problem Problem 13(c)

Internal problem ID [12346]
Internal ﬁle name [OUTPUT/10999_Monday_October_02_2023_02_47_47_AM_5883988/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number: Problem 13(c).
ODE order: 3.
ODE degree: 1.

\[ \boxed {y^{\prime \prime \prime }-y^{\prime \prime }+4 y^{\prime }-4 y=8 \,{\mathrm e}^{2 t}-5 \,{\mathrm e}^{t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 3] \end {align*}

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime \prime }\right ) &= s^3 Y(s) - y''(0) - s y'(0) - s^2 y \left (0\right ) \end {align*}

The given ode becomes an algebraic equation in the Laplace domain \[ s^{3} Y \left (s \right )-y^{\prime \prime }\left (0\right )-s y^{\prime }\left (0\right )-s^{2} y \left (0\right )-s^{2} Y \left (s \right )+y^{\prime }\left (0\right )+s y \left (0\right )+4 s Y \left (s \right )-4 y \left (0\right )-4 Y \left (s \right ) = \frac {3 s +2}{\left (s -2\right ) \left (s -1\right )}\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=2\\ y^{\prime }\left (0\right )&=0\\ y^{\prime \prime }\left (0\right )&=3 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \[ s^{3} Y \left (s \right )-11-2 s^{2}-s^{2} Y \left (s \right )+2 s +4 s Y \left (s \right )-4 Y \left (s \right ) = \frac {3 s +2}{\left (s -2\right ) \left (s -1\right )} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {2 s^{4}-8 s^{3}+21 s^{2}-34 s +24}{\left (s -2\right ) \left (s -1\right ) \left (s^{3}-s^{2}+4 s -4\right )} \] Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{\left (s -1\right )^{2}}+\frac {1}{s -1}+\frac {1}{s -2}+\frac {i}{2 s -4 i}-\frac {i}{2 \left (s +2 i\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{\left (s -1\right )^{2}}\right ) &= -{\mathrm e}^{t} t\\ \mathcal {L}^{-1}\left (\frac {1}{s -1}\right ) &= {\mathrm e}^{t}\\ \mathcal {L}^{-1}\left (\frac {1}{s -2}\right ) &= {\mathrm e}^{2 t}\\ \mathcal {L}^{-1}\left (\frac {i}{2 s -4 i}\right ) &= \frac {i {\mathrm e}^{2 i t}}{2}\\ \mathcal {L}^{-1}\left (-\frac {i}{2 \left (s +2 i\right )}\right ) &= -\frac {i {\mathrm e}^{-2 i t}}{2} \end {align*}

Adding the above results and simplifying gives \[ y=-\sin \left (2 t \right )+{\mathrm e}^{2 t}-{\mathrm e}^{t} \left (t -1\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\sin \left (2 t \right )+{\mathrm e}^{2 t}-{\mathrm e}^{t} \left (t -1\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\sin \left (2 t \right )+{\mathrm e}^{2 t}-{\mathrm e}^{t} \left (t -1\right ) \] Veriﬁed OK.

4.39.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime \prime }-\frac {d}{d t}y^{\prime }+4 y^{\prime }-4 y=8 \,{\mathrm e}^{2 t}-5 \,{\mathrm e}^{t}, y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0, \left (\frac {d}{d t}y^{\prime }\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3\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 )=8 \,{\mathrm e}^{2 t}-5 \,{\mathrm e}^{t}+y_{3}\left (t \right )-4 y_{2}\left (t \right )+4 y_{1}\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 )=8 \,{\mathrm e}^{2 t}-5 \,{\mathrm e}^{t}+y_{3}\left (t \right )-4 y_{2}\left (t \right )+4 y_{1}\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 \\ 4 & -4 & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 8 \,{\mathrm e}^{2 t}-5 \,{\mathrm e}^{t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ 8 \,{\mathrm e}^{2 t}-5 \,{\mathrm e}^{t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & -4 & 1 \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 [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 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}^{t}\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 [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} t}\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 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 \\ {} & {} & \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right )\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\cos \left (2 t \right )}{4}+\frac {\mathrm {I} \sin \left (2 t \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \cos \left (2 t \right )-\mathrm {I} \sin \left (2 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 )=\left [\begin {array}{c} -\frac {\cos \left (2 t \right )}{4} \\ \frac {\sin \left (2 t \right )}{2} \\ \cos \left (2 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )=\left [\begin {array}{c} \frac {\sin \left (2 t \right )}{4} \\ \frac {\cos \left (2 t \right )}{2} \\ -\sin \left (2 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} {\mathrm e}^{t} & -\frac {\cos \left (2 t \right )}{4} & \frac {\sin \left (2 t \right )}{4} \\ {\mathrm e}^{t} & \frac {\sin \left (2 t \right )}{2} & \frac {\cos \left (2 t \right )}{2} \\ {\mathrm e}^{t} & \cos \left (2 t \right ) & -\sin \left (2 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} {\mathrm e}^{t} & -\frac {\cos \left (2 t \right )}{4} & \frac {\sin \left (2 t \right )}{4} \\ {\mathrm e}^{t} & \frac {\sin \left (2 t \right )}{2} & \frac {\cos \left (2 t \right )}{2} \\ {\mathrm e}^{t} & \cos \left (2 t \right ) & -\sin \left (2 t \right ) \end {array}\right ]\cdot \left [\begin {array}{ccc} 1 & -\frac {1}{4} & 0 \\ 1 & 0 & \frac {1}{2} \\ 1 & 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} \frac {4 \,{\mathrm e}^{t}}{5}+\frac {\cos \left (2 t \right )}{5}-\frac {2 \sin \left (2 t \right )}{5} & \frac {\sin \left (2 t \right )}{2} & \frac {{\mathrm e}^{t}}{5}-\frac {\cos \left (2 t \right )}{5}-\frac {\sin \left (2 t \right )}{10} \\ \frac {4 \,{\mathrm e}^{t}}{5}-\frac {2 \sin \left (2 t \right )}{5}-\frac {4 \cos \left (2 t \right )}{5} & \cos \left (2 t \right ) & \frac {{\mathrm e}^{t}}{5}+\frac {2 \sin \left (2 t \right )}{5}-\frac {\cos \left (2 t \right )}{5} \\ \frac {4 \,{\mathrm e}^{t}}{5}-\frac {4 \cos \left (2 t \right )}{5}+\frac {8 \sin \left (2 t \right )}{5} & -2 \sin \left (2 t \right ) & \frac {{\mathrm e}^{t}}{5}+\frac {4 \cos \left (2 t \right )}{5}+\frac {2 \sin \left (2 t \right )}{5} \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 {\sin \left (2 t \right )}{10}+\frac {\cos \left (2 t \right )}{5}-{\mathrm e}^{t} t +{\mathrm e}^{2 t}-\frac {6 \,{\mathrm e}^{t}}{5} \\ \frac {\cos \left (2 t \right )}{5}-\frac {2 \sin \left (2 t \right )}{5}-{\mathrm e}^{t} t +2 \,{\mathrm e}^{2 t}-\frac {11 \,{\mathrm e}^{t}}{5} \\ -\frac {2 \sin \left (2 t \right )}{5}-\frac {4 \cos \left (2 t \right )}{5}-{\mathrm e}^{t} t +4 \,{\mathrm e}^{2 t}-\frac {16 \,{\mathrm e}^{t}}{5} \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 {\sin \left (2 t \right )}{10}+\frac {\cos \left (2 t \right )}{5}-{\mathrm e}^{t} t +{\mathrm e}^{2 t}-\frac {6 \,{\mathrm e}^{t}}{5} \\ \frac {\cos \left (2 t \right )}{5}-\frac {2 \sin \left (2 t \right )}{5}-{\mathrm e}^{t} t +2 \,{\mathrm e}^{2 t}-\frac {11 \,{\mathrm e}^{t}}{5} \\ -\frac {2 \sin \left (2 t \right )}{5}-\frac {4 \cos \left (2 t \right )}{5}-{\mathrm e}^{t} t +4 \,{\mathrm e}^{2 t}-\frac {16 \,{\mathrm e}^{t}}{5} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (-5 c_{2} +4\right ) \cos \left (2 t \right )}{20}+\frac {\left (5 c_{3} +2\right ) \sin \left (2 t \right )}{20}+{\mathrm e}^{2 t}+\frac {\left (-6-5 t +5 c_{1} \right ) {\mathrm e}^{t}}{5} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=-\frac {c_{2}}{4}+c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\left (-5 c_{2} +4\right ) \sin \left (2 t \right )}{10}+\frac {\left (5 c_{3} +2\right ) \cos \left (2 t \right )}{10}+2 \,{\mathrm e}^{2 t}-{\mathrm e}^{t}+\frac {\left (-6-5 t +5 c_{1} \right ) {\mathrm e}^{t}}{5} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {c_{3}}{2}+c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-\frac {\left (-5 c_{2} +4\right ) \cos \left (2 t \right )}{5}-\frac {\left (5 c_{3} +2\right ) \sin \left (2 t \right )}{5}+4 \,{\mathrm e}^{2 t}-2 \,{\mathrm e}^{t}+\frac {\left (-6-5 t +5 c_{1} \right ) {\mathrm e}^{t}}{5} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d t}y^{\prime }\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3 \\ {} & {} & 3=c_{2} +c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {11}{5}, c_{2} =\frac {4}{5}, c_{3} =-\frac {22}{5}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{t} t -\sin \left (2 t \right )+{\mathrm e}^{2 t}+{\mathrm e}^{t} \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`

\[ y \left (t \right ) = -t \,{\mathrm e}^{t}+{\mathrm e}^{2 t}+{\mathrm e}^{t}-\sin \left (2 t \right ) \]

4.39 problem Problem 13(c)

4.39.1 Maple step by step solution