11.10 problem 10

Internal problem ID [11783]
Internal file name [OUTPUT/11793_Thursday_April_11_2024_08_49_43_PM_4198183/index.tex]

Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi. 2004.
Section: Chapter 4, Section 4.3. The method of undetermined coefficients. Exercises page 151
Problem number: 10.
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]]

\[ \boxed {y^{\prime \prime \prime }+2 y^{\prime \prime }-3 y^{\prime }-10 y=8 x \,{\mathrm e}^{-2 x}} \] 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 \[ y^{\prime \prime \prime }+2 y^{\prime \prime }-3 y^{\prime }-10 y = 0 \] The characteristic equation is \[ \lambda ^{3}+2 \lambda ^{2}-3 \lambda -10 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2-i\\ \lambda _3 &= -2+i \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-2-i\right ) x} c_{2} +{\mathrm e}^{\left (-2+i\right ) x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{2 x} \\ y_2 &= {\mathrm e}^{\left (-2-i\right ) x} \\ y_3 &= {\mathrm e}^{\left (-2+i\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+2 y^{\prime \prime }-3 y^{\prime }-10 y = 8 x \,{\mathrm e}^{-2 x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 8 x \,{\mathrm e}^{-2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{-2 x}, {\mathrm e}^{-2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{\left (-2-i\right ) x}, {\mathrm e}^{\left (-2+i\right ) x}, {\mathrm e}^{2 x}\} \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{-2 x}+A_{2} {\mathrm e}^{-2 x} \] 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 \[ A_{1} {\mathrm e}^{-2 x}-4 A_{1} x \,{\mathrm e}^{-2 x}-4 A_{2} {\mathrm e}^{-2 x} = 8 x \,{\mathrm e}^{-2 x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -2, A_{2} = -{\frac {1}{2}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -2 x \,{\mathrm e}^{-2 x}-\frac {{\mathrm e}^{-2 x}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-2-i\right ) x} c_{2} +{\mathrm e}^{\left (-2+i\right ) x} c_{3}\right ) + \left (-2 x \,{\mathrm e}^{-2 x}-\frac {{\mathrm e}^{-2 x}}{2}\right ) \\ \end{align*}

Summary

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

Verification of solutions

\[ y = c_{1} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-2-i\right ) x} c_{2} +{\mathrm e}^{\left (-2+i\right ) x} c_{3} -2 x \,{\mathrm e}^{-2 x}-\frac {{\mathrm e}^{-2 x}}{2} \] Verified OK.

11.10.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+2 y^{\prime \prime }-3 y^{\prime }-10 y=8 x \,{\mathrm e}^{-2 x} \\ \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 x \,{\mathrm e}^{-2 x}-2 y_{3}\left (x \right )+3 y_{2}\left (x \right )+10 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 x \,{\mathrm e}^{-2 x}-2 y_{3}\left (x \right )+3 y_{2}\left (x \right )+10 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 \\ 10 & 3 & -2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 8 x \,{\mathrm e}^{-2 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 8 x \,{\mathrm e}^{-2 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 10 & 3 & -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 [-2-\mathrm {I}, \left [\begin {array}{c} \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}+\frac {\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ], \left [-2+\mathrm {I}, \left [\begin {array}{c} \frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}-\frac {\mathrm {I}}{5} \\ 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 complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2-\mathrm {I}, \left [\begin {array}{c} \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}+\frac {\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-2-\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}+\frac {\mathrm {I}}{5} \\ 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 \\ {} & {} & {\mathrm e}^{-2 x}\cdot \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}+\frac {\mathrm {I}}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \left (\frac {3}{25}-\frac {4 \,\mathrm {I}}{25}\right ) \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \left (-\frac {2}{5}+\frac {\mathrm {I}}{5}\right ) \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \cos \left (x \right )-\mathrm {I} \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {3 \cos \left (x \right )}{25}-\frac {4 \sin \left (x \right )}{25} \\ -\frac {2 \cos \left (x \right )}{5}+\frac {\sin \left (x \right )}{5} \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} -\frac {3 \sin \left (x \right )}{25}-\frac {4 \cos \left (x \right )}{25} \\ \frac {2 \sin \left (x \right )}{5}+\frac {\cos \left (x \right )}{5} \\ -\sin \left (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}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\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}{ccc} \frac {{\mathrm e}^{2 x}}{4} & {\mathrm e}^{-2 x} \left (\frac {3 \cos \left (x \right )}{25}-\frac {4 \sin \left (x \right )}{25}\right ) & {\mathrm e}^{-2 x} \left (-\frac {3 \sin \left (x \right )}{25}-\frac {4 \cos \left (x \right )}{25}\right ) \\ \frac {{\mathrm e}^{2 x}}{2} & {\mathrm e}^{-2 x} \left (-\frac {2 \cos \left (x \right )}{5}+\frac {\sin \left (x \right )}{5}\right ) & {\mathrm e}^{-2 x} \left (\frac {2 \sin \left (x \right )}{5}+\frac {\cos \left (x \right )}{5}\right ) \\ {\mathrm e}^{2 x} & \cos \left (x \right ) {\mathrm e}^{-2 x} & -{\mathrm e}^{-2 x} \sin \left (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}{ccc} \frac {{\mathrm e}^{2 x}}{4} & {\mathrm e}^{-2 x} \left (\frac {3 \cos \left (x \right )}{25}-\frac {4 \sin \left (x \right )}{25}\right ) & {\mathrm e}^{-2 x} \left (-\frac {3 \sin \left (x \right )}{25}-\frac {4 \cos \left (x \right )}{25}\right ) \\ \frac {{\mathrm e}^{2 x}}{2} & {\mathrm e}^{-2 x} \left (-\frac {2 \cos \left (x \right )}{5}+\frac {\sin \left (x \right )}{5}\right ) & {\mathrm e}^{-2 x} \left (\frac {2 \sin \left (x \right )}{5}+\frac {\cos \left (x \right )}{5}\right ) \\ {\mathrm e}^{2 x} & \cos \left (x \right ) {\mathrm e}^{-2 x} & -{\mathrm e}^{-2 x} \sin \left (x \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{4} & \frac {3}{25} & -\frac {4}{25} \\ \frac {1}{2} & -\frac {2}{5} & \frac {1}{5} \\ 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}{ccc} \frac {2 \left (6 \cos \left (x \right )+7 \sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {5 \,{\mathrm e}^{2 x}}{17} & \frac {\left (-4 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {4 \,{\mathrm e}^{2 x}}{17} & \frac {\left (-\cos \left (x \right )-4 \sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {{\mathrm e}^{2 x}}{17} \\ \frac {10 \left (-\cos \left (x \right )-4 \sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {10 \,{\mathrm e}^{2 x}}{17} & \frac {\left (9 \cos \left (x \right )+2 \sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {8 \,{\mathrm e}^{2 x}}{17} & \frac {\left (-2 \cos \left (x \right )+9 \sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {2 \,{\mathrm e}^{2 x}}{17} \\ \frac {10 \left (-2 \cos \left (x \right )+9 \sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {20 \,{\mathrm e}^{2 x}}{17} & \frac {\left (-16 \cos \left (x \right )-13 \sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {16 \,{\mathrm e}^{2 x}}{17} & \frac {\left (13 \cos \left (x \right )-16 \sin \left (x \right )\right ) {\mathrm e}^{-2 x}}{17}+\frac {4 \,{\mathrm e}^{2 x}}{17} \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 ({\mathrm e}^{4 x}+16 \cos \left (x \right )+64 \sin \left (x \right )-68 x -17\right )}{34} \\ \frac {{\mathrm e}^{-2 x} \left ({\mathrm e}^{4 x}+16 \cos \left (x \right )-72 \sin \left (x \right )+68 x -17\right )}{17} \\ \frac {2 \,{\mathrm e}^{-2 x} \left ({\mathrm e}^{4 x}-52 \cos \left (x \right )+64 \sin \left (x \right )-68 x +51\right )}{17} \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}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+\left [\begin {array}{c} \frac {{\mathrm e}^{-2 x} \left ({\mathrm e}^{4 x}+16 \cos \left (x \right )+64 \sin \left (x \right )-68 x -17\right )}{34} \\ \frac {{\mathrm e}^{-2 x} \left ({\mathrm e}^{4 x}+16 \cos \left (x \right )-72 \sin \left (x \right )+68 x -17\right )}{17} \\ \frac {2 \,{\mathrm e}^{-2 x} \left ({\mathrm e}^{4 x}-52 \cos \left (x \right )+64 \sin \left (x \right )-68 x +51\right )}{17} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (204 c_{2} -272 c_{3} +800\right ) \cos \left (x \right )+\left (-272 c_{2} -204 c_{3} +3200\right ) \sin \left (x \right )-3400 x -850\right ) {\mathrm e}^{-2 x}}{1700}+\frac {\left (425 c_{1} +50\right ) {\mathrm e}^{2 x}}{1700} \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: 32

dsolve(diff(y(x),x$3)+2*diff(y(x),x$2)-3*diff(y(x),x)-10*y(x)=8*x*exp(-2*x),y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.007 (sec). Leaf size: 40

DSolve[y'''[x]+2*y''[x]-3*y'[x]-10*y[x]==8*x*Exp[-2*x],y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{2} e^{-2 x} \left (-4 x+2 c_3 e^{4 x}+2 c_2 \cos (x)+2 c_1 \sin (x)-1\right ) \]