11.11 problem 11

11.11.1 Maple step by step solution

Internal problem ID [11784]
Internal file name [OUTPUT/11794_Thursday_April_11_2024_08_49_43_PM_74629381/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: 11.
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]]

\[ \boxed {y^{\prime \prime \prime }+y^{\prime \prime }+3 y^{\prime }-5 y=5 \sin \left (2 x \right )+10 x^{2}+3 x +7} \] 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 \[ y^{\prime \prime \prime }+y^{\prime \prime }+3 y^{\prime }-5 y = 0 \] The characteristic equation is \[ \lambda ^{3}+\lambda ^{2}+3 \lambda -5 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -1-2 i\\ \lambda _3 &= -1+2 i \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-1+2 i\right ) x} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{\left (-1+2 i\right ) x} \\ y_3 &= {\mathrm e}^{\left (-1-2 i\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+y^{\prime \prime }+3 y^{\prime }-5 y = 5 \sin \left (2 x \right )+10 x^{2}+3 x +7 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 5 \sin \left (2 x \right )+10 x^{2}+3 x +7 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (2 x \right ), \sin \left (2 x \right )\}, \{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{\left (-1-2 i\right ) x}, {\mathrm e}^{\left (-1+2 i\right ) x}\} \] 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. \[ y_p = A_{1} \cos \left (2 x \right )+A_{2} \sin \left (2 x \right )+A_{3}+A_{4} x +A_{5} x^{2} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\}\) 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 \[ 2 A_{1} \sin \left (2 x \right )-2 A_{2} \cos \left (2 x \right )-9 A_{1} \cos \left (2 x \right )-9 A_{2} \sin \left (2 x \right )+2 A_{5}+3 A_{4}+6 A_{5} x -5 A_{3}-5 A_{4} x -5 A_{5} x^{2} = 5 \sin \left (2 x \right )+10 x^{2}+3 x +7 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {2}{17}}, A_{2} = -{\frac {9}{17}}, A_{3} = -4, A_{4} = -3, A_{5} = -2\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {2 \cos \left (2 x \right )}{17}-\frac {9 \sin \left (2 x \right )}{17}-4-3 x -2 x^{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-1+2 i\right ) x} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) x} c_{3}\right ) + \left (\frac {2 \cos \left (2 x \right )}{17}-\frac {9 \sin \left (2 x \right )}{17}-4-3 x -2 x^{2}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-1+2 i\right ) x} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) x} c_{3} +\frac {2 \cos \left (2 x \right )}{17}-\frac {9 \sin \left (2 x \right )}{17}-4-3 x -2 x^{2} \\ \end{align*}

Verification of solutions

\[ y = c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-1+2 i\right ) x} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) x} c_{3} +\frac {2 \cos \left (2 x \right )}{17}-\frac {9 \sin \left (2 x \right )}{17}-4-3 x -2 x^{2} \] Verified OK.

11.11.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+y^{\prime \prime }+3 y^{\prime }-5 y=5 \sin \left (2 x \right )+10 x^{2}+3 x +7 \\ \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 )=10 x^{2}+5 \sin \left (2 x \right )-y_{3}\left (x \right )-3 y_{2}\left (x \right )+5 y_{1}\left (x \right )+3 x +7 \\ & {} & \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 )=10 x^{2}+5 \sin \left (2 x \right )-y_{3}\left (x \right )-3 y_{2}\left (x \right )+5 y_{1}\left (x \right )+3 x +7\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 \\ 5 & -3 & -1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 5 \sin \left (2 x \right )+10 x^{2}+3 x +7 \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 5 \sin \left (2 x \right )+10 x^{2}+3 x +7 \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & -3 & -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 [-1-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ], \left [-1+2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}-\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\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}}_{1}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-1-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-1-2 \,\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\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}^{-x}\cdot \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right )\cdot \left [\begin {array}{c} -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-x}\cdot \left [\begin {array}{c} \left (-\frac {3}{25}-\frac {4 \,\mathrm {I}}{25}\right ) \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \left (-\frac {1}{5}+\frac {2 \,\mathrm {I}}{5}\right ) \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \cos \left (2 x \right )-\mathrm {I} \sin \left (2 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}^{-x}\cdot \left [\begin {array}{c} -\frac {3 \cos \left (2 x \right )}{25}-\frac {4 \sin \left (2 x \right )}{25} \\ -\frac {\cos \left (2 x \right )}{5}+\frac {2 \sin \left (2 x \right )}{5} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {3 \sin \left (2 x \right )}{25}-\frac {4 \cos \left (2 x \right )}{25} \\ \frac {\sin \left (2 x \right )}{5}+\frac {2 \cos \left (2 x \right )}{5} \\ -\sin \left (2 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} {\mathrm e}^{x} & {\mathrm e}^{-x} \left (-\frac {3 \cos \left (2 x \right )}{25}-\frac {4 \sin \left (2 x \right )}{25}\right ) & {\mathrm e}^{-x} \left (\frac {3 \sin \left (2 x \right )}{25}-\frac {4 \cos \left (2 x \right )}{25}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \left (-\frac {\cos \left (2 x \right )}{5}+\frac {2 \sin \left (2 x \right )}{5}\right ) & {\mathrm e}^{-x} \left (\frac {\sin \left (2 x \right )}{5}+\frac {2 \cos \left (2 x \right )}{5}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \cos \left (2 x \right ) & -{\mathrm e}^{-x} \sin \left (2 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} {\mathrm e}^{x} & {\mathrm e}^{-x} \left (-\frac {3 \cos \left (2 x \right )}{25}-\frac {4 \sin \left (2 x \right )}{25}\right ) & {\mathrm e}^{-x} \left (\frac {3 \sin \left (2 x \right )}{25}-\frac {4 \cos \left (2 x \right )}{25}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \left (-\frac {\cos \left (2 x \right )}{5}+\frac {2 \sin \left (2 x \right )}{5}\right ) & {\mathrm e}^{-x} \left (\frac {\sin \left (2 x \right )}{5}+\frac {2 \cos \left (2 x \right )}{5}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \cos \left (2 x \right ) & -{\mathrm e}^{-x} \sin \left (2 x \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & -\frac {3}{25} & -\frac {4}{25} \\ 1 & -\frac {1}{5} & \frac {2}{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 {\left (-\sin \left (2 x \right )+3 \cos \left (2 x \right )\right ) {\mathrm e}^{-x}}{8}+\frac {5 \,{\mathrm e}^{x}}{8} & \frac {\left (\sin \left (2 x \right )-\cos \left (2 x \right )\right ) {\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x}}{4} & \frac {\left (-\sin \left (2 x \right )-\cos \left (2 x \right )\right ) {\mathrm e}^{-x}}{8}+\frac {{\mathrm e}^{x}}{8} \\ \frac {\left (-5 \sin \left (2 x \right )-5 \cos \left (2 x \right )\right ) {\mathrm e}^{-x}}{8}+\frac {5 \,{\mathrm e}^{x}}{8} & \frac {\left (\sin \left (2 x \right )+3 \cos \left (2 x \right )\right ) {\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x}}{4} & \frac {\left (3 \sin \left (2 x \right )-\cos \left (2 x \right )\right ) {\mathrm e}^{-x}}{8}+\frac {{\mathrm e}^{x}}{8} \\ \frac {5 \left (3 \sin \left (2 x \right )-\cos \left (2 x \right )\right ) {\mathrm e}^{-x}}{8}+\frac {5 \,{\mathrm e}^{x}}{8} & \frac {\left (-\cos \left (2 x \right )-7 \sin \left (2 x \right )\right ) {\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x}}{4} & \frac {\left (7 \cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-x}}{8}+\frac {{\mathrm e}^{x}}{8} \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 (-4 \,{\mathrm e}^{-x}+4\right ) \cos \left (2 x \right )}{34}-2 x^{2}-\frac {{\mathrm e}^{-x} \sin \left (2 x \right )}{34}-3 x +4 \,{\mathrm e}^{x}-\frac {9 \sin \left (2 x \right )}{17}-4 \\ \frac {\left (2 \,{\mathrm e}^{-x}-36\right ) \cos \left (2 x \right )}{34}+\frac {9 \,{\mathrm e}^{-x} \sin \left (2 x \right )}{34}-4 x +4 \,{\mathrm e}^{x}-\frac {4 \sin \left (2 x \right )}{17}-3 \\ -4+\frac {8 \left (-1+{\mathrm e}^{-x}\right ) \cos \left (2 x \right )}{17}-\frac {13 \,{\mathrm e}^{-x} \sin \left (2 x \right )}{34}+4 \,{\mathrm e}^{x}+\frac {36 \sin \left (2 x \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 {\left (-4 \,{\mathrm e}^{-x}+4\right ) \cos \left (2 x \right )}{34}-2 x^{2}-\frac {{\mathrm e}^{-x} \sin \left (2 x \right )}{34}-3 x +4 \,{\mathrm e}^{x}-\frac {9 \sin \left (2 x \right )}{17}-4 \\ \frac {\left (2 \,{\mathrm e}^{-x}-36\right ) \cos \left (2 x \right )}{34}+\frac {9 \,{\mathrm e}^{-x} \sin \left (2 x \right )}{34}-4 x +4 \,{\mathrm e}^{x}-\frac {4 \sin \left (2 x \right )}{17}-3 \\ -4+\frac {8 \left (-1+{\mathrm e}^{-x}\right ) \cos \left (2 x \right )}{17}-\frac {13 \,{\mathrm e}^{-x} \sin \left (2 x \right )}{34}+4 \,{\mathrm e}^{x}+\frac {36 \sin \left (2 x \right )}{17} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-4+\frac {\left (\left (-50-51 c_{2} -68 c_{3} \right ) \cos \left (2 x \right )-68 \left (c_{2} -\frac {3 c_{3}}{4}+\frac {25}{136}\right ) \sin \left (2 x \right )\right ) {\mathrm e}^{-x}}{425}+\frac {2 \cos \left (2 x \right )}{17}-\frac {9 \sin \left (2 x \right )}{17}+\left (c_{1} +4\right ) {\mathrm e}^{x}-2 x^{2}-3 x \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: 49

dsolve(diff(y(x),x$3)+diff(y(x),x$2)+3*diff(y(x),x)-5*y(x)=5*sin(2*x)+10*x^2+3*x+7,y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\left (17 c_{3} {\mathrm e}^{-x}-9\right ) \sin \left (2 x \right )}{17}+c_{2} {\mathrm e}^{-x} \cos \left (2 x \right )-2 x^{2}+c_{1} {\mathrm e}^{x}-3 x +\frac {2 \cos \left (2 x \right )}{17}-4 \]

Solution by Mathematica

Time used: 0.419 (sec). Leaf size: 55

DSolve[y'''[x]+y''[x]+3*y'[x]-5*y[x]==5*Sin[2*x]+10*x^2+3*x+7,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to -2 x^2-3 x+c_3 e^x+\left (\frac {2}{17}+c_2 e^{-x}\right ) \cos (2 x)+\left (-\frac {9}{17}+c_1 e^{-x}\right ) \sin (2 x)-4 \]