Internal problem ID [2830]
Internal file name [OUTPUT/2322_Sunday_June_05_2022_02_59_06_AM_37579000/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 8, Linear differential equations of order n. Section 8.10, Chapter review. page
575
Problem number: Problem 22.
ODE order: 3.
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
[[_3rd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime \prime }+9 y^{\prime \prime }+24 y^{\prime }+16 y=8 \,{\mathrm e}^{-x}+1} \] 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 \[ y^{\prime \prime \prime }+9 y^{\prime \prime }+24 y^{\prime }+16 y = 0 \] The characteristic equation is \[ \lambda ^{3}+9 \lambda ^{2}+24 \lambda +16 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= -4\\ \lambda _3 &= -4 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{-4 x}+x \,{\mathrm e}^{-4 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= {\mathrm e}^{-4 x} \\ y_3 &= x \,{\mathrm e}^{-4 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+9 y^{\prime \prime }+24 y^{\prime }+16 y = 8 \,{\mathrm e}^{-x}+1 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 8 \,{\mathrm e}^{-x}+1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1\}, \{{\mathrm e}^{-x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{-4 x}, {\mathrm e}^{-4 x}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{1\}, \{x \,{\mathrm e}^{-x}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{1}+A_{2} x \,{\mathrm e}^{-x} \] The unknowns \(\{A_{1}, A_{2}\}\) 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 \[ 9 A_{2} {\mathrm e}^{-x}+16 A_{1} = 8 \,{\mathrm e}^{-x}+1 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{16}}, A_{2} = {\frac {8}{9}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {1}{16}+\frac {8 x \,{\mathrm e}^{-x}}{9} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{-4 x}+x \,{\mathrm e}^{-4 x} c_{3}\right ) + \left (\frac {1}{16}+\frac {8 x \,{\mathrm e}^{-x}}{9}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{3} x +c_{2} \right ) {\mathrm e}^{-4 x}+c_{1} {\mathrm e}^{-x}+\frac {1}{16}+\frac {8 x \,{\mathrm e}^{-x}}{9} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{3} x +c_{2} \right ) {\mathrm e}^{-4 x}+c_{1} {\mathrm e}^{-x}+\frac {1}{16}+\frac {8 x \,{\mathrm e}^{-x}}{9} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{3} x +c_{2} \right ) {\mathrm e}^{-4 x}+c_{1} {\mathrm e}^{-x}+\frac {1}{16}+\frac {8 x \,{\mathrm e}^{-x}}{9} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+9 y^{\prime \prime }+24 y^{\prime }+16 y=8 \,{\mathrm e}^{-x}+1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\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 {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=8 \,{\mathrm e}^{-x}+1-9 y_{3}\left (x \right )-24 y_{2}\left (x \right )-16 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_{3}^{\prime }\left (x \right )=8 \,{\mathrm e}^{-x}+1-9 y_{3}\left (x \right )-24 y_{2}\left (x \right )-16 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 ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -16 & -24 & -9 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 8 \,{\mathrm e}^{-x}+1 \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 8 \,{\mathrm e}^{-x}+1 \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -16 & -24 & -9 \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 [-4, \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]\right ], \left [-4, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-4, \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -4 \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-4 x}\cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{4} \\ 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 =-4\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\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 =-4 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\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}}_{2}\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} -4 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -16 & -24 & -9 \end {array}\right ]-\left (-4\right )\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{64} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -4 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-4 x}\cdot \left (x \cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]+\left [\begin {array}{c} \frac {1}{64} \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\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}{ccc} \frac {{\mathrm e}^{-4 x}}{16} & {\mathrm e}^{-4 x} \left (\frac {x}{16}+\frac {1}{64}\right ) & {\mathrm e}^{-x} \\ -\frac {{\mathrm e}^{-4 x}}{4} & -\frac {x \,{\mathrm e}^{-4 x}}{4} & -{\mathrm e}^{-x} \\ {\mathrm e}^{-4 x} & x \,{\mathrm e}^{-4 x} & {\mathrm e}^{-x} \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}{ccc} \frac {{\mathrm e}^{-4 x}}{16} & {\mathrm e}^{-4 x} \left (\frac {x}{16}+\frac {1}{64}\right ) & {\mathrm e}^{-x} \\ -\frac {{\mathrm e}^{-4 x}}{4} & -\frac {x \,{\mathrm e}^{-4 x}}{4} & -{\mathrm e}^{-x} \\ {\mathrm e}^{-4 x} & x \,{\mathrm e}^{-4 x} & {\mathrm e}^{-x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{16} & \frac {1}{64} & 1 \\ -\frac {1}{4} & 0 & -1 \\ 1 & 0 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-4 x} \left (4 x +1\right ) & \frac {4 \,{\mathrm e}^{-4 x}}{3}+5 x \,{\mathrm e}^{-4 x}-\frac {4 \,{\mathrm e}^{-x}}{3} & \frac {{\mathrm e}^{-4 x}}{3}+x \,{\mathrm e}^{-4 x}-\frac {{\mathrm e}^{-x}}{3} \\ -16 x \,{\mathrm e}^{-4 x} & -\frac {{\mathrm e}^{-4 x}}{3}-20 x \,{\mathrm e}^{-4 x}+\frac {4 \,{\mathrm e}^{-x}}{3} & -\frac {{\mathrm e}^{-4 x}}{3}-4 x \,{\mathrm e}^{-4 x}+\frac {{\mathrm e}^{-x}}{3} \\ 64 x \,{\mathrm e}^{-4 x} & \frac {4 \,{\mathrm e}^{-4 x}}{3}+80 x \,{\mathrm e}^{-4 x}-\frac {4 \,{\mathrm e}^{-x}}{3} & \frac {4 \,{\mathrm e}^{-4 x}}{3}+16 x \,{\mathrm e}^{-4 x}-\frac {{\mathrm e}^{-x}}{3} \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 {3}{16}+\frac {\left (-420 x -277\right ) {\mathrm e}^{-4 x}}{144}+\frac {\left (19-24 x \right ) {\mathrm e}^{-x}}{9} \\ \frac {\left (105 x +43\right ) {\mathrm e}^{-4 x}}{9}+\frac {{\mathrm e}^{-x} \left (24 x -43\right )}{9} \\ 1+\frac {4 \left (-105 x -43\right ) {\mathrm e}^{-4 x}}{9}+\frac {\left (-24 x +163\right ) {\mathrm e}^{-x}}{9} \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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} -\frac {3}{16}+\frac {\left (-420 x -277\right ) {\mathrm e}^{-4 x}}{144}+\frac {\left (19-24 x \right ) {\mathrm e}^{-x}}{9} \\ \frac {\left (105 x +43\right ) {\mathrm e}^{-4 x}}{9}+\frac {{\mathrm e}^{-x} \left (24 x -43\right )}{9} \\ 1+\frac {4 \left (-105 x -43\right ) {\mathrm e}^{-4 x}}{9}+\frac {\left (-24 x +163\right ) {\mathrm e}^{-x}}{9} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (36 c_{2} -1680\right ) x +36 c_{1} +9 c_{2} -1108\right ) {\mathrm e}^{-4 x}}{576}-\frac {3}{16}+\frac {\left (-1536 x +576 c_{3} +1216\right ) {\mathrm e}^{-x}}{576} \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 3; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 30
dsolve(diff(y(x),x$3)+9*diff(y(x),x$2)+24*diff(y(x),x)+16*y(x)=8*exp(-x)+1,y(x), singsol=all)
\[ y \left (x \right ) = \frac {1}{16}+\frac {\left (-16+24 x +27 c_{2} \right ) {\mathrm e}^{-x}}{27}+\left (c_{3} x +c_{1} \right ) {\mathrm e}^{-4 x} \]
✓ Solution by Mathematica
Time used: 0.076 (sec). Leaf size: 39
DSolve[y'''[x]+9*y''[x]+24*y'[x]+16*y[x]==8*Exp[-x]+1,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-4 x} (c_2 x+c_1)+e^{-x} \left (\frac {8 x}{9}-\frac {16}{27}+c_3\right )+\frac {1}{16} \]