Internal problem ID [1500]
Internal file name [OUTPUT/1501_Sunday_June_05_2022_02_19_48_AM_67624774/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 3.
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]]
\[ \boxed {4 y^{\prime \prime \prime }+8 y^{\prime \prime }-y^{\prime }-2 y=-{\mathrm e}^{x} \left (6 x^{2}+45 x +4\right )} \] 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 \[ 4 y^{\prime \prime \prime }+8 y^{\prime \prime }-y^{\prime }-2 y = 0 \] The characteristic equation is \[ 4 \lambda ^{3}+8 \lambda ^{2}-\lambda -2 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -{\frac {1}{2}}\\ \lambda _2 &= {\frac {1}{2}}\\ \lambda _3 &= -2 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 x} \\ y_2 &= {\mathrm e}^{-\frac {x}{2}} \\ y_3 &= {\mathrm e}^{\frac {x}{2}} \\ \end{align*} Now the particular solution to the given ODE is found \[ 4 y^{\prime \prime \prime }+8 y^{\prime \prime }-y^{\prime }-2 y = -{\mathrm e}^{x} \left (6 x^{2}+45 x +4\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ -{\mathrm e}^{x} \left (6 x^{2}+45 x +4\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{-2 x}, {\mathrm e}^{-\frac {x}{2}}, {\mathrm e}^{\frac {x}{2}}\right \} \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{x}+A_{2} x^{2} {\mathrm e}^{x}+A_{3} {\mathrm e}^{x} \] The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) 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 \[ 27 A_{1} {\mathrm e}^{x}+9 A_{1} x \,{\mathrm e}^{x}+40 A_{2} {\mathrm e}^{x}+54 A_{2} x \,{\mathrm e}^{x}+9 A_{2} x^{2} {\mathrm e}^{x}+9 A_{3} {\mathrm e}^{x} = -{\mathrm e}^{x} \left (6 x^{2}+45 x +4\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -1, A_{2} = -{\frac {2}{3}}, A_{3} = {\frac {149}{27}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -x \,{\mathrm e}^{x}-\frac {2 x^{2} {\mathrm e}^{x}}{3}+\frac {149 \,{\mathrm e}^{x}}{27} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}} c_{3}\right ) + \left (-x \,{\mathrm e}^{x}-\frac {2 x^{2} {\mathrm e}^{x}}{3}+\frac {149 \,{\mathrm e}^{x}}{27}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}} c_{3} -x \,{\mathrm e}^{x}-\frac {2 x^{2} {\mathrm e}^{x}}{3}+\frac {149 \,{\mathrm e}^{x}}{27} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}} c_{3} -x \,{\mathrm e}^{x}-\frac {2 x^{2} {\mathrm e}^{x}}{3}+\frac {149 \,{\mathrm e}^{x}}{27} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime \prime }+8 y^{\prime \prime }-y^{\prime }-2 y=-{\mathrm e}^{x} \left (6 x^{2}+45 x +4\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 }=\frac {y}{2}-2 y^{\prime \prime }+\frac {y^{\prime }}{4}-\frac {3 x^{2} {\mathrm e}^{x}}{2}-\frac {45 x \,{\mathrm e}^{x}}{4}-{\mathrm e}^{x} \\ \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 }+2 y^{\prime \prime }-\frac {y^{\prime }}{4}-\frac {y}{2}=-\frac {{\mathrm e}^{x} \left (6 x^{2}+45 x +4\right )}{4} \\ \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 )=-\frac {3 x^{2} {\mathrm e}^{x}}{2}-\frac {45 x \,{\mathrm e}^{x}}{4}-{\mathrm e}^{x}-2 y_{3}\left (x \right )+\frac {y_{2}\left (x \right )}{4}+\frac {y_{1}\left (x \right )}{2} \\ & {} & \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 )=-\frac {3 x^{2} {\mathrm e}^{x}}{2}-\frac {45 x \,{\mathrm e}^{x}}{4}-{\mathrm e}^{x}-2 y_{3}\left (x \right )+\frac {y_{2}\left (x \right )}{4}+\frac {y_{1}\left (x \right )}{2}\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 \\ \frac {1}{2} & \frac {1}{4} & -2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ -\frac {3 x^{2} {\mathrm e}^{x}}{2}-\frac {45 x \,{\mathrm e}^{x}}{4}-{\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 \\ -\frac {3 x^{2} {\mathrm e}^{x}}{2}-\frac {45 x \,{\mathrm e}^{x}}{4}-{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {1}{2} & \frac {1}{4} & -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 )+{\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}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} \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}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ 2 \\ 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}+c_{2} {\moverset {\rightarrow }{y}}_{2}+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}^{-2 x}}{4} & 4 \,{\mathrm e}^{-\frac {x}{2}} & 4 \,{\mathrm e}^{\frac {x}{2}} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -2 \,{\mathrm e}^{-\frac {x}{2}} & 2 \,{\mathrm e}^{\frac {x}{2}} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{\frac {x}{2}} \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}^{-2 x}}{4} & 4 \,{\mathrm e}^{-\frac {x}{2}} & 4 \,{\mathrm e}^{\frac {x}{2}} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -2 \,{\mathrm e}^{-\frac {x}{2}} & 2 \,{\mathrm e}^{\frac {x}{2}} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{\frac {x}{2}} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{4} & 4 & 4 \\ -\frac {1}{2} & -2 & 2 \\ 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {\left (6 \,{\mathrm e}^{\frac {5 x}{2}}+10 \,{\mathrm e}^{\frac {3 x}{2}}-1\right ) {\mathrm e}^{-2 x}}{15} & -{\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}} & \frac {2 \left (3 \,{\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}+2\right ) {\mathrm e}^{-2 x}}{15} \\ \frac {\left (3 \,{\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}+2\right ) {\mathrm e}^{-2 x}}{15} & \frac {{\mathrm e}^{-\frac {x}{2}}}{2}+\frac {{\mathrm e}^{\frac {x}{2}}}{2} & \frac {\left (3 \,{\mathrm e}^{\frac {5 x}{2}}+5 \,{\mathrm e}^{\frac {3 x}{2}}-8\right ) {\mathrm e}^{-2 x}}{15} \\ \frac {\left (3 \,{\mathrm e}^{\frac {5 x}{2}}+5 \,{\mathrm e}^{\frac {3 x}{2}}-8\right ) {\mathrm e}^{-2 x}}{30} & -\frac {{\mathrm e}^{-\frac {x}{2}}}{4}+\frac {{\mathrm e}^{\frac {x}{2}}}{4} & \frac {\left (3 \,{\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}+32\right ) {\mathrm e}^{-2 x}}{30} \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 {\left (90 x^{2} {\mathrm e}^{3 x}+135 x \,{\mathrm e}^{3 x}-745 \,{\mathrm e}^{3 x}+1026 \,{\mathrm e}^{\frac {5 x}{2}}-310 \,{\mathrm e}^{\frac {3 x}{2}}+29\right ) {\mathrm e}^{-2 x}}{135} \\ -\frac {\left (90 x^{2} {\mathrm e}^{3 x}+315 x \,{\mathrm e}^{3 x}-610 \,{\mathrm e}^{3 x}+513 \,{\mathrm e}^{\frac {5 x}{2}}+155 \,{\mathrm e}^{\frac {3 x}{2}}-58\right ) {\mathrm e}^{-2 x}}{135} \\ -\frac {\left (180 x^{2} {\mathrm e}^{3 x}+990 x \,{\mathrm e}^{3 x}-590 \,{\mathrm e}^{3 x}+513 \,{\mathrm e}^{\frac {5 x}{2}}-155 \,{\mathrm e}^{\frac {3 x}{2}}+232\right ) {\mathrm e}^{-2 x}}{270} \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 [\begin {array}{c} -\frac {\left (90 x^{2} {\mathrm e}^{3 x}+135 x \,{\mathrm e}^{3 x}-745 \,{\mathrm e}^{3 x}+1026 \,{\mathrm e}^{\frac {5 x}{2}}-310 \,{\mathrm e}^{\frac {3 x}{2}}+29\right ) {\mathrm e}^{-2 x}}{135} \\ -\frac {\left (90 x^{2} {\mathrm e}^{3 x}+315 x \,{\mathrm e}^{3 x}-610 \,{\mathrm e}^{3 x}+513 \,{\mathrm e}^{\frac {5 x}{2}}+155 \,{\mathrm e}^{\frac {3 x}{2}}-58\right ) {\mathrm e}^{-2 x}}{135} \\ -\frac {\left (180 x^{2} {\mathrm e}^{3 x}+990 x \,{\mathrm e}^{3 x}-590 \,{\mathrm e}^{3 x}+513 \,{\mathrm e}^{\frac {5 x}{2}}-155 \,{\mathrm e}^{\frac {3 x}{2}}+232\right ) {\mathrm e}^{-2 x}}{270} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {2 \left (\left (-6 c_{2} -\frac {31}{9}\right ) {\mathrm e}^{\frac {3 x}{2}}+\left (-6 c_{3} +\frac {57}{5}\right ) {\mathrm e}^{\frac {5 x}{2}}+\left (x^{2}+\frac {3}{2} x -\frac {149}{18}\right ) {\mathrm e}^{3 x}-\frac {3 c_{1}}{8}+\frac {29}{90}\right ) {\mathrm e}^{-2 x}}{3} \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: 50
dsolve(4*diff(y(x),x$3)+8*diff(y(x),x$2)-diff(y(x),x)-2*y(x)=-exp(x)*(4+45*x+6*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-18 x^{2} {\mathrm e}^{3 x}-27 x \,{\mathrm e}^{3 x}+149 \,{\mathrm e}^{3 x}+27 c_{3} {\mathrm e}^{\frac {5 x}{2}}+27 c_{2} {\mathrm e}^{\frac {3 x}{2}}+27 c_{1} \right ) {\mathrm e}^{-2 x}}{27} \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 52
DSolve[4*y'''[x]+8*y''[x]-y'[x]-2*y[x]==-Exp[x]*(4+45*x+6*x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x \left (-\frac {2 x^2}{3}-x+\frac {149}{27}\right )+c_1 e^{-x/2}+c_2 e^{x/2}+c_3 e^{-2 x} \]