Internal problem ID [1550]
Internal file name [OUTPUT/1551_Sunday_June_05_2022_02_22_00_AM_73103628/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 9.3. Undetermined
Coefficients for Higher Order Equations. Page 495
Problem number: section 9.3, problem 53.
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, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime }+y^{\prime \prime }-y^{\prime }-y=4 \,{\mathrm e}^{-x} \left (1-6 x \right )-2 x \cos \left (x \right )+2 \left (x +1\right ) \sin \left (x \right )} \] 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 \[ y^{\prime \prime \prime }+y^{\prime \prime }-y^{\prime }-y = 0 \] The characteristic equation is \[ \lambda ^{3}+\lambda ^{2}-\lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -1\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+x \,{\mathrm e}^{-x} c_{2} +{\mathrm e}^{x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= x \,{\mathrm e}^{-x} \\ y_3 &= {\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+y^{\prime \prime }-y^{\prime }-y = 4 \,{\mathrm e}^{-x} \left (1-6 x \right )-2 x \cos \left (x \right )+2 \left (x +1\right ) \sin \left (x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 4 \,{\mathrm e}^{-x} \left (1-6 x \right )-2 x \cos \left (x \right )+2 \left (x +1\right ) \sin \left (x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{-x}, {\mathrm e}^{-x}\}, \{x \cos \left (x \right ), \sin \left (x \right ) x, \cos \left (x \right ), \sin \left (x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{-x}, {\mathrm e}^{x}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{-x}, x^{2} {\mathrm e}^{-x}\}, \{x \cos \left (x \right ), \sin \left (x \right ) x, \cos \left (x \right ), \sin \left (x \right )\}] \] Since \(x \,{\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-x}, {\mathrm e}^{-x} x^{3}\}, \{x \cos \left (x \right ), \sin \left (x \right ) x, \cos \left (x \right ), \sin \left (x \right )\}] \] 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. \[ y_p = A_{1} x^{2} {\mathrm e}^{-x}+A_{2} {\mathrm e}^{-x} x^{3}+A_{3} x \cos \left (x \right )+A_{4} \sin \left (x \right ) x +A_{5} \cos \left (x \right )+A_{6} \sin \left (x \right ) \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) 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 \[ -12 A_{2} {\mathrm e}^{-x} x -2 A_{5} \cos \left (x \right )-2 A_{6} \sin \left (x \right )+2 A_{5} \sin \left (x \right )-2 A_{6} \cos \left (x \right )-2 A_{3} \sin \left (x \right )+2 A_{4} \cos \left (x \right )+6 A_{2} {\mathrm e}^{-x}-2 A_{3} x \cos \left (x \right )-2 A_{4} \sin \left (x \right ) x -4 A_{3} \cos \left (x \right )+2 A_{3} x \sin \left (x \right )-2 A_{4} \cos \left (x \right ) x -4 A_{4} \sin \left (x \right )-4 A_{1} {\mathrm e}^{-x} = 4 \,{\mathrm e}^{-x} \left (1-6 x \right )-2 x \cos \left (x \right )+2 \left (x +1\right ) \sin \left (x \right ) \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 2, A_{2} = 2, A_{3} = 1, A_{4} = 0, A_{5} = 0, A_{6} = -2] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = 2 x^{2} {\mathrm e}^{-x}+2 \,{\mathrm e}^{-x} x^{3}+x \cos \left (x \right )-2 \sin \left (x \right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+x \,{\mathrm e}^{-x} c_{2} +{\mathrm e}^{x} c_{3}\right ) + \left (2 x^{2} {\mathrm e}^{-x}+2 \,{\mathrm e}^{-x} x^{3}+x \cos \left (x \right )-2 \sin \left (x \right )\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{x} c_{3} +2 x^{2} {\mathrm e}^{-x}+2 \,{\mathrm e}^{-x} x^{3}+x \cos \left (x \right )-2 \sin \left (x \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{x} c_{3} +2 x^{2} {\mathrm e}^{-x}+2 \,{\mathrm e}^{-x} x^{3}+x \cos \left (x \right )-2 \sin \left (x \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{x} c_{3} +2 x^{2} {\mathrm e}^{-x}+2 \,{\mathrm e}^{-x} x^{3}+x \cos \left (x \right )-2 \sin \left (x \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+y^{\prime \prime }-y^{\prime }-y=4 \,{\mathrm e}^{-x} \left (1-6 x \right )-2 x \cos \left (x \right )+2 \left (x +1\right ) \sin \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=y-24 x \,{\mathrm e}^{-x}+2 \sin \left (x \right ) x -2 x \cos \left (x \right )+4 \,{\mathrm e}^{-x}+2 \sin \left (x \right )-y^{\prime \prime }+y^{\prime } \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+y^{\prime \prime }-y^{\prime }-y=-24 x \,{\mathrm e}^{-x}-2 x \cos \left (x \right )+2 \sin \left (x \right ) x +4 \,{\mathrm e}^{-x}+2 \sin \left (x \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 {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=-24 x \,{\mathrm e}^{-x}+2 \sin \left (x \right ) x -2 x \cos \left (x \right )+4 \,{\mathrm e}^{-x}+2 \sin \left (x \right )-y_{3}\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_{3}^{\prime }\left (x \right )=-24 x \,{\mathrm e}^{-x}+2 \sin \left (x \right ) x -2 x \cos \left (x \right )+4 \,{\mathrm e}^{-x}+2 \sin \left (x \right )-y_{3}\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 ) \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 \\ 1 & 1 & -1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ -24 x \,{\mathrm e}^{-x}-2 x \cos \left (x \right )+2 \sin \left (x \right ) x +4 \,{\mathrm e}^{-x}+2 \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ -24 x \,{\mathrm e}^{-x}-2 x \cos \left (x \right )+2 \sin \left (x \right ) x +4 \,{\mathrm e}^{-x}+2 \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & -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 )+{\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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [-1, \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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 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}}_{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 =-1 \\ \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} -1 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & -1 \end {array}\right ]-\left (-1\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} 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 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-x}\cdot \left (x \cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} 1 \\ 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} {\mathrm e}^{-x} & \left (x +1\right ) {\mathrm e}^{-x} & {\mathrm e}^{x} \\ -{\mathrm e}^{-x} & -x \,{\mathrm e}^{-x} & {\mathrm e}^{x} \\ {\mathrm e}^{-x} & x \,{\mathrm e}^{-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} {\mathrm e}^{-x} & \left (x +1\right ) {\mathrm e}^{-x} & {\mathrm e}^{x} \\ -{\mathrm e}^{-x} & -x \,{\mathrm e}^{-x} & {\mathrm e}^{x} \\ {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & {\mathrm e}^{x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & 1 & 1 \\ -1 & 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} \left (x +1\right ) {\mathrm e}^{-x} & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & -\frac {{\mathrm e}^{-x}}{2}-x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{x}}{2} \\ -x \,{\mathrm e}^{-x} & \frac {{\mathrm e}^{x}}{2}+\frac {{\mathrm e}^{-x}}{2} & -\frac {{\mathrm e}^{-x}}{2}+x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{x}}{2} \\ x \,{\mathrm e}^{-x} & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & \frac {{\mathrm e}^{-x}}{2}-x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{x}}{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} \left (4 x^{3}+4 x^{2}+4 x +1\right ) {\mathrm e}^{-x}+2 x \cos \left (x \right )-{\mathrm e}^{x}-4 \sin \left (x \right ) \\ \left (-4 x^{3}+8 x^{2}+4 x +3\right ) {\mathrm e}^{-x}-2 \sin \left (x \right ) x -2 \cos \left (x \right )-{\mathrm e}^{x} \\ \left (4 x^{3}-8 x^{2}+8 x +1\right ) {\mathrm e}^{-x}-{\mathrm e}^{x}-2 \sin \left (x \right ) \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} \left (4 x^{3}+4 x^{2}+4 x +1\right ) {\mathrm e}^{-x}+2 x \cos \left (x \right )-{\mathrm e}^{x}-4 \sin \left (x \right ) \\ \left (-4 x^{3}+8 x^{2}+4 x +3\right ) {\mathrm e}^{-x}-2 \sin \left (x \right ) x -2 \cos \left (x \right )-{\mathrm e}^{x} \\ \left (4 x^{3}-8 x^{2}+8 x +1\right ) {\mathrm e}^{-x}-{\mathrm e}^{x}-2 \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\left (4 x^{3}+4 x^{2}+\left (c_{2} +4\right ) x +c_{1} +c_{2} +1\right ) {\mathrm e}^{-x}+\left (c_{3} -1\right ) {\mathrm e}^{x}+2 x \cos \left (x \right )-4 \sin \left (x \right ) \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: 0.016 (sec). Leaf size: 40
dsolve(1*diff(y(x),x$3)+1*diff(y(x),x$2)-1*diff(y(x),x)-1*y(x)=4*exp(-x)*(1-6*x)-2*x*cos(x)+2*(1+x)*sin(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (2 x^{3}+2 x^{2}+\left (c_{3} +2\right ) x +c_{2} +1\right ) {\mathrm e}^{-x}+x \cos \left (x \right )+{\mathrm e}^{x} c_{1} -2 \sin \left (x \right ) \]
✓ Solution by Mathematica
Time used: 0.627 (sec). Leaf size: 54
DSolve[1*y'''[x]+1*y''[x]-1*y'[x]-1*y[x]==4*Exp[-x]*(1-6*x)-2*x*Cos[x]+2*(1+x)*Sin[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x} \left (2 x^3+2 x^2+2 x-2 e^x \sin (x)+e^x x \cos (x)+c_2 x+c_3 e^{2 x}+1+c_1\right ) \]