left (-3+4 i\right ) c_{3} +\frac {c_{4}}{4}+\left (-3-4 i\right ) c_{5}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -c_{1} {\mathrm e}^{-t}+\left (11+2 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} c_{3} -\frac {c_{4} {\mathrm e}^{-\frac {t}{2}}}{8}+\left (11-2 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} c_{5} \end {align*}
substituting \(y^{\prime \prime \prime } = -{\frac {385}{8}}\) and \(t = 0\) in the above gives \begin {align*} -{\frac {385}{8}} = -c_{1} +\left (11+2 i\right ) c_{3} -\frac {c_{4}}{8}+\left (11-2 i\right ) c_{5}\tag {4A} \end {align*}
Taking four derivatives of the solution gives \begin {align*} y^{\prime \prime \prime \prime } = c_{1} {\mathrm e}^{-t}+\left (-7-24 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +\frac {c_{4} {\mathrm e}^{-\frac {t}{2}}}{16}+\left (-7+24 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} c_{5} \end {align*}
substituting \(y^{\prime \prime \prime \prime } = {\frac {1217}{16}}\) and \(t = 0\) in the above gives \begin {align*} {\frac {1217}{16}} = c_{1} +\left (-7-24 i\right ) c_{3} +\frac {c_{4}}{16}+\left (-7+24 i\right ) c_{5}\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}&=-2+i\\ c_{4}&=1\\ c_{5}&=-2-i \end {align*}
Substituting these values back in above solution results in \begin {align*} y = i {\mathrm e}^{\left (-1-2 i\right ) t}-i {\mathrm e}^{\left (-1+2 i\right ) t}-2 \,{\mathrm e}^{\left (-1-2 i\right ) t}+{\mathrm e}^{-\frac {t}{2}}-2 \,{\mathrm e}^{\left (-1+2 i\right ) t} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (-2+i\right ) {\mathrm e}^{\left (-1-2 i\right ) t}+\left (-2-i\right ) {\mathrm e}^{\left (-1+2 i\right ) t}+{\mathrm e}^{-\frac {t}{2}} \\ \end{align*}
Verification of solutions
\[ y = \left (-2+i\right ) {\mathrm e}^{\left (-1-2 i\right ) t}+\left (-2-i\right ) {\mathrm e}^{\left (-1+2 i\right ) t}+{\mathrm e}^{-\frac {t}{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 y^{\left (5\right )}+7 y^{\prime \prime \prime \prime }+17 y^{\prime \prime \prime }+17 y^{\prime \prime }+5 y^{\prime }=0, y \left (0\right )=-3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {15}{2}, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {17}{4}, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-\frac {385}{8}, y^{\prime \prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {1217}{16}\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 {7 y^{\prime \prime \prime \prime }}{2}-\frac {17 y^{\prime \prime \prime }}{2}-\frac {17 y^{\prime \prime }}{2}-\frac {5 y^{\prime }}{2} \\ \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 {7 y^{\prime \prime \prime \prime }}{2}+\frac {17 y^{\prime \prime \prime }}{2}+\frac {17 y^{\prime \prime }}{2}+\frac {5 y^{\prime }}{2}=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 {7 y_{5}\left (t \right )}{2}-\frac {17 y_{4}\left (t \right )}{2}-\frac {17 y_{3}\left (t \right )}{2}-\frac {5 y_{2}\left (t \right )}{2} \\ & {} & \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 {7 y_{5}\left (t \right )}{2}-\frac {17 y_{4}\left (t \right )}{2}-\frac {17 y_{3}\left (t \right )}{2}-\frac {5 y_{2}\left (t \right )}{2}\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 {5}{2} & -\frac {17}{2} & -\frac {17}{2} & -\frac {7}{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 {5}{2} & -\frac {17}{2} & -\frac {17}{2} & -\frac {7}{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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} 16 \\ -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [-1-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ], \left [-1+2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {7}{625}-\frac {24 \,\mathrm {I}}{625} \\ \frac {11}{125}+\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}-\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ]\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}}_{1}={\mathrm e}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end {array}\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}}_{2}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \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 [-1-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-1-2 \,\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} -\frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 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}^{-t}\cdot \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right )\cdot \left [\begin {array}{c} -\frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-t}\cdot \left [\begin {array}{c} \left (-\frac {7}{625}+\frac {24 \,\mathrm {I}}{625}\right ) \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \left (\frac {11}{125}-\frac {2 \,\mathrm {I}}{125}\right ) \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \left (-\frac {3}{25}-\frac {4 \,\mathrm {I}}{25}\right ) \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \left (-\frac {1}{5}+\frac {2 \,\mathrm {I}}{5}\right ) \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \cos \left (2 t \right )-\mathrm {I} \sin \left (2 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}^{-t}\cdot \left [\begin {array}{c} -\frac {7 \cos \left (2 t \right )}{625}+\frac {24 \sin \left (2 t \right )}{625} \\ \frac {11 \cos \left (2 t \right )}{125}-\frac {2 \sin \left (2 t \right )}{125} \\ -\frac {3 \cos \left (2 t \right )}{25}-\frac {4 \sin \left (2 t \right )}{25} \\ -\frac {\cos \left (2 t \right )}{5}+\frac {2 \sin \left (2 t \right )}{5} \\ \cos \left (2 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{5}\left (t \right )={\mathrm e}^{-t}\cdot \left [\begin {array}{c} \frac {7 \sin \left (2 t \right )}{625}+\frac {24 \cos \left (2 t \right )}{625} \\ -\frac {11 \sin \left (2 t \right )}{125}-\frac {2 \cos \left (2 t \right )}{125} \\ \frac {3 \sin \left (2 t \right )}{25}-\frac {4 \cos \left (2 t \right )}{25} \\ \frac {\sin \left (2 t \right )}{5}+\frac {2 \cos \left (2 t \right )}{5} \\ -\sin \left (2 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_{1} {\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}^{-t}\cdot \left [\begin {array}{c} -\frac {7 \cos \left (2 t \right )}{625}+\frac {24 \sin \left (2 t \right )}{625} \\ \frac {11 \cos \left (2 t \right )}{125}-\frac {2 \sin \left (2 t \right )}{125} \\ -\frac {3 \cos \left (2 t \right )}{25}-\frac {4 \sin \left (2 t \right )}{25} \\ -\frac {\cos \left (2 t \right )}{5}+\frac {2 \sin \left (2 t \right )}{5} \\ \cos \left (2 t \right ) \end {array}\right ]+c_{5} {\mathrm e}^{-t}\cdot \left [\begin {array}{c} \frac {7 \sin \left (2 t \right )}{625}+\frac {24 \cos \left (2 t \right )}{625} \\ -\frac {11 \sin \left (2 t \right )}{125}-\frac {2 \cos \left (2 t \right )}{125} \\ \frac {3 \sin \left (2 t \right )}{25}-\frac {4 \cos \left (2 t \right )}{25} \\ \frac {\sin \left (2 t \right )}{5}+\frac {2 \cos \left (2 t \right )}{5} \\ -\sin \left (2 t \right ) \end {array}\right ]+\left [\begin {array}{c} c_{2} \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=16 \,{\mathrm e}^{-\frac {t}{2}} c_{3} +\frac {\left (\left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )+\left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )+625 c_{1} \right ) {\mathrm e}^{-t}}{625}+c_{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=-3 \\ {} & {} & -3=16 c_{3} -\frac {7 c_{4}}{625}+\frac {24 c_{5}}{625}+c_{1} +c_{2} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-8 \,{\mathrm e}^{-\frac {t}{2}} c_{3} +\frac {\left (-2 \left (-7 c_{4} +24 c_{5} \right ) \sin \left (2 t \right )+2 \left (24 c_{4} +7 c_{5} \right ) \cos \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}-\frac {\left (\left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )+\left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )+625 c_{1} \right ) {\mathrm e}^{-t}}{625} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {15}{2} \\ {} & {} & \frac {15}{2}=-8 c_{3} +\frac {11 c_{4}}{125}-\frac {2 c_{5}}{125}-c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=4 \,{\mathrm e}^{-\frac {t}{2}} c_{3} +\frac {\left (-4 \left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )-4 \left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}-\frac {2 \left (-2 \left (-7 c_{4} +24 c_{5} \right ) \sin \left (2 t \right )+2 \left (24 c_{4} +7 c_{5} \right ) \cos \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}+\frac {\left (\left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )+\left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )+625 c_{1} \right ) {\mathrm e}^{-t}}{625} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {17}{4} \\ {} & {} & \frac {17}{4}=4 c_{3} -\frac {3 c_{4}}{25}-\frac {4 c_{5}}{25}+c_{1} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-2 \,{\mathrm e}^{-\frac {t}{2}} c_{3} +\frac {\left (8 \left (-7 c_{4} +24 c_{5} \right ) \sin \left (2 t \right )-8 \left (24 c_{4} +7 c_{5} \right ) \cos \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}-\frac {3 \left (-4 \left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )-4 \left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}+\frac {3 \left (-2 \left (-7 c_{4} +24 c_{5} \right ) \sin \left (2 t \right )+2 \left (24 c_{4} +7 c_{5} \right ) \cos \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}-\frac {\left (\left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )+\left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )+625 c_{1} \right ) {\mathrm e}^{-t}}{625} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-\frac {385}{8} \\ {} & {} & -\frac {385}{8}=-2 c_{3} -\frac {c_{4}}{5}+\frac {2 c_{5}}{5}-c_{1} \\ \bullet & {} & \textrm {Calculate the 4th derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }={\mathrm e}^{-\frac {t}{2}} c_{3} +\frac {\left (16 \left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )+16 \left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}-\frac {4 \left (8 \left (-7 c_{4} +24 c_{5} \right ) \sin \left (2 t \right )-8 \left (24 c_{4} +7 c_{5} \right ) \cos \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}+\frac {6 \left (-4 \left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )-4 \left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}-\frac {4 \left (-2 \left (-7 c_{4} +24 c_{5} \right ) \sin \left (2 t \right )+2 \left (24 c_{4} +7 c_{5} \right ) \cos \left (2 t \right )\right ) {\mathrm e}^{-t}}{625}+\frac {\left (\left (-7 c_{4} +24 c_{5} \right ) \cos \left (2 t \right )+\left (24 c_{4} +7 c_{5} \right ) \sin \left (2 t \right )+625 c_{1} \right ) {\mathrm e}^{-t}}{625} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {1217}{16} \\ {} & {} & \frac {1217}{16}=c_{1} +c_{3} +c_{4} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =0, c_{2} =0, c_{3} =\frac {1}{16}, c_{4} =76, c_{5} =-82\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-\frac {t}{2}}+\left (-4 \cos \left (2 t \right )+2 \sin \left (2 t \right )\right ) {\mathrm e}^{-t} \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: 29
dsolve([2*diff(y(t),t$5)+7*diff(y(t),t$4)+17*diff(y(t),t$3)+17*diff(y(t),t$2)+5*diff(y(t),t)=0,y(0) = -3, D(y)(0) = 15/2, (D@@2)(y)(0) = 17/4, (D@@3)(y)(0) = -385/8, (D@@4)(y)(0) = 1217/16],y(t), singsol=all)
\[ y \left (t \right ) = \left (2 \sin \left (2 t \right )-4 \cos \left (2 t \right )\right ) {\mathrm e}^{-t}+{\mathrm e}^{-\frac {t}{2}} \]
✓ Solution by Mathematica
Time used: 0.298 (sec). Leaf size: 31
DSolve[{2*y'''''[t]+7*y''''[t]+17*y'''[t]+17*y''[t]+5*y'[t]==0,{y[0]==-3,y'[0]==15/2,y''[0]==17/4,y'''[0]==-385/8,y''''[0]==1217/16}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{-t} \left (e^{t/2}+2 \sin (2 t)-4 \cos (2 t)\right ) \]