18.26 problem section 9.2, problem 26

Internal problem ID [1490]
Internal file name [OUTPUT/1491_Sunday_June_05_2022_02_19_32_AM_68701485/index.tex]

Book: Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.2. constant coefficient. Page 483
Problem number: section 9.2, problem 26.
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, _missing_x]]

\[ \boxed {y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-2 y^{\prime \prime }-8 y^{\prime }-8 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 5, y^{\prime }\left (0\right ) = -2, y^{\prime \prime }\left (0\right ) = 6, y^{\prime \prime \prime }\left (0\right ) = 8] \end {align*}

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

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-1-i\right ) x} c_{3} +{\mathrm e}^{\left (-1+i\right ) x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-2 x}\\ y_2 &= {\mathrm e}^{2 x}\\ y_3 &= {\mathrm e}^{\left (-1-i\right ) x}\\ y_4 &= {\mathrm e}^{\left (-1+i\right ) 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}^{-2 x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-1-i\right ) x} c_{3} +{\mathrm e}^{\left (-1+i\right ) x} c_{4} \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 \(y = 5\) and \(x = 0\) in the above gives \begin {align*} 5 = c_{1} +c_{2} +c_{3} +c_{4}\tag {1A} \end {align*}

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

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

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

substituting \(y^{\prime \prime } = 6\) and \(x = 0\) in the above gives \begin {align*} 6 = 2 c_{3} i-2 c_{4} i+4 c_{1} +4 c_{2}\tag {3A} \end {align*}

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

substituting \(y^{\prime \prime \prime } = 8\) and \(x = 0\) in the above gives \begin {align*} 8 = -8 c_{1} +8 c_{2} +\left (2-2 i\right ) c_{3} +\left (2+2 i\right ) c_{4}\tag {4A} \end {align*}

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

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

Summary

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

Verification of solutions

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

18.26.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-2 y^{\prime \prime }-8 y^{\prime }-8 y=0, y \left (0\right )=5, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-2, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=6, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=8\right ] \\ \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 (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 {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=-2 y_{4}\left (x \right )+2 y_{3}\left (x \right )+8 y_{2}\left (x \right )+8 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_{4}^{\prime }\left (x \right )=-2 y_{4}\left (x \right )+2 y_{3}\left (x \right )+8 y_{2}\left (x \right )+8 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 ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 8 & 8 & 2 & -2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \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 \\ 8 & 8 & 2 & -2 \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 [-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} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1-\mathrm {I}, \left [\begin {array}{c} \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [-1+\mathrm {I}, \left [\begin {array}{c} \frac {1}{4}-\frac {\mathrm {I}}{4} \\ \frac {\mathrm {I}}{2} \\ -\frac {1}{2}-\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-1-\mathrm {I}, \left [\begin {array}{c} \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-1-\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 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}^{-x}\cdot \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-x}\cdot \left [\begin {array}{c} \left (\frac {1}{4}+\frac {\mathrm {I}}{4}\right ) \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ -\frac {\mathrm {I}}{2} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \left (-\frac {1}{2}+\frac {\mathrm {I}}{2}\right ) \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}}_{3}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {\sin \left (x \right )}{4}+\frac {\cos \left (x \right )}{4} \\ -\frac {\sin \left (x \right )}{2} \\ -\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \\ -\frac {\cos \left (x \right )}{2} \\ \frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ -\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}+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {\sin \left (x \right )}{4}+\frac {\cos \left (x \right )}{4} \\ -\frac {\sin \left (x \right )}{2} \\ -\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ \cos \left (x \right ) \end {array}\right ]+c_{4} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \\ -\frac {\cos \left (x \right )}{2} \\ \frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ -\sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 x} \left (c_{2} {\mathrm e}^{4 x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}-c_{1} \right )}{8} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=5 \\ {} & {} & 5=\frac {c_{2}}{8}+\frac {c_{3}}{4}+\frac {c_{4}}{4}-\frac {c_{1}}{8} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {{\mathrm e}^{-2 x} \left (c_{2} {\mathrm e}^{4 x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}-c_{1} \right )}{4}+\frac {{\mathrm e}^{-2 x} \left (4 c_{2} {\mathrm e}^{4 x}+\left (-\left (2 c_{3} +2 c_{4} \right ) \sin \left (x \right )+2 \left (c_{3} -c_{4} \right ) \cos \left (x \right )\right ) {\mathrm e}^{x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}\right )}{8} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-2 \\ {} & {} & -2=\frac {c_{2}}{4}-\frac {c_{4}}{2}+\frac {c_{1}}{4} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {{\mathrm e}^{-2 x} \left (c_{2} {\mathrm e}^{4 x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}-c_{1} \right )}{2}-\frac {{\mathrm e}^{-2 x} \left (4 c_{2} {\mathrm e}^{4 x}+\left (-\left (2 c_{3} +2 c_{4} \right ) \sin \left (x \right )+2 \left (c_{3} -c_{4} \right ) \cos \left (x \right )\right ) {\mathrm e}^{x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}\right )}{2}+\frac {{\mathrm e}^{-2 x} \left (16 c_{2} {\mathrm e}^{4 x}+\left (-\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )-2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}+2 \left (-\left (2 c_{3} +2 c_{4} \right ) \sin \left (x \right )+2 \left (c_{3} -c_{4} \right ) \cos \left (x \right )\right ) {\mathrm e}^{x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}\right )}{8} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=6 \\ {} & {} & 6=\frac {c_{2}}{2}-\frac {c_{3}}{2}+\frac {c_{4}}{2}-\frac {c_{1}}{2} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-{\mathrm e}^{-2 x} \left (c_{2} {\mathrm e}^{4 x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}-c_{1} \right )+\frac {3 \,{\mathrm e}^{-2 x} \left (4 c_{2} {\mathrm e}^{4 x}+\left (-\left (2 c_{3} +2 c_{4} \right ) \sin \left (x \right )+2 \left (c_{3} -c_{4} \right ) \cos \left (x \right )\right ) {\mathrm e}^{x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}\right )}{2}-\frac {3 \,{\mathrm e}^{-2 x} \left (16 c_{2} {\mathrm e}^{4 x}+\left (-\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )-2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}+2 \left (-\left (2 c_{3} +2 c_{4} \right ) \sin \left (x \right )+2 \left (c_{3} -c_{4} \right ) \cos \left (x \right )\right ) {\mathrm e}^{x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}\right )}{4}+\frac {{\mathrm e}^{-2 x} \left (64 c_{2} {\mathrm e}^{4 x}+\left (\left (2 c_{3} +2 c_{4} \right ) \sin \left (x \right )-2 \left (c_{3} -c_{4} \right ) \cos \left (x \right )\right ) {\mathrm e}^{x}+3 \left (-\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )-2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}+3 \left (-\left (2 c_{3} +2 c_{4} \right ) \sin \left (x \right )+2 \left (c_{3} -c_{4} \right ) \cos \left (x \right )\right ) {\mathrm e}^{x}+\left (\left (2 c_{3} +2 c_{4} \right ) \cos \left (x \right )+2 \left (c_{3} -c_{4} \right ) \sin \left (x \right )\right ) {\mathrm e}^{x}\right )}{8} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=8 \\ {} & {} & 8=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-8, c_{2} =8, c_{3} =8, c_{4} =4\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left ({\mathrm e}^{4 x}+\left (3 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{x}+1\right ) {\mathrm e}^{-2 x} \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$4)+2*diff(y(x),x$3)-2*diff(y(x),x$2)-8*diff(y(x),x)-8*y(x)=0,y(0) = 5, D(y)(0) = -2, (D@@2)(y)(0) = 6, (D@@3)(y)(0) = 8],y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 31

DSolve[{y''''[x]+2*y'''[x]-2*y''[x]-8*y'[x]-8*y[x]==0,{y[0]==5,y'[0]==-2,y''[0]==6,y'''[0]==8}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-2 x} \left (e^{4 x}+e^x \sin (x)+3 e^x \cos (x)+1\right ) \]