Internal problem ID [12360]
Internal file name [OUTPUT/11013_Monday_October_02_2023_02_48_03_AM_40015569/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 6. Introduction to Systems of ODEs. Problems page 408
Problem number: Problem 2(f).
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, _missing_y]]
\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime \prime }+4 y^{\prime }=\sin \left (t \right )} \] 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 \[ y^{\prime \prime \prime }-2 y^{\prime \prime }+4 y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{3}-2 \lambda ^{2}+4 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 1+i \sqrt {3}\\ \lambda _3 &= 1-i \sqrt {3} \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} +{\mathrm e}^{\left (1+i \sqrt {3}\right ) t} c_{2} +{\mathrm e}^{\left (1-i \sqrt {3}\right ) t} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{\left (1+i \sqrt {3}\right ) t} \\ y_3 &= {\mathrm e}^{\left (1-i \sqrt {3}\right ) t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-2 y^{\prime \prime }+4 y^{\prime } = \sin \left (t \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ \sin \left (t \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (t \right ), \sin \left (t \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{1, {\mathrm e}^{\left (1-i \sqrt {3}\right ) t}, {\mathrm e}^{\left (1+i \sqrt {3}\right ) t}\right \} \] 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. \[ y_p = A_{1} \cos \left (t \right )+A_{2} \sin \left (t \right ) \] 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 \[ -3 A_{1} \sin \left (t \right )+3 A_{2} \cos \left (t \right )+2 A_{1} \cos \left (t \right )+2 A_{2} \sin \left (t \right ) = \sin \left (t \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {3}{13}}, A_{2} = {\frac {2}{13}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {3 \cos \left (t \right )}{13}+\frac {2 \sin \left (t \right )}{13} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +{\mathrm e}^{\left (1+i \sqrt {3}\right ) t} c_{2} +{\mathrm e}^{\left (1-i \sqrt {3}\right ) t} c_{3}\right ) + \left (-\frac {3 \cos \left (t \right )}{13}+\frac {2 \sin \left (t \right )}{13}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +{\mathrm e}^{\left (1+i \sqrt {3}\right ) t} c_{2} +{\mathrm e}^{\left (1-i \sqrt {3}\right ) t} c_{3} -\frac {3 \cos \left (t \right )}{13}+\frac {2 \sin \left (t \right )}{13} \\ \end{align*}
Verification of solutions
\[ y = c_{1} +{\mathrm e}^{\left (1+i \sqrt {3}\right ) t} c_{2} +{\mathrm e}^{\left (1-i \sqrt {3}\right ) t} c_{3} -\frac {3 \cos \left (t \right )}{13}+\frac {2 \sin \left (t \right )}{13} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime \prime }-2 \frac {d}{d t}y^{\prime }+4 y^{\prime }=\sin \left (t \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d t}y^{\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 (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=\frac {d}{d t}y^{\prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (t \right )=\sin \left (t \right )+2 y_{3}\left (t \right )-4 y_{2}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=y_{1}^{\prime }\left (t \right ), y_{3}\left (t \right )=y_{2}^{\prime }\left (t \right ), y_{3}^{\prime }\left (t \right )=\sin \left (t \right )+2 y_{3}\left (t \right )-4 y_{2}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -4 & 2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ \sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ \sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -4 & 2 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\left (t \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [1-\mathrm {I} \sqrt {3}, \left [\begin {array}{c} \frac {1}{\left (1-\mathrm {I} \sqrt {3}\right )^{2}} \\ \frac {1}{1-\mathrm {I} \sqrt {3}} \\ 1 \end {array}\right ]\right ], \left [1+\mathrm {I} \sqrt {3}, \left [\begin {array}{c} \frac {1}{\left (1+\mathrm {I} \sqrt {3}\right )^{2}} \\ \frac {1}{1+\mathrm {I} \sqrt {3}} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [1-\mathrm {I} \sqrt {3}, \left [\begin {array}{c} \frac {1}{\left (1-\mathrm {I} \sqrt {3}\right )^{2}} \\ \frac {1}{1-\mathrm {I} \sqrt {3}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (1-\mathrm {I} \sqrt {3}\right ) t}\cdot \left [\begin {array}{c} \frac {1}{\left (1-\mathrm {I} \sqrt {3}\right )^{2}} \\ \frac {1}{1-\mathrm {I} \sqrt {3}} \\ 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}^{t}\cdot \left (\cos \left (\sqrt {3}\, t \right )-\mathrm {I} \sin \left (\sqrt {3}\, t \right )\right )\cdot \left [\begin {array}{c} \frac {1}{\left (1-\mathrm {I} \sqrt {3}\right )^{2}} \\ \frac {1}{1-\mathrm {I} \sqrt {3}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{t}\cdot \left [\begin {array}{c} \frac {\cos \left (\sqrt {3}\, t \right )-\mathrm {I} \sin \left (\sqrt {3}\, t \right )}{\left (1-\mathrm {I} \sqrt {3}\right )^{2}} \\ \frac {\cos \left (\sqrt {3}\, t \right )-\mathrm {I} \sin \left (\sqrt {3}\, t \right )}{1-\mathrm {I} \sqrt {3}} \\ \cos \left (\sqrt {3}\, t \right )-\mathrm {I} \sin \left (\sqrt {3}\, t \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 (t \right )={\mathrm e}^{t}\cdot \left [\begin {array}{c} -\frac {\cos \left (\sqrt {3}\, t \right )}{8}+\frac {\sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{8} \\ \frac {\cos \left (\sqrt {3}\, t \right )}{4}+\frac {\sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{4} \\ \cos \left (\sqrt {3}\, t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{t}\cdot \left [\begin {array}{c} \frac {\cos \left (\sqrt {3}\, t \right ) \sqrt {3}}{8}+\frac {\sin \left (\sqrt {3}\, t \right )}{8} \\ \frac {\cos \left (\sqrt {3}\, t \right ) \sqrt {3}}{4}-\frac {\sin \left (\sqrt {3}\, t \right )}{4} \\ -\sin \left (\sqrt {3}\, t \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 (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+{\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (t \right )=\left [\begin {array}{ccc} 1 & {\mathrm e}^{t} \left (-\frac {\cos \left (\sqrt {3}\, t \right )}{8}+\frac {\sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{8}\right ) & {\mathrm e}^{t} \left (\frac {\cos \left (\sqrt {3}\, t \right ) \sqrt {3}}{8}+\frac {\sin \left (\sqrt {3}\, t \right )}{8}\right ) \\ 0 & {\mathrm e}^{t} \left (\frac {\cos \left (\sqrt {3}\, t \right )}{4}+\frac {\sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{4}\right ) & {\mathrm e}^{t} \left (\frac {\cos \left (\sqrt {3}\, t \right ) \sqrt {3}}{4}-\frac {\sin \left (\sqrt {3}\, t \right )}{4}\right ) \\ 0 & {\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right ) & -{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (t \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (t \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 (t \right )=\phi \left (t \right )\cdot \phi \left (0\right )^{-1} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} 1 & {\mathrm e}^{t} \left (-\frac {\cos \left (\sqrt {3}\, t \right )}{8}+\frac {\sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{8}\right ) & {\mathrm e}^{t} \left (\frac {\cos \left (\sqrt {3}\, t \right ) \sqrt {3}}{8}+\frac {\sin \left (\sqrt {3}\, t \right )}{8}\right ) \\ 0 & {\mathrm e}^{t} \left (\frac {\cos \left (\sqrt {3}\, t \right )}{4}+\frac {\sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{4}\right ) & {\mathrm e}^{t} \left (\frac {\cos \left (\sqrt {3}\, t \right ) \sqrt {3}}{4}-\frac {\sin \left (\sqrt {3}\, t \right )}{4}\right ) \\ 0 & {\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right ) & -{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \end {array}\right ]\cdot \left [\begin {array}{ccc} 1 & -\frac {1}{8} & \frac {\sqrt {3}}{8} \\ 0 & \frac {1}{4} & \frac {\sqrt {3}}{4} \\ 0 & 1 & 0 \end {array}\right ]^{-1} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} 1 & \frac {{\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{2}+\frac {{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{6}-\frac {1}{2} & -\frac {{\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{4}+\frac {{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{12}+\frac {1}{4} \\ 0 & \frac {{\mathrm e}^{t} \left (\cos \left (\sqrt {3}\, t \right ) \sqrt {3}-\sin \left (\sqrt {3}\, t \right )\right ) \sqrt {3}}{3} & \frac {{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{3} \\ 0 & -\frac {4 \,{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{3} & \frac {{\mathrm e}^{t} \left (\sin \left (\sqrt {3}\, t \right ) \sqrt {3}+3 \cos \left (\sqrt {3}\, t \right )\right )}{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 (t \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (t \right )=\Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \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 (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )={\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=\Phi \left (t \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (t \right )=\int _{0}^{t}\Phi \left (s \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot \left (\int _{0}^{t}\Phi \left (s \right )^{-1}\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 (t \right )=\left [\begin {array}{c} \frac {1}{4}-\frac {7 \,{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{156}-\frac {3 \cos \left (t \right )}{13}+\frac {2 \sin \left (t \right )}{13}-\frac {{\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{52} \\ -\frac {2 \,{\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{13}+\frac {2 \cos \left (t \right )}{13}+\frac {3 \sin \left (t \right )}{13}-\frac {{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{39} \\ \frac {5 \,{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{39}-\frac {3 \,{\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{13}+\frac {3 \cos \left (t \right )}{13}-\frac {2 \sin \left (t \right )}{13} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+\left [\begin {array}{c} \frac {1}{4}-\frac {7 \,{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{156}-\frac {3 \cos \left (t \right )}{13}+\frac {2 \sin \left (t \right )}{13}-\frac {{\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{52} \\ -\frac {2 \,{\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{13}+\frac {2 \cos \left (t \right )}{13}+\frac {3 \sin \left (t \right )}{13}-\frac {{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{39} \\ \frac {5 \,{\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}}{39}-\frac {3 \,{\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{13}+\frac {3 \cos \left (t \right )}{13}-\frac {2 \sin \left (t \right )}{13} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {\left (-c_{3} \sqrt {3}+c_{2} +\frac {2}{13}\right ) {\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )}{8}+\frac {\left (\left (c_{2} -\frac {14}{39}\right ) \sqrt {3}+c_{3} \right ) {\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right )}{8}+c_{1} -\frac {3 \cos \left (t \right )}{13}+\frac {2 \sin \left (t \right )}{13}+\frac {1}{4} \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] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = 2*(diff(_b(_a), _a))-4*_b(_a)+sin(_a), _b(_a)` *** Sublevel 2 *** Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] -> Try solving first the homogeneous part of the ODE checking if the LODE has constant coefficients <- constant coefficients successful <- solving first the homogeneous part of the ODE successful <- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 49
dsolve(diff(y(t),t$3)=2*diff(y(t),t$2)-4*diff(y(t),t)+sin(t),y(t), singsol=all)
\[ y \left (t \right ) = \frac {{\mathrm e}^{t} \left (-c_{2} \sqrt {3}+c_{1} \right ) \cos \left (\sqrt {3}\, t \right )}{4}+\frac {{\mathrm e}^{t} \left (\sqrt {3}\, c_{1} +c_{2} \right ) \sin \left (\sqrt {3}\, t \right )}{4}+c_{3} -\frac {3 \cos \left (t \right )}{13}+\frac {2 \sin \left (t \right )}{13} \]
✓ Solution by Mathematica
Time used: 1.636 (sec). Leaf size: 82
DSolve[y'''[t]==2*y''[t]-4*y'[t]+Sin[t],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{52} \left (8 \sin (t)-12 \cos (t)-13 \left (\sqrt {3} c_1-c_2\right ) e^t \cos \left (\sqrt {3} t\right )+13 c_1 e^t \sin \left (\sqrt {3} t\right )+13 \sqrt {3} c_2 e^t \sin \left (\sqrt {3} t\right )\right )+c_3 \]