Internal problem ID [9801]
Internal file name [OUTPUT/8744_Monday_June_06_2022_05_22_50_AM_79086207/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1475.
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]]
\[ \boxed {4 y^{\prime \prime \prime }-8 y^{\prime \prime }-11 y^{\prime }-3 y=-18 \,{\mathrm e}^{x}} \] 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 \[ 4 y^{\prime \prime \prime }-8 y^{\prime \prime }-11 y^{\prime }-3 y = 0 \] The characteristic equation is \[ 4 \lambda ^{3}-8 \lambda ^{2}-11 \lambda -3 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 3\\ \lambda _2 &= -{\frac {1}{2}}\\ \lambda _3 &= -{\frac {1}{2}} \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{3 x}+c_{2} {\mathrm e}^{-\frac {x}{2}}+x \,{\mathrm e}^{-\frac {x}{2}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{3 x} \\ y_2 &= {\mathrm e}^{-\frac {x}{2}} \\ y_3 &= x \,{\mathrm e}^{-\frac {x}{2}} \\ \end{align*} Now the particular solution to the given ODE is found \[ 4 y^{\prime \prime \prime }-8 y^{\prime \prime }-11 y^{\prime }-3 y = -18 \,{\mathrm e}^{x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ -18 \,{\mathrm e}^{x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{x \,{\mathrm e}^{-\frac {x}{2}}, {\mathrm e}^{3 x}, {\mathrm e}^{-\frac {x}{2}}\right \} \] 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. \[ y_p = A_{1} {\mathrm e}^{x} \] 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 \[ -18 A_{1} {\mathrm e}^{x} = -18 \,{\mathrm e}^{x} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = {\mathrm e}^{x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{3 x}+c_{2} {\mathrm e}^{-\frac {x}{2}}+x \,{\mathrm e}^{-\frac {x}{2}} c_{3}\right ) + \left ({\mathrm e}^{x}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{3} x +c_{2} \right ) {\mathrm e}^{-\frac {x}{2}}+c_{1} {\mathrm e}^{3 x}+{\mathrm e}^{x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{3} x +c_{2} \right ) {\mathrm e}^{-\frac {x}{2}}+c_{1} {\mathrm e}^{3 x}+{\mathrm e}^{x} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{3} x +c_{2} \right ) {\mathrm e}^{-\frac {x}{2}}+c_{1} {\mathrm e}^{3 x}+{\mathrm e}^{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime \prime }-8 y^{\prime \prime }-11 y^{\prime }-3 y=-18 \,{\mathrm e}^{x} \\ \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 }=2 y^{\prime \prime }+\frac {11 y^{\prime }}{4}+\frac {3 y}{4}-\frac {9 \,{\mathrm e}^{x}}{2} \\ \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 {11 y^{\prime }}{4}-\frac {3 y}{4}=-\frac {9 \,{\mathrm e}^{x}}{2} \\ \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 {9 \,{\mathrm e}^{x}}{2}+2 y_{3}\left (x \right )+\frac {11 y_{2}\left (x \right )}{4}+\frac {3 y_{1}\left (x \right )}{4} \\ & {} & \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 {9 \,{\mathrm e}^{x}}{2}+2 y_{3}\left (x \right )+\frac {11 y_{2}\left (x \right )}{4}+\frac {3 y_{1}\left (x \right )}{4}\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 {3}{4} & \frac {11}{4} & 2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ -\frac {9 \,{\mathrm e}^{x}}{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ -\frac {9 \,{\mathrm e}^{x}}{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {3}{4} & \frac {11}{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 [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -\frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ -2 \\ 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 =-\frac {1}{2}\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 =-\frac {1}{2} \\ \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} -\frac {1}{2} \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {3}{4} & \frac {11}{4} & 2 \end {array}\right ]--\frac {1}{2}\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} 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 8 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -\frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left (x \cdot \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} 8 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 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} 4 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} \left (4 x +8\right ) & \frac {{\mathrm e}^{3 x}}{9} \\ -2 \,{\mathrm e}^{-\frac {x}{2}} & -2 x \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-\frac {x}{2}} & x \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{3 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} 4 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} \left (4 x +8\right ) & \frac {{\mathrm e}^{3 x}}{9} \\ -2 \,{\mathrm e}^{-\frac {x}{2}} & -2 x \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-\frac {x}{2}} & x \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{3 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 4 & 8 & \frac {1}{9} \\ -2 & 0 & \frac {1}{3} \\ 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} \frac {\left (x +2\right ) {\mathrm e}^{-\frac {x}{2}}}{2} & \frac {\left (35 x -2\right ) {\mathrm e}^{-\frac {x}{2}}}{42}+\frac {{\mathrm e}^{3 x}}{21} & \frac {\left (-7 x -2\right ) {\mathrm e}^{-\frac {x}{2}}}{21}+\frac {2 \,{\mathrm e}^{3 x}}{21} \\ -\frac {x \,{\mathrm e}^{-\frac {x}{2}}}{4} & \frac {6 \,{\mathrm e}^{-\frac {x}{2}}}{7}-\frac {5 x \,{\mathrm e}^{-\frac {x}{2}}}{12}+\frac {{\mathrm e}^{3 x}}{7} & -\frac {2 \,{\mathrm e}^{-\frac {x}{2}}}{7}+\frac {x \,{\mathrm e}^{-\frac {x}{2}}}{6}+\frac {2 \,{\mathrm e}^{3 x}}{7} \\ \frac {x \,{\mathrm e}^{-\frac {x}{2}}}{8} & -\frac {3 \,{\mathrm e}^{-\frac {x}{2}}}{7}+\frac {5 x \,{\mathrm e}^{-\frac {x}{2}}}{24}+\frac {3 \,{\mathrm e}^{3 x}}{7} & \frac {\left (-7 x +12\right ) {\mathrm e}^{-\frac {x}{2}}}{84}+\frac {6 \,{\mathrm e}^{3 x}}{7} \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} -x \,{\mathrm e}^{-\frac {x}{2}}+\frac {7 \,{\mathrm e}^{x}}{6}-\frac {20 \,{\mathrm e}^{-\frac {x}{2}}}{21}-\frac {3 \,{\mathrm e}^{3 x}}{14} \\ \frac {\left (21 x -22\right ) {\mathrm e}^{-\frac {x}{2}}}{42}+\frac {7 \,{\mathrm e}^{x}}{6}-\frac {9 \,{\mathrm e}^{3 x}}{14} \\ \frac {\left (-21 x +22\right ) {\mathrm e}^{-\frac {x}{2}}}{84}+\frac {5 \,{\mathrm e}^{x}}{3}-\frac {27 \,{\mathrm e}^{3 x}}{14} \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} -x \,{\mathrm e}^{-\frac {x}{2}}+\frac {7 \,{\mathrm e}^{x}}{6}-\frac {20 \,{\mathrm e}^{-\frac {x}{2}}}{21}-\frac {3 \,{\mathrm e}^{3 x}}{14} \\ \frac {\left (21 x -22\right ) {\mathrm e}^{-\frac {x}{2}}}{42}+\frac {7 \,{\mathrm e}^{x}}{6}-\frac {9 \,{\mathrm e}^{3 x}}{14} \\ \frac {\left (-21 x +22\right ) {\mathrm e}^{-\frac {x}{2}}}{84}+\frac {5 \,{\mathrm e}^{x}}{3}-\frac {27 \,{\mathrm e}^{3 x}}{14} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (-20+84 c_{2} \left (x +2\right )-21 x +84 c_{1} \right ) {\mathrm e}^{-\frac {x}{2}}}{21}+\frac {\left (14 c_{3} -27\right ) {\mathrm e}^{3 x}}{126}+\frac {7 \,{\mathrm e}^{x}}{6} \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: 23
dsolve(4*diff(diff(diff(y(x),x),x),x)-8*diff(diff(y(x),x),x)-11*diff(y(x),x)-3*y(x)+18*exp(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (c_{3} x +c_{2} \right ) {\mathrm e}^{-\frac {x}{2}}+c_{1} {\mathrm e}^{3 x}+{\mathrm e}^{x} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 37
DSolve[18*E^x - 3*y[x] - 11*y'[x] - 8*y''[x] + 4*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x/2} \left (e^{3 x/2}+c_2 x+c_3 e^{7 x/2}+c_1\right ) \]