9.2 problem 16.1 (ii)

Internal problem ID [12044]
Internal file name [OUTPUT/10697_Sunday_September_03_2023_12_36_41_PM_80785119/index.tex]

Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section: Chapter 16, Higher order linear equations with constant coefficients. Exercises page 153
Problem number: 16.1 (ii).
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]]

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

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (1-\sqrt {3}\right ) x} c_{2} +{\mathrm e}^{\left (1+\sqrt {3}\right ) x} c_{3} \] 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-\sqrt {3}\right ) x} \\ y_3 &= {\mathrm e}^{\left (1+\sqrt {3}\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-3 y^{\prime \prime }+2 y = \sin \left (x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ \sin \left (x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (x \right ), \sin \left (x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{x}, {\mathrm e}^{\left (1-\sqrt {3}\right ) x}, {\mathrm e}^{\left (1+\sqrt {3}\right ) x}\right \} \] 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. \[ y_p = A_{1} \cos \left (x \right )+A_{2} \sin \left (x \right ) \] 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 \[ A_{1} \sin \left (x \right )-A_{2} \cos \left (x \right )+5 A_{1} \cos \left (x \right )+5 A_{2} \sin \left (x \right ) = \sin \left (x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{26}}, A_{2} = {\frac {5}{26}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {\cos \left (x \right )}{26}+\frac {5 \sin \left (x \right )}{26} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (1-\sqrt {3}\right ) x} c_{2} +{\mathrm e}^{\left (1+\sqrt {3}\right ) x} c_{3}\right ) + \left (\frac {\cos \left (x \right )}{26}+\frac {5 \sin \left (x \right )}{26}\right ) \\ \end{align*} Which simplifies to \[ y = c_{1} {\mathrm e}^{x}+{\mathrm e}^{-\left (\sqrt {3}-1\right ) x} c_{2} +{\mathrm e}^{\left (1+\sqrt {3}\right ) x} c_{3} +\frac {\cos \left (x \right )}{26}+\frac {5 \sin \left (x \right )}{26} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{x}+{\mathrm e}^{-\left (\sqrt {3}-1\right ) x} c_{2} +{\mathrm e}^{\left (1+\sqrt {3}\right ) x} c_{3} +\frac {\cos \left (x \right )}{26}+\frac {5 \sin \left (x \right )}{26} \\ \end{align*}

Verification of solutions

\[ y = c_{1} {\mathrm e}^{x}+{\mathrm e}^{-\left (\sqrt {3}-1\right ) x} c_{2} +{\mathrm e}^{\left (1+\sqrt {3}\right ) x} c_{3} +\frac {\cos \left (x \right )}{26}+\frac {5 \sin \left (x \right )}{26} \] Verified OK.

9.2.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-3 y^{\prime \prime }+2 y=\sin \left (x \right ) \\ \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 )=\sin \left (x \right )+3 y_{3}\left (x \right )-2 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 )=\sin \left (x \right )+3 y_{3}\left (x \right )-2 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 \\ -2 & 0 & 3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ \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 \\ \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 \\ -2 & 0 & 3 \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-\sqrt {3}, \left [\begin {array}{c} \frac {1}{\left (1-\sqrt {3}\right )^{2}} \\ \frac {1}{1-\sqrt {3}} \\ 1 \end {array}\right ]\right ], \left [1+\sqrt {3}, \left [\begin {array}{c} \frac {1}{\left (1+\sqrt {3}\right )^{2}} \\ \frac {1}{1+\sqrt {3}} \\ 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 eigenpair}\hspace {3pt} \\ {} & {} & \left [1-\sqrt {3}, \left [\begin {array}{c} \frac {1}{\left (1-\sqrt {3}\right )^{2}} \\ \frac {1}{1-\sqrt {3}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{\left (1-\sqrt {3}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (1-\sqrt {3}\right )^{2}} \\ \frac {1}{1-\sqrt {3}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1+\sqrt {3}, \left [\begin {array}{c} \frac {1}{\left (1+\sqrt {3}\right )^{2}} \\ \frac {1}{1+\sqrt {3}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{\left (1+\sqrt {3}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (1+\sqrt {3}\right )^{2}} \\ \frac {1}{1+\sqrt {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}+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} {\mathrm e}^{x} & \frac {{\mathrm e}^{\left (1-\sqrt {3}\right ) x}}{\left (1-\sqrt {3}\right )^{2}} & \frac {{\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{\left (1+\sqrt {3}\right )^{2}} \\ {\mathrm e}^{x} & \frac {{\mathrm e}^{\left (1-\sqrt {3}\right ) x}}{1-\sqrt {3}} & \frac {{\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{1+\sqrt {3}} \\ {\mathrm e}^{x} & {\mathrm e}^{\left (1-\sqrt {3}\right ) x} & {\mathrm e}^{\left (1+\sqrt {3}\right ) 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} & \frac {{\mathrm e}^{\left (1-\sqrt {3}\right ) x}}{\left (1-\sqrt {3}\right )^{2}} & \frac {{\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{\left (1+\sqrt {3}\right )^{2}} \\ {\mathrm e}^{x} & \frac {{\mathrm e}^{\left (1-\sqrt {3}\right ) x}}{1-\sqrt {3}} & \frac {{\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{1+\sqrt {3}} \\ {\mathrm e}^{x} & {\mathrm e}^{\left (1-\sqrt {3}\right ) x} & {\mathrm e}^{\left (1+\sqrt {3}\right ) x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & \frac {1}{\left (1-\sqrt {3}\right )^{2}} & \frac {1}{\left (1+\sqrt {3}\right )^{2}} \\ 1 & \frac {1}{1-\sqrt {3}} & \frac {1}{1+\sqrt {3}} \\ 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 (1+\sqrt {3}\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{6}+\frac {\left (1-\sqrt {3}\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{6}+\frac {2 \,{\mathrm e}^{x}}{3} & \frac {\left (-\sqrt {3}-2\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{6}+\frac {{\mathrm e}^{\left (1+\sqrt {3}\right ) x} \left (\sqrt {3}-2\right )}{6}+\frac {2 \,{\mathrm e}^{x}}{3} & -\frac {{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{6}+\frac {{\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{6} \\ \frac {2 \,{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{3}-\frac {{\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{3} & \frac {\left (1+\sqrt {3}\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{6}+\frac {\left (1-\sqrt {3}\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{6}+\frac {2 \,{\mathrm e}^{x}}{3} & \frac {\left (1-\sqrt {3}\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{6}+\frac {\left (1+\sqrt {3}\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{6}-\frac {{\mathrm e}^{x}}{3} \\ \frac {\left (\sqrt {3}-1\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{3}+\frac {\left (-1-\sqrt {3}\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{3}+\frac {2 \,{\mathrm e}^{x}}{3} & \frac {2 \,{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{3}-\frac {{\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{3} & \frac {\left (2-\sqrt {3}\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{3}+\frac {\left (2+\sqrt {3}\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{3}-\frac {{\mathrm e}^{x}}{3} \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 (5+2 \sqrt {3}\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{78}+\frac {\left (5-2 \sqrt {3}\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{78}+\frac {\cos \left (x \right )}{26}-\frac {{\mathrm e}^{x}}{6}+\frac {5 \sin \left (x \right )}{26} \\ \frac {\left (-3 \sqrt {3}-1\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{78}+\frac {\left (3 \sqrt {3}-1\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{78}+\frac {5 \cos \left (x \right )}{26}-\frac {{\mathrm e}^{x}}{6}-\frac {\sin \left (x \right )}{26} \\ \frac {\left (-2 \sqrt {3}+8\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{78}+\frac {\left (2 \sqrt {3}+8\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{78}-\frac {\cos \left (x \right )}{26}-\frac {{\mathrm e}^{x}}{6}-\frac {5 \sin \left (x \right )}{26} \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 (5+2 \sqrt {3}\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{78}+\frac {\left (5-2 \sqrt {3}\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{78}+\frac {\cos \left (x \right )}{26}-\frac {{\mathrm e}^{x}}{6}+\frac {5 \sin \left (x \right )}{26} \\ \frac {\left (-3 \sqrt {3}-1\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{78}+\frac {\left (3 \sqrt {3}-1\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{78}+\frac {5 \cos \left (x \right )}{26}-\frac {{\mathrm e}^{x}}{6}-\frac {\sin \left (x \right )}{26} \\ \frac {\left (-2 \sqrt {3}+8\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{78}+\frac {\left (2 \sqrt {3}+8\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{78}-\frac {\cos \left (x \right )}{26}-\frac {{\mathrm e}^{x}}{6}-\frac {5 \sin \left (x \right )}{26} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (39 c_{2} +2\right ) \sqrt {3}+78 c_{2} +5\right ) {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}}{78}+\frac {\left (\left (-39 c_{3} -2\right ) \sqrt {3}+78 c_{3} +5\right ) {\mathrm e}^{\left (1+\sqrt {3}\right ) x}}{78}+\frac {\left (-1+6 c_{1} \right ) {\mathrm e}^{x}}{6}+\frac {\cos \left (x \right )}{26}+\frac {5 \sin \left (x \right )}{26} \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: 38

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

\[ y \left (x \right ) = c_{3} {\mathrm e}^{-\left (\sqrt {3}-1\right ) x}+c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{\left (1+\sqrt {3}\right ) x}+\frac {5 \sin \left (x \right )}{26}+\frac {\cos \left (x \right )}{26} \]

✓ Solution by Mathematica

Time used: 0.206 (sec). Leaf size: 49

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

\[ y(x)\to \frac {1}{26} \left (5 \sin (x)+\cos (x)+26 e^x \left (c_1 e^{-\sqrt {3} x}+c_2 e^{\sqrt {3} x}+c_3\right )\right ) \]