Internal problem ID [5981]
Internal file name [OUTPUT/5229_Sunday_June_05_2022_03_28_00_PM_96542066/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 2. Linear equations with constant coefficients. Page 79
Problem number: 2(c).
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]]
\[ \boxed {y^{\left (5\right )}-y^{\prime \prime \prime \prime }-y^{\prime }+y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime \prime }\left (0\right ) = 0] \end {align*}
The characteristic equation is \[ \lambda ^{5}-\lambda ^{4}-\lambda +1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= i\\ \lambda _3 &= -i\\ \lambda _4 &= 1\\ \lambda _5 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+x \,{\mathrm e}^{x} c_{3} +{\mathrm e}^{-i x} c_{4} +{\mathrm e}^{i x} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{x}\\ y_3 &= x \,{\mathrm e}^{x}\\ y_4 &= {\mathrm e}^{-i x}\\ y_5 &= {\mathrm e}^{i x} \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}^{-x}+c_{2} {\mathrm e}^{x}+x \,{\mathrm e}^{x} c_{3} +{\mathrm e}^{-i x} c_{4} +{\mathrm e}^{i x} 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 \(y = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{1} +c_{2} +c_{4} +c_{5}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{x} c_{3} +x \,{\mathrm e}^{x} c_{3} -i {\mathrm e}^{-i x} c_{4} +i {\mathrm e}^{i x} c_{5} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = -c_{4} i+c_{5} i-c_{1} +c_{2} +c_{3}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+2 \,{\mathrm e}^{x} c_{3} +x \,{\mathrm e}^{x} c_{3} -{\mathrm e}^{-i x} c_{4} -{\mathrm e}^{i x} c_{5} \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{2} +2 c_{3} -c_{4} -c_{5}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{x} c_{3} +x \,{\mathrm e}^{x} c_{3} +i {\mathrm e}^{-i x} c_{4} -i {\mathrm e}^{i x} c_{5} \end {align*}
substituting \(y^{\prime \prime \prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{4} i-c_{5} i-c_{1} +c_{2} +3 c_{3}\tag {4A} \end {align*}
Taking four derivatives of the solution gives \begin {align*} y^{\prime \prime \prime \prime } = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+4 \,{\mathrm e}^{x} c_{3} +x \,{\mathrm e}^{x} c_{3} +{\mathrm e}^{-i x} c_{4} +{\mathrm e}^{i x} c_{5} \end {align*}
substituting \(y^{\prime \prime \prime \prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{2} +4 c_{3} +c_{4} +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}&={\frac {1}{8}}\\ c_{2}&={\frac {5}{8}}\\ c_{3}&=-{\frac {1}{4}}\\ c_{4}&=\frac {1}{8}-\frac {i}{8}\\ c_{5}&=\frac {1}{8}+\frac {i}{8} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {{\mathrm e}^{-x}}{8}+\frac {5 \,{\mathrm e}^{x}}{8}-\frac {x \,{\mathrm e}^{x}}{4}+\frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-x}}{8}+\frac {5 \,{\mathrm e}^{x}}{8}-\frac {x \,{\mathrm e}^{x}}{4}+\frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-x}}{8}+\frac {5 \,{\mathrm e}^{x}}{8}-\frac {x \,{\mathrm e}^{x}}{4}+\frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\left (5\right )}-y^{\prime \prime \prime \prime }-y^{\prime }+y=0, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, y^{\prime \prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 5 \\ {} & {} & y^{\left (5\right )} \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{5}\left (x \right ) \\ {} & {} & y_{5}\left (x \right )=y^{\prime \prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{5}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{5}^{\prime }\left (x \right )=y_{5}\left (x \right )+y_{2}\left (x \right )-y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{5}\left (x \right )=y_{4}^{\prime }\left (x \right ), y_{5}^{\prime }\left (x \right )=y_{5}\left (x \right )+y_{2}\left (x \right )-y_{1}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \\ y_{4}\left (x \right ) \\ y_{5}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \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 \\ -1 & 1 & 0 & 0 & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \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 \\ -1 & 1 & 0 & 0 & 1 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \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 [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [\mathrm {-I}, \left [\begin {array}{c} 1 \\ \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I}, \left [\begin {array}{c} 1 \\ \mathrm {I} \\ -1 \\ \mathrm {-I} \\ 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}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} 1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 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 =1\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \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 =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\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 x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\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}}_{3}\left (x \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} 1 \\ {} & {} & \left (\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ -1 & 1 & 0 & 0 & 1 \end {array}\right ]-1\cdot \left [\begin {array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} 1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{x}\cdot \left (x \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I}, \left [\begin {array}{c} 1 \\ \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} x}\cdot \left [\begin {array}{c} 1 \\ \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 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 \\ {} & {} & \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} 1 \\ \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} \cos \left (x \right )-\mathrm {I} \sin \left (x \right ) \\ \mathrm {-I} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ -\cos \left (x \right )+\mathrm {I} \sin \left (x \right ) \\ \mathrm {I} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \cos \left (x \right )-\mathrm {I} \sin \left (x \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 (x \right )=\left [\begin {array}{c} \cos \left (x \right ) \\ -\sin \left (x \right ) \\ -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{5}\left (x \right )=\left [\begin {array}{c} -\sin \left (x \right ) \\ -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \\ -\sin \left (x \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}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right )+c_{5} {\moverset {\rightarrow }{y}}_{5}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}={\mathrm e}^{-x} c_{1} \cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+{\mathrm e}^{x} c_{3} \cdot \left (x \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right )+\left [\begin {array}{c} c_{4} \cos \left (x \right )-c_{5} \sin \left (x \right ) \\ -c_{4} \sin \left (x \right )-c_{5} \cos \left (x \right ) \\ -c_{4} \cos \left (x \right )+c_{5} \sin \left (x \right ) \\ c_{4} \sin \left (x \right )+c_{5} \cos \left (x \right ) \\ c_{4} \cos \left (x \right )-c_{5} \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-x} c_{1} +\left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{x}+c_{4} \cos \left (x \right )-c_{5} \sin \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} -c_{3} +c_{2} +c_{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-{\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{3} +\left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{x}-c_{4} \sin \left (x \right )-c_{5} \cos \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-c_{1} +c_{2} -c_{5} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }={\mathrm e}^{-x} c_{1} +2 \,{\mathrm e}^{x} c_{3} +\left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{x}-c_{4} \cos \left (x \right )+c_{5} \sin \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +c_{3} +c_{2} -c_{4} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-{\mathrm e}^{-x} c_{1} +3 \,{\mathrm e}^{x} c_{3} +\left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{x}+c_{4} \sin \left (x \right )+c_{5} \cos \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-c_{1} +2 c_{3} +c_{2} +c_{5} \\ \bullet & {} & \textrm {Calculate the 4th derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }={\mathrm e}^{-x} c_{1} +4 \,{\mathrm e}^{x} c_{3} +\left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{x}+c_{4} \cos \left (x \right )-c_{5} \sin \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +3 c_{3} +c_{2} +c_{4} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {1}{8}, c_{2} =\frac {3}{8}, c_{3} =-\frac {1}{4}, c_{4} =\frac {1}{4}, c_{5} =\frac {1}{4}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-x}}{8}+\frac {\left (-2 x +5\right ) {\mathrm e}^{x}}{8}+\frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \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: 0.047 (sec). Leaf size: 28
dsolve([diff(y(x),x$5)-diff(y(x),x$4)-diff(y(x),x)+y(x)=0,y(0) = 1, D(y)(0) = 0, (D@@2)(y)(0) = 0, (D@@3)(y)(0) = 0, (D@@4)(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-x}}{8}+\frac {\left (-2 x +5\right ) {\mathrm e}^{x}}{8}+\frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 34
DSolve[{y'''''[x]-y''''[x]-y'[x]+y[x]==0,{y[0]==1,y'[0]==0,y''[0]==0,y'''[0]==0,y''''[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{8} \left (-2 e^x x+e^{-x}+5 e^x-2 \sin (x)+2 \cos (x)\right ) \]