problem Problem 14(b)

Internal problem ID [12349]
Internal ﬁle name [OUTPUT/11002_Monday_October_02_2023_02_47_49_AM_8359695/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 14(b).
ODE order: 4.
ODE degree: 1.

\[ \boxed {y^{\prime \prime \prime \prime }-16 y=32 \operatorname {Heaviside}\left (t \right )-32 \operatorname {Heaviside}\left (t -\pi \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 0] \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 )\\ \mathcal {L}\left (y^{\prime \prime \prime \prime }\right ) &= s^4 Y(s) - y'''(0) - s y''(0) - s^2 y'(0)- s^3 y \left (0\right ) \end {align*}

The given ode becomes an algebraic equation in the Laplace domain \[ s^{4} Y \left (s \right )-y^{\prime \prime \prime }\left (0\right )-s y^{\prime \prime }\left (0\right )-s^{2} y^{\prime }\left (0\right )-s^{3} y \left (0\right )-16 Y \left (s \right ) = \frac {32-32 \,{\mathrm e}^{-s \pi }}{s}\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=0\\ y^{\prime \prime }\left (0\right )&=0\\ y^{\prime \prime \prime }\left (0\right )&=0 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \[ s^{4} Y \left (s \right )-16 Y \left (s \right ) = \frac {32-32 \,{\mathrm e}^{-s \pi }}{s} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = -\frac {32 \left (-1+{\mathrm e}^{-s \pi }\right )}{s \left (s^{4}-16\right )} \] Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {32 \left (-1+{\mathrm e}^{-s \pi }\right )}{s \left (s^{4}-16\right )}\right )\\ &= \left (2 \cos \left (t \right )^{2}-3\right ) \operatorname {Heaviside}\left (-t +\pi \right )-\operatorname {Heaviside}\left (t -\pi \right ) \cosh \left (2 t -2 \pi \right )+\cosh \left (2 t \right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \cosh \left (2 t \right )+2 \cos \left (t \right )^{2}-3 & t <\pi \\ \cosh \left (2 \pi \right )-2 & t =\pi \\ \cosh \left (2 t \right )-\cosh \left (2 t -2 \pi \right ) & \pi

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} \cosh \left (2 t \right )+2 \cos \left (t \right )^{2}-3 & t <\pi \\ \cosh \left (2 \pi \right )-2 & t &=\pi \\ \cosh \left (2 t \right )-\cosh \left (2 t -2 \pi \right ) & \pi

Veriﬁcation of solutions

\[ y = \left \{\begin {array}{cc} \cosh \left (2 t \right )+2 \cos \left (t \right )^{2}-3 & t <\pi \\ \cosh \left (2 \pi \right )-2 & t =\pi \\ \cosh \left (2 t \right )-\cosh \left (2 t -2 \pi \right ) & \pi

4.42.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}\frac {d}{d t}y^{\prime \prime }-16 y=32 \mathit {Heaviside}\left (t \right )-32 \mathit {Heaviside}\left (t -\pi \right ), y \left (0\right )=0, 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 \}}}}=0, \left (\frac {d}{d t}y^{\prime \prime }\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d t}\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 {Define new variable}\hspace {3pt} y_{4}\left (t \right ) \\ {} & {} & y_{4}\left (t \right )=\frac {d}{d t}y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (t \right )=32 \mathit {Heaviside}\left (t \right )-32 \mathit {Heaviside}\left (t -\pi \right )+16 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_{4}\left (t \right )=y_{3}^{\prime }\left (t \right ), y_{4}^{\prime }\left (t \right )=32 \mathit {Heaviside}\left (t \right )-32 \mathit {Heaviside}\left (t -\pi \right )+16 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 ) \\ y_{4}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 16 & 0 & 0 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 32 \mathit {Heaviside}\left (t \right )-32 \mathit {Heaviside}\left (t -\pi \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 32 \mathit {Heaviside}\left (t \right )-32 \mathit {Heaviside}\left (t -\pi \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 16 & 0 & 0 & 0 \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 [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \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 t}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \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 {\mathrm {I}}{8} \\ -\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 {\mathrm {I}}{8} \\ -\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 {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ -\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}}_{3}\left (t \right )=\left [\begin {array}{c} -\frac {\sin \left (2 t \right )}{8} \\ -\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}}_{4}\left (t \right )=\left [\begin {array}{c} -\frac {\cos \left (2 t \right )}{8} \\ \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}+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\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}{cccc} -\frac {{\mathrm e}^{-2 t}}{8} & \frac {{\mathrm e}^{2 t}}{8} & -\frac {\sin \left (2 t \right )}{8} & -\frac {\cos \left (2 t \right )}{8} \\ \frac {{\mathrm e}^{-2 t}}{4} & \frac {{\mathrm e}^{2 t}}{4} & -\frac {\cos \left (2 t \right )}{4} & \frac {\sin \left (2 t \right )}{4} \\ -\frac {{\mathrm e}^{-2 t}}{2} & \frac {{\mathrm e}^{2 t}}{2} & \frac {\sin \left (2 t \right )}{2} & \frac {\cos \left (2 t \right )}{2} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{2 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 \frac {1}{\phi \left (0\right )} \\ {} & \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}{cccc} -\frac {{\mathrm e}^{-2 t}}{8} & \frac {{\mathrm e}^{2 t}}{8} & -\frac {\sin \left (2 t \right )}{8} & -\frac {\cos \left (2 t \right )}{8} \\ \frac {{\mathrm e}^{-2 t}}{4} & \frac {{\mathrm e}^{2 t}}{4} & -\frac {\cos \left (2 t \right )}{4} & \frac {\sin \left (2 t \right )}{4} \\ -\frac {{\mathrm e}^{-2 t}}{2} & \frac {{\mathrm e}^{2 t}}{2} & \frac {\sin \left (2 t \right )}{2} & \frac {\cos \left (2 t \right )}{2} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{2 t} & \cos \left (2 t \right ) & -\sin \left (2 t \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -\frac {1}{8} & \frac {1}{8} & 0 & -\frac {1}{8} \\ \frac {1}{4} & \frac {1}{4} & -\frac {1}{4} & 0 \\ -\frac {1}{2} & \frac {1}{2} & 0 & \frac {1}{2} \\ 1 & 1 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{cccc} \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} & \frac {{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t}}{16}-\frac {\cos \left (2 t \right )}{8} & -\frac {{\mathrm e}^{-2 t}}{32}+\frac {{\mathrm e}^{2 t}}{32}-\frac {\sin \left (2 t \right )}{16} \\ -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} & \frac {{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t}}{16}-\frac {\cos \left (2 t \right )}{8} \\ {\mathrm e}^{-2 t}+{\mathrm e}^{2 t}-2 \cos \left (2 t \right ) & -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} \\ -2 \,{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{2 t}+4 \sin \left (2 t \right ) & {\mathrm e}^{-2 t}+{\mathrm e}^{2 t}-2 \cos \left (2 t \right ) & -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} \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 )=\frac {1}{\Phi \left (t \right )}\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}\frac {1}{\Phi \left (s \right )}\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}\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 (t \right )=\left [\begin {array}{c} \frac {{\mathrm e}^{-2 t -2 \pi } \left (-2 \left (\cos \left (2 t \right )-2\right ) \left (\mathit {Heaviside}\left (t -\pi \right )-\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{2 t +2 \pi }+{\mathrm e}^{4 t +2 \pi } \mathit {Heaviside}\left (t \right )+\left (-{\mathrm e}^{4 t}-{\mathrm e}^{4 \pi }\right ) \mathit {Heaviside}\left (t -\pi \right )+{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t \right )\right )}{2} \\ {\mathrm e}^{-2 t -2 \pi } \left (2 \sin \left (2 t \right ) \left (\mathit {Heaviside}\left (t -\pi \right )-\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{2 t +2 \pi }+{\mathrm e}^{4 t +2 \pi } \mathit {Heaviside}\left (t \right )+\left (-{\mathrm e}^{4 t}+{\mathrm e}^{4 \pi }\right ) \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t \right )\right ) \\ 2 \,{\mathrm e}^{-2 t -2 \pi } \left (2 \cos \left (2 t \right ) \left (\mathit {Heaviside}\left (t -\pi \right )-\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{2 t +2 \pi }+{\mathrm e}^{4 t +2 \pi } \mathit {Heaviside}\left (t \right )+\left (-{\mathrm e}^{4 t}-{\mathrm e}^{4 \pi }\right ) \mathit {Heaviside}\left (t -\pi \right )+{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t \right )\right ) \\ 4 \,{\mathrm e}^{-2 t -2 \pi } \left (-2 \sin \left (2 t \right ) \left (\mathit {Heaviside}\left (t -\pi \right )-\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{2 t +2 \pi }+{\mathrm e}^{4 t +2 \pi } \mathit {Heaviside}\left (t \right )+\left (-{\mathrm e}^{4 t}+{\mathrm e}^{4 \pi }\right ) \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t \right )\right ) \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}+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right )+\left [\begin {array}{c} \frac {{\mathrm e}^{-2 t -2 \pi } \left (-2 \left (\cos \left (2 t \right )-2\right ) \left (\mathit {Heaviside}\left (t -\pi \right )-\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{2 t +2 \pi }+{\mathrm e}^{4 t +2 \pi } \mathit {Heaviside}\left (t \right )+\left (-{\mathrm e}^{4 t}-{\mathrm e}^{4 \pi }\right ) \mathit {Heaviside}\left (t -\pi \right )+{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t \right )\right )}{2} \\ {\mathrm e}^{-2 t -2 \pi } \left (2 \sin \left (2 t \right ) \left (\mathit {Heaviside}\left (t -\pi \right )-\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{2 t +2 \pi }+{\mathrm e}^{4 t +2 \pi } \mathit {Heaviside}\left (t \right )+\left (-{\mathrm e}^{4 t}+{\mathrm e}^{4 \pi }\right ) \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t \right )\right ) \\ 2 \,{\mathrm e}^{-2 t -2 \pi } \left (2 \cos \left (2 t \right ) \left (\mathit {Heaviside}\left (t -\pi \right )-\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{2 t +2 \pi }+{\mathrm e}^{4 t +2 \pi } \mathit {Heaviside}\left (t \right )+\left (-{\mathrm e}^{4 t}-{\mathrm e}^{4 \pi }\right ) \mathit {Heaviside}\left (t -\pi \right )+{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t \right )\right ) \\ 4 \,{\mathrm e}^{-2 t -2 \pi } \left (-2 \sin \left (2 t \right ) \left (\mathit {Heaviside}\left (t -\pi \right )-\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{2 t +2 \pi }+{\mathrm e}^{4 t +2 \pi } \mathit {Heaviside}\left (t \right )+\left (-{\mathrm e}^{4 t}+{\mathrm e}^{4 \pi }\right ) \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t \right )\right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {\mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }}{2}-\frac {\mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{2 t -2 \pi }}{2}+\left (-\cos \left (2 t \right )+2\right ) \mathit {Heaviside}\left (t -\pi \right )+\frac {\left (8 \mathit {Heaviside}\left (t \right )-c_{4} \right ) \cos \left (2 t \right )}{8}+\frac {\left (4 \mathit {Heaviside}\left (t \right )-c_{1} \right ) {\mathrm e}^{-2 t}}{8}+\frac {\left (4 \mathit {Heaviside}\left (t \right )+c_{2} \right ) {\mathrm e}^{2 t}}{8}-\frac {c_{3} \sin \left (2 t \right )}{8}-2 \mathit {Heaviside}\left (t \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-\frac {c_{4}}{8}-\frac {c_{1}}{8}+\frac {c_{2}}{8} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\mathit {Dirac}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }}{2}+\mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }-\frac {\mathit {Dirac}\left (t -\pi \right ) {\mathrm e}^{2 t -2 \pi }}{2}-\mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{2 t -2 \pi }+2 \sin \left (2 t \right ) \mathit {Heaviside}\left (t -\pi \right )+\left (-\cos \left (2 t \right )+2\right ) \mathit {Dirac}\left (t -\pi \right )+\mathit {Dirac}\left (t \right ) \cos \left (2 t \right )-\frac {\left (8 \mathit {Heaviside}\left (t \right )-c_{4} \right ) \sin \left (2 t \right )}{4}+\frac {\mathit {Dirac}\left (t \right ) {\mathrm e}^{-2 t}}{2}-\frac {\left (4 \mathit {Heaviside}\left (t \right )-c_{1} \right ) {\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t} \mathit {Dirac}\left (t \right )}{2}+\frac {\left (4 \mathit {Heaviside}\left (t \right )+c_{2} \right ) {\mathrm e}^{2 t}}{4}-\frac {c_{3} \cos \left (2 t \right )}{4}-2 \mathit {Dirac}\left (t \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {c_{1}}{4}+\frac {c_{2}}{4}-\frac {c_{3}}{4} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-2 \mathit {Dirac}\left (1, t\right )+4 \cos \left (2 t \right ) \mathit {Heaviside}\left (t -\pi \right )+2 \,{\mathrm e}^{2 t} \mathit {Dirac}\left (t \right )+\frac {{\mathrm e}^{2 t} \mathit {Dirac}\left (1, t\right )}{2}+\frac {c_{3} \sin \left (2 t \right )}{2}-2 \mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{2 t -2 \pi }-2 \mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }-\frac {\left (8 \mathit {Heaviside}\left (t \right )-c_{4} \right ) \cos \left (2 t \right )}{2}+\frac {\left (4 \mathit {Heaviside}\left (t \right )-c_{1} \right ) {\mathrm e}^{-2 t}}{2}+\frac {\left (4 \mathit {Heaviside}\left (t \right )+c_{2} \right ) {\mathrm e}^{2 t}}{2}+2 \mathit {Dirac}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }-2 \mathit {Dirac}\left (t -\pi \right ) {\mathrm e}^{2 t -2 \pi }-2 \mathit {Dirac}\left (t \right ) {\mathrm e}^{-2 t}-\frac {\mathit {Dirac}\left (1, t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }}{2}-\frac {\mathit {Dirac}\left (1, t -\pi \right ) {\mathrm e}^{2 t -2 \pi }}{2}+4 \sin \left (2 t \right ) \mathit {Dirac}\left (t -\pi \right )+\left (-\cos \left (2 t \right )+2\right ) \mathit {Dirac}\left (1, t -\pi \right )+\mathit {Dirac}\left (1, t\right ) \cos \left (2 t \right )-4 \mathit {Dirac}\left (t \right ) \sin \left (2 t \right )+\frac {\mathit {Dirac}\left (1, t\right ) {\mathrm e}^{-2 t}}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d t}y^{\prime }\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {c_{4}}{2}-\frac {c_{1}}{2}+\frac {c_{2}}{2} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime \prime }=-2 \mathit {Dirac}\left (2, t\right )-8 \sin \left (2 t \right ) \mathit {Heaviside}\left (t -\pi \right )+6 \,{\mathrm e}^{2 t} \mathit {Dirac}\left (t \right )+3 \,{\mathrm e}^{2 t} \mathit {Dirac}\left (1, t\right )+\frac {{\mathrm e}^{2 t} \mathit {Dirac}\left (2, t\right )}{2}+c_{3} \cos \left (2 t \right )-4 \mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{2 t -2 \pi }+4 \mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }-\left (4 \mathit {Heaviside}\left (t \right )-c_{1} \right ) {\mathrm e}^{-2 t}+\left (4 \mathit {Heaviside}\left (t \right )+c_{2} \right ) {\mathrm e}^{2 t}+\left (8 \mathit {Heaviside}\left (t \right )-c_{4} \right ) \sin \left (2 t \right )-6 \mathit {Dirac}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }-6 \mathit {Dirac}\left (t -\pi \right ) {\mathrm e}^{2 t -2 \pi }-12 \mathit {Dirac}\left (t \right ) \cos \left (2 t \right )+6 \mathit {Dirac}\left (t \right ) {\mathrm e}^{-2 t}+3 \mathit {Dirac}\left (1, t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }-3 \mathit {Dirac}\left (1, t -\pi \right ) {\mathrm e}^{2 t -2 \pi }-3 \mathit {Dirac}\left (1, t\right ) {\mathrm e}^{-2 t}+12 \cos \left (2 t \right ) \mathit {Dirac}\left (t -\pi \right )-\frac {\mathit {Dirac}\left (2, t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }}{2}-\frac {\mathit {Dirac}\left (2, t -\pi \right ) {\mathrm e}^{2 t -2 \pi }}{2}+6 \sin \left (2 t \right ) \mathit {Dirac}\left (1, t -\pi \right )+\left (-\cos \left (2 t \right )+2\right ) \mathit {Dirac}\left (2, t -\pi \right )+\mathit {Dirac}\left (2, t\right ) \cos \left (2 t \right )-6 \mathit {Dirac}\left (1, t\right ) \sin \left (2 t \right )+\frac {\mathit {Dirac}\left (2, t\right ) {\mathrm e}^{-2 t}}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d t}y^{\prime \prime }\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =0, c_{2} =0, c_{3} =0, c_{4} =0\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }}{2}-\frac {\mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{2 t -2 \pi }}{2}+\left (-\cos \left (2 t \right )+2\right ) \mathit {Heaviside}\left (t -\pi \right )+\mathit {Heaviside}\left (t \right ) \left (\cos \left (2 t \right )+\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-2\right ) \end {array} \]

Maple trace

`Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 4; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 4; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`

\[ y \left (t \right ) = -\operatorname {Heaviside}\left (t -\pi \right ) \cosh \left (2 t -2 \pi \right )+\left (-\cos \left (2 t \right )+2\right ) \operatorname {Heaviside}\left (t -\pi \right )+\cos \left (2 t \right )+\cosh \left (2 t \right )-2 \]

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} e^{-2 (t+\pi )} \left (-1+e^{2 \pi }\right ) \left (-e^{2 \pi }+e^{4 t}\right ) & t>\pi \\ \frac {1}{2} \left (2 \cos (2 t)+e^{-2 t}+e^{2 t}-4\right ) & 0\leq t\leq \pi \\ \end {array} \\ \end {array} \]

4.42 problem Problem 14(b)

4.42.1 Maple step by step solution