problem Problem 11

Internal problem ID [12173]
Internal ﬁle name [OUTPUT/10826_Thursday_September_21_2023_05_47_33_AM_71755707/index.tex]

Book: Diﬀerential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS, MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER. Problems page 172
Problem number: Problem 11.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime \prime }-16 y=x^{2}-{\mathrm e}^{x}} \] 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 \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{-2 i x} c_{3} +{\mathrm e}^{2 i x} c_{4} \] 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}^{-2 i x} \\ y_4 &= {\mathrm e}^{2 i x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-16 y = x^{2}-{\mathrm e}^{x} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ x^{2}-{\mathrm e}^{x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x}\}, \{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{-2 x}, {\mathrm e}^{2 x}, {\mathrm e}^{-2 i x}, {\mathrm e}^{2 i x}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{x}+A_{2}+A_{3} x +A_{4} x^{2} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}\}\) 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 \[ -15 A_{1} {\mathrm e}^{x}-16 A_{2}-16 A_{3} x -16 A_{4} x^{2} = x^{2}-{\mathrm e}^{x} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = {\frac {1}{15}}, A_{2} = 0, A_{3} = 0, A_{4} = -{\frac {1}{16}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{x}}{15}-\frac {x^{2}}{16} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{-2 i x} c_{3} +{\mathrm e}^{2 i x} c_{4}\right ) + \left (\frac {{\mathrm e}^{x}}{15}-\frac {x^{2}}{16}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{-2 i x} c_{3} +{\mathrm e}^{2 i x} c_{4} +\frac {{\mathrm e}^{x}}{15}-\frac {x^{2}}{16} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{-2 i x} c_{3} +{\mathrm e}^{2 i x} c_{4} +\frac {{\mathrm e}^{x}}{15}-\frac {x^{2}}{16} \] Veriﬁed OK.

2.11.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }-16 y=x^{2}-{\mathrm e}^{x} \\ \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 )=x^{2}+16 y_{1}\left (x \right )-{\mathrm e}^{x} \\ & {} & \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 )=x^{2}+16 y_{1}\left (x \right )-{\mathrm e}^{x}\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 \\ 16 & 0 & 0 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ x^{2}-{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ x^{2}-{\mathrm e}^{x} \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 (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \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 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 [-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} x}\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 x \right )-\mathrm {I} \sin \left (2 x \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 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ -\frac {\cos \left (2 x \right )}{4}+\frac {\mathrm {I} \sin \left (2 x \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \cos \left (2 x \right )-\mathrm {I} \sin \left (2 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 )=\left [\begin {array}{c} -\frac {\sin \left (2 x \right )}{8} \\ -\frac {\cos \left (2 x \right )}{4} \\ \frac {\sin \left (2 x \right )}{2} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )=\left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{8} \\ \frac {\sin \left (2 x \right )}{4} \\ \frac {\cos \left (2 x \right )}{2} \\ -\sin \left (2 x \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 (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=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 )+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{cccc} -\frac {{\mathrm e}^{-2 x}}{8} & \frac {{\mathrm e}^{2 x}}{8} & -\frac {\sin \left (2 x \right )}{8} & -\frac {\cos \left (2 x \right )}{8} \\ \frac {{\mathrm e}^{-2 x}}{4} & \frac {{\mathrm e}^{2 x}}{4} & -\frac {\cos \left (2 x \right )}{4} & \frac {\sin \left (2 x \right )}{4} \\ -\frac {{\mathrm e}^{-2 x}}{2} & \frac {{\mathrm e}^{2 x}}{2} & \frac {\sin \left (2 x \right )}{2} & \frac {\cos \left (2 x \right )}{2} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{2 x} & \cos \left (2 x \right ) & -\sin \left (2 x \right ) \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \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 (x \right )=\phi \left (x \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{cccc} -\frac {{\mathrm e}^{-2 x}}{8} & \frac {{\mathrm e}^{2 x}}{8} & -\frac {\sin \left (2 x \right )}{8} & -\frac {\cos \left (2 x \right )}{8} \\ \frac {{\mathrm e}^{-2 x}}{4} & \frac {{\mathrm e}^{2 x}}{4} & -\frac {\cos \left (2 x \right )}{4} & \frac {\sin \left (2 x \right )}{4} \\ -\frac {{\mathrm e}^{-2 x}}{2} & \frac {{\mathrm e}^{2 x}}{2} & \frac {\sin \left (2 x \right )}{2} & \frac {\cos \left (2 x \right )}{2} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{2 x} & \cos \left (2 x \right ) & -\sin \left (2 x \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 (x \right )=\left [\begin {array}{cccc} \frac {{\mathrm e}^{-2 x}}{4}+\frac {{\mathrm e}^{2 x}}{4}+\frac {\cos \left (2 x \right )}{2} & -\frac {{\mathrm e}^{-2 x}}{8}+\frac {{\mathrm e}^{2 x}}{8}+\frac {\sin \left (2 x \right )}{4} & \frac {{\mathrm e}^{-2 x}}{16}+\frac {{\mathrm e}^{2 x}}{16}-\frac {\cos \left (2 x \right )}{8} & -\frac {{\mathrm e}^{-2 x}}{32}+\frac {{\mathrm e}^{2 x}}{32}-\frac {\sin \left (2 x \right )}{16} \\ -\frac {{\mathrm e}^{-2 x}}{2}+\frac {{\mathrm e}^{2 x}}{2}-\sin \left (2 x \right ) & \frac {{\mathrm e}^{-2 x}}{4}+\frac {{\mathrm e}^{2 x}}{4}+\frac {\cos \left (2 x \right )}{2} & -\frac {{\mathrm e}^{-2 x}}{8}+\frac {{\mathrm e}^{2 x}}{8}+\frac {\sin \left (2 x \right )}{4} & \frac {{\mathrm e}^{-2 x}}{16}+\frac {{\mathrm e}^{2 x}}{16}-\frac {\cos \left (2 x \right )}{8} \\ {\mathrm e}^{-2 x}+{\mathrm e}^{2 x}-2 \cos \left (2 x \right ) & -\frac {{\mathrm e}^{-2 x}}{2}+\frac {{\mathrm e}^{2 x}}{2}-\sin \left (2 x \right ) & \frac {{\mathrm e}^{-2 x}}{4}+\frac {{\mathrm e}^{2 x}}{4}+\frac {\cos \left (2 x \right )}{2} & -\frac {{\mathrm e}^{-2 x}}{8}+\frac {{\mathrm e}^{2 x}}{8}+\frac {\sin \left (2 x \right )}{4} \\ -2 \,{\mathrm e}^{-2 x}+2 \,{\mathrm e}^{2 x}+4 \sin \left (2 x \right ) & {\mathrm e}^{-2 x}+{\mathrm e}^{2 x}-2 \cos \left (2 x \right ) & -\frac {{\mathrm e}^{-2 x}}{2}+\frac {{\mathrm e}^{2 x}}{2}-\sin \left (2 x \right ) & \frac {{\mathrm e}^{-2 x}}{4}+\frac {{\mathrm e}^{2 x}}{4}+\frac {\cos \left (2 x \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 (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \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 (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\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 (x \right )=\left [\begin {array}{c} -\frac {{\mathrm e}^{-2 x} \left (\left (x^{2}+\frac {13 \cos \left (2 x \right )}{20}+\frac {\sin \left (2 x \right )}{5}\right ) {\mathrm e}^{2 x}-\frac {16 \,{\mathrm e}^{3 x}}{15}+\frac {3 \,{\mathrm e}^{4 x}}{8}+\frac {1}{24}\right )}{16} \\ -\frac {\left (\left (x +\frac {\cos \left (2 x \right )}{5}-\frac {13 \sin \left (2 x \right )}{20}\right ) {\mathrm e}^{2 x}-\frac {8 \,{\mathrm e}^{3 x}}{15}+\frac {3 \,{\mathrm e}^{4 x}}{8}-\frac {1}{24}\right ) {\mathrm e}^{-2 x}}{8} \\ \frac {{\mathrm e}^{-2 x} \left (\left (-10+13 \cos \left (2 x \right )+4 \sin \left (2 x \right )\right ) {\mathrm e}^{2 x}+\frac {16 \,{\mathrm e}^{3 x}}{3}-\frac {15 \,{\mathrm e}^{4 x}}{2}-\frac {5}{6}\right )}{80} \\ \frac {\left (24 \,{\mathrm e}^{2 x} \cos \left (2 x \right )-78 \,{\mathrm e}^{2 x} \sin \left (2 x \right )-45 \,{\mathrm e}^{4 x}+16 \,{\mathrm e}^{3 x}+5\right ) {\mathrm e}^{-2 x}}{240} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=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 )+\left [\begin {array}{c} -\frac {{\mathrm e}^{-2 x} \left (\left (x^{2}+\frac {13 \cos \left (2 x \right )}{20}+\frac {\sin \left (2 x \right )}{5}\right ) {\mathrm e}^{2 x}-\frac {16 \,{\mathrm e}^{3 x}}{15}+\frac {3 \,{\mathrm e}^{4 x}}{8}+\frac {1}{24}\right )}{16} \\ -\frac {\left (\left (x +\frac {\cos \left (2 x \right )}{5}-\frac {13 \sin \left (2 x \right )}{20}\right ) {\mathrm e}^{2 x}-\frac {8 \,{\mathrm e}^{3 x}}{15}+\frac {3 \,{\mathrm e}^{4 x}}{8}-\frac {1}{24}\right ) {\mathrm e}^{-2 x}}{8} \\ \frac {{\mathrm e}^{-2 x} \left (\left (-10+13 \cos \left (2 x \right )+4 \sin \left (2 x \right )\right ) {\mathrm e}^{2 x}+\frac {16 \,{\mathrm e}^{3 x}}{3}-\frac {15 \,{\mathrm e}^{4 x}}{2}-\frac {5}{6}\right )}{80} \\ \frac {\left (24 \,{\mathrm e}^{2 x} \cos \left (2 x \right )-78 \,{\mathrm e}^{2 x} \sin \left (2 x \right )-45 \,{\mathrm e}^{4 x}+16 \,{\mathrm e}^{3 x}+5\right ) {\mathrm e}^{-2 x}}{240} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {\left (\left (\left (2 c_{4} +\frac {13}{20}\right ) \cos \left (2 x \right )+\left (2 c_{3} +\frac {1}{5}\right ) \sin \left (2 x \right )+x^{2}\right ) {\mathrm e}^{2 x}+\left (-2 c_{2} +\frac {3}{8}\right ) {\mathrm e}^{4 x}+2 c_{1} -\frac {16 \,{\mathrm e}^{3 x}}{15}+\frac {1}{24}\right ) {\mathrm e}^{-2 x}}{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`

\[ y \left (x \right ) = -\frac {\left (\left (\left (-16 c_{1} +\frac {1}{4}\right ) \cos \left (2 x \right )+x^{2}-16 c_{4} \sin \left (2 x \right )\right ) {\mathrm e}^{2 x}-16 c_{3} {\mathrm e}^{4 x}-16 c_{2} -\frac {16 \,{\mathrm e}^{3 x}}{15}\right ) {\mathrm e}^{-2 x}}{16} \]

\[ y(x)\to -\frac {x^2}{16}+\frac {e^x}{15}+c_1 e^{2 x}+c_3 e^{-2 x}+c_2 \cos (2 x)+c_4 \sin (2 x) \]

2.11 problem Problem 11

2.11.1 Maple step by step solution