problem 2

Internal problem ID [14676]
Internal ﬁle name [OUTPUT/14356_Wednesday_April_03_2024_02_17_36_PM_46820025/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: 2.
ODE order: 4.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime \prime }-16 y=1} \] 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 \[ y^{\prime \prime \prime \prime }-16 y = 0 \] The characteristic equation is \[ \lambda ^{4}-16 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2\\ \lambda _3 &= 2 i\\ \lambda _4 &= -2 i \end {align*}

Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{-2 i t} c_{3} +{\mathrm e}^{2 i t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 t} \\ y_2 &= {\mathrm e}^{2 t} \\ y_3 &= {\mathrm e}^{-2 i t} \\ y_4 &= {\mathrm e}^{2 i t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-16 y = 1 \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ 1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{-2 t}, {\mathrm e}^{2 t}, {\mathrm e}^{-2 i t}, {\mathrm e}^{2 i t}\} \] 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. \[ y_p = A_{1} \] 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 \[ -16 A_{1} = 1 \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {1}{16}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -{\frac {1}{16}} \] 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}+{\mathrm e}^{-2 i t} c_{3} +{\mathrm e}^{2 i t} c_{4}\right ) + \left (-{\frac {1}{16}}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{-2 i t} c_{3} +{\mathrm e}^{2 i t} c_{4} -\frac {1}{16} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{-2 i t} c_{3} +{\mathrm e}^{2 i t} c_{4} -\frac {1}{16} \] Veriﬁed OK.

14.2.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }-16 y=1 \\ \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 )=1+16 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 )=1+16 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 \\ 16 & 0 & 0 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 1 \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 \\ 16 & 0 & 0 & 0 \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} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ -\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 t}\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 t}\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 [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} t}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \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 \\ {} & {} & \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ -\frac {\cos \left (2 t \right )}{4}+\frac {\mathrm {I} \sin \left (2 t \right )}{4} \\ \frac {\mathrm {I}}{2} \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}}_{3}\left (t \right )=\left [\begin {array}{c} -\frac {\sin \left (2 t \right )}{8} \\ -\frac {\cos \left (2 t \right )}{4} \\ \frac {\sin \left (2 t \right )}{2} \\ \cos \left (2 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (t \right )=\left [\begin {array}{c} -\frac {\cos \left (2 t \right )}{8} \\ \frac {\sin \left (2 t \right )}{4} \\ \frac {\cos \left (2 t \right )}{2} \\ -\sin \left (2 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}+c_{2} {\moverset {\rightarrow }{y}}_{2}+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} & \frac {{\mathrm e}^{2 t}}{8} & -\frac {\sin \left (2 t \right )}{8} & -\frac {\cos \left (2 t \right )}{8} \\ \frac {{\mathrm e}^{-2 t}}{4} & \frac {{\mathrm e}^{2 t}}{4} & -\frac {\cos \left (2 t \right )}{4} & \frac {\sin \left (2 t \right )}{4} \\ -\frac {{\mathrm e}^{-2 t}}{2} & \frac {{\mathrm e}^{2 t}}{2} & \frac {\sin \left (2 t \right )}{2} & \frac {\cos \left (2 t \right )}{2} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{2 t} & \cos \left (2 t \right ) & -\sin \left (2 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} & \frac {{\mathrm e}^{2 t}}{8} & -\frac {\sin \left (2 t \right )}{8} & -\frac {\cos \left (2 t \right )}{8} \\ \frac {{\mathrm e}^{-2 t}}{4} & \frac {{\mathrm e}^{2 t}}{4} & -\frac {\cos \left (2 t \right )}{4} & \frac {\sin \left (2 t \right )}{4} \\ -\frac {{\mathrm e}^{-2 t}}{2} & \frac {{\mathrm e}^{2 t}}{2} & \frac {\sin \left (2 t \right )}{2} & \frac {\cos \left (2 t \right )}{2} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{2 t} & \cos \left (2 t \right ) & -\sin \left (2 t \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -\frac {1}{8} & \frac {1}{8} & 0 & -\frac {1}{8} \\ \frac {1}{4} & \frac {1}{4} & -\frac {1}{4} & 0 \\ -\frac {1}{2} & \frac {1}{2} & 0 & \frac {1}{2} \\ 1 & 1 & 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} \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} & \frac {{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t}}{16}-\frac {\cos \left (2 t \right )}{8} & -\frac {{\mathrm e}^{-2 t}}{32}+\frac {{\mathrm e}^{2 t}}{32}-\frac {\sin \left (2 t \right )}{16} \\ -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} & \frac {{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t}}{16}-\frac {\cos \left (2 t \right )}{8} \\ {\mathrm e}^{-2 t}+{\mathrm e}^{2 t}-2 \cos \left (2 t \right ) & -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} \\ -2 \,{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{2 t}+4 \sin \left (2 t \right ) & {\mathrm e}^{-2 t}+{\mathrm e}^{2 t}-2 \cos \left (2 t \right ) & -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} \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}^{2 t}}{64}+\frac {\cos \left (2 t \right )}{32}-\frac {1}{16}+\frac {{\mathrm e}^{-2 t}}{64} \\ -\frac {{\mathrm e}^{-2 t}}{32}+\frac {{\mathrm e}^{2 t}}{32}-\frac {\sin \left (2 t \right )}{16} \\ \frac {{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t}}{16}-\frac {\cos \left (2 t \right )}{8} \\ -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} \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}+c_{2} {\moverset {\rightarrow }{y}}_{2}+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}^{2 t}}{64}+\frac {\cos \left (2 t \right )}{32}-\frac {1}{16}+\frac {{\mathrm e}^{-2 t}}{64} \\ -\frac {{\mathrm e}^{-2 t}}{32}+\frac {{\mathrm e}^{2 t}}{32}-\frac {\sin \left (2 t \right )}{16} \\ \frac {{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t}}{16}-\frac {\cos \left (2 t \right )}{8} \\ -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (-8 c_{4} +2\right ) \cos \left (2 t \right )}{64}+\frac {\left (-8 c_{1} +1\right ) {\mathrm e}^{-2 t}}{64}+\frac {\left (8 c_{2} +1\right ) {\mathrm e}^{2 t}}{64}-\frac {c_{3} \sin \left (2 t \right )}{8}-\frac {1}{16} \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`

\[ y \left (t \right ) = -\frac {1}{16}+\cos \left (2 t \right ) c_{1} +c_{2} {\mathrm e}^{-2 t}+c_{3} {\mathrm e}^{2 t}+c_{4} \sin \left (2 t \right ) \]

\[ y(t)\to c_1 e^{2 t}+c_3 e^{-2 t}+c_2 \cos (2 t)+c_4 \sin (2 t)-\frac {1}{16} \]

14.2 problem 2

14.2.1 Maple step by step solution