problem 40

Internal problem ID [6665]
Internal ﬁle name [OUTPUT/5913_Sunday_June_05_2022_04_01_20_PM_70541589/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. 7.2.2 TRANSFORMS OF DERIVATIVES Page 289
Problem number: 40.
ODE order: 3.
ODE degree: 1.

\[ \boxed {y^{\prime \prime \prime }+2 y^{\prime \prime }-y^{\prime }-2 y=\sin \left (3 t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 1] \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 \[ s^{3} Y \left (s \right )-y^{\prime \prime }\left (0\right )-s y^{\prime }\left (0\right )-s^{2} y \left (0\right )+2 s^{2} Y \left (s \right )-2 y^{\prime }\left (0\right )-2 s y \left (0\right )-s Y \left (s \right )+y \left (0\right )-2 Y \left (s \right ) = \frac {3}{s^{2}+9}\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=0\\ y^{\prime \prime }\left (0\right )&=1 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \[ s^{3} Y \left (s \right )-1+2 s^{2} Y \left (s \right )-s Y \left (s \right )-2 Y \left (s \right ) = \frac {3}{s^{2}+9} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {s^{2}+12}{\left (s^{2}+9\right ) \left (s^{3}+2 s^{2}-s -2\right )} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {13}{60 \left (s -1\right )}+\frac {\frac {3}{260}+\frac {i}{130}}{s -3 i}+\frac {\frac {3}{260}-\frac {i}{130}}{s +3 i}+\frac {16}{39 \left (s +2\right )}-\frac {13}{20 \left (s +1\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {13}{60 \left (s -1\right )}\right ) &= \frac {13 \,{\mathrm e}^{t}}{60}\\ \mathcal {L}^{-1}\left (\frac {\frac {3}{260}+\frac {i}{130}}{s -3 i}\right ) &= \left (\frac {3}{260}+\frac {i}{130}\right ) {\mathrm e}^{3 i t}\\ \mathcal {L}^{-1}\left (\frac {\frac {3}{260}-\frac {i}{130}}{s +3 i}\right ) &= \left (\frac {3}{260}-\frac {i}{130}\right ) {\mathrm e}^{-3 i t}\\ \mathcal {L}^{-1}\left (\frac {16}{39 \left (s +2\right )}\right ) &= \frac {16 \,{\mathrm e}^{-2 t}}{39}\\ \mathcal {L}^{-1}\left (-\frac {13}{20 \left (s +1\right )}\right ) &= -\frac {13 \,{\mathrm e}^{-t}}{20} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {3 \cos \left (3 t \right )}{130}-\frac {\sin \left (3 t \right )}{65}-\frac {13 \cosh \left (t \right )}{30}+\frac {13 \sinh \left (t \right )}{15}+\frac {16 \,{\mathrm e}^{-2 t}}{39} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {3 \cos \left (3 t \right )}{130}-\frac {\sin \left (3 t \right )}{65}-\frac {13 \cosh \left (t \right )}{30}+\frac {13 \sinh \left (t \right )}{15}+\frac {16 \,{\mathrm e}^{-2 t}}{39} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {3 \cos \left (3 t \right )}{130}-\frac {\sin \left (3 t \right )}{65}-\frac {13 \cosh \left (t \right )}{30}+\frac {13 \sinh \left (t \right )}{15}+\frac {16 \,{\mathrm e}^{-2 t}}{39} \] Veriﬁed OK.

5.10.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }+2 y^{\prime \prime }-y^{\prime }-2 y=\sin \left (3 t \right ), y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\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 (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 )=y^{\prime \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 (3 t \right )-2 y_{3}\left (t \right )+y_{2}\left (t \right )+2 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 )=\sin \left (3 t \right )-2 y_{3}\left (t \right )+y_{2}\left (t \right )+2 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 \\ 2 & 1 & -2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ \sin \left (3 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 (3 t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 1 & -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 [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} \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}{4} \\ -\frac {1}{2} \\ 1 \end {array}\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}}_{2}={\mathrm e}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\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}}_{3}={\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 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 (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}+{\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} \frac {{\mathrm e}^{-2 t}}{4} & {\mathrm e}^{-t} & {\mathrm e}^{t} \\ -\frac {{\mathrm e}^{-2 t}}{2} & -{\mathrm e}^{-t} & {\mathrm e}^{t} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{-t} & {\mathrm e}^{t} \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}{ccc} \frac {{\mathrm e}^{-2 t}}{4} & {\mathrm e}^{-t} & {\mathrm e}^{t} \\ -\frac {{\mathrm e}^{-2 t}}{2} & -{\mathrm e}^{-t} & {\mathrm e}^{t} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{-t} & {\mathrm e}^{t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{4} & 1 & 1 \\ -\frac {1}{2} & -1 & 1 \\ 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} \frac {\left ({\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{3} & -\frac {{\mathrm e}^{-t}}{2}+\frac {{\mathrm e}^{t}}{2} & \frac {\left ({\mathrm e}^{3 t}-3 \,{\mathrm e}^{t}+2\right ) {\mathrm e}^{-2 t}}{6} \\ \frac {\left ({\mathrm e}^{3 t}-3 \,{\mathrm e}^{t}+2\right ) {\mathrm e}^{-2 t}}{3} & \frac {{\mathrm e}^{-t}}{2}+\frac {{\mathrm e}^{t}}{2} & \frac {\left ({\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-4\right ) {\mathrm e}^{-2 t}}{6} \\ \frac {\left ({\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-4\right ) {\mathrm e}^{-2 t}}{3} & -\frac {{\mathrm e}^{-t}}{2}+\frac {{\mathrm e}^{t}}{2} & \frac {\left ({\mathrm e}^{3 t}-3 \,{\mathrm e}^{t}+8\right ) {\mathrm e}^{-2 t}}{6} \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 {\left (4 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-13 \,{\mathrm e}^{3 t}-6 \,{\mathrm e}^{2 t} \cos \left (3 t \right )+39 \,{\mathrm e}^{t}-20\right ) {\mathrm e}^{-2 t}}{260} \\ -\frac {\left (18 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-13 \,{\mathrm e}^{3 t}+12 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-39 \,{\mathrm e}^{t}+40\right ) {\mathrm e}^{-2 t}}{260} \\ \frac {\left (36 \,{\mathrm e}^{2 t} \sin \left (3 t \right )+13 \,{\mathrm e}^{3 t}-54 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-39 \,{\mathrm e}^{t}+80\right ) {\mathrm e}^{-2 t}}{260} \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 [\begin {array}{c} -\frac {\left (4 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-13 \,{\mathrm e}^{3 t}-6 \,{\mathrm e}^{2 t} \cos \left (3 t \right )+39 \,{\mathrm e}^{t}-20\right ) {\mathrm e}^{-2 t}}{260} \\ -\frac {\left (18 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-13 \,{\mathrm e}^{3 t}+12 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-39 \,{\mathrm e}^{t}+40\right ) {\mathrm e}^{-2 t}}{260} \\ \frac {\left (36 \,{\mathrm e}^{2 t} \sin \left (3 t \right )+13 \,{\mathrm e}^{3 t}-54 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-39 \,{\mathrm e}^{t}+80\right ) {\mathrm e}^{-2 t}}{260} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {\left (-260 c_{3} {\mathrm e}^{3 t}+4 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-13 \,{\mathrm e}^{3 t}-6 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-260 c_{2} {\mathrm e}^{t}+39 \,{\mathrm e}^{t}-65 c_{1} -20\right ) {\mathrm e}^{-2 t}}{260} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{3} +c_{2} +\frac {c_{1}}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\left (-780 c_{3} {\mathrm e}^{3 t}+26 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-39 \,{\mathrm e}^{3 t}-260 c_{2} {\mathrm e}^{t}+39 \,{\mathrm e}^{t}\right ) {\mathrm e}^{-2 t}}{260}+\frac {\left (-260 c_{3} {\mathrm e}^{3 t}+4 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-13 \,{\mathrm e}^{3 t}-6 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-260 c_{2} {\mathrm e}^{t}+39 \,{\mathrm e}^{t}-65 c_{1} -20\right ) {\mathrm e}^{-2 t}}{130} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{3} -c_{2} -\frac {c_{1}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\left (-2340 c_{3} {\mathrm e}^{3 t}+52 \,{\mathrm e}^{2 t} \sin \left (3 t \right )+78 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-117 \,{\mathrm e}^{3 t}-260 c_{2} {\mathrm e}^{t}+39 \,{\mathrm e}^{t}\right ) {\mathrm e}^{-2 t}}{260}+\frac {\left (-780 c_{3} {\mathrm e}^{3 t}+26 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-39 \,{\mathrm e}^{3 t}-260 c_{2} {\mathrm e}^{t}+39 \,{\mathrm e}^{t}\right ) {\mathrm e}^{-2 t}}{65}-\frac {\left (-260 c_{3} {\mathrm e}^{3 t}+4 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-13 \,{\mathrm e}^{3 t}-6 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-260 c_{2} {\mathrm e}^{t}+39 \,{\mathrm e}^{t}-65 c_{1} -20\right ) {\mathrm e}^{-2 t}}{65} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=c_{3} +c_{2} +c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {4}{3}, c_{2} =-\frac {1}{2}, c_{3} =\frac {1}{6}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\left (12 \,{\mathrm e}^{2 t} \sin \left (3 t \right )-169 \,{\mathrm e}^{3 t}-18 \,{\mathrm e}^{2 t} \cos \left (3 t \right )+507 \,{\mathrm e}^{t}-320\right ) {\mathrm e}^{-2 t}}{780} \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`

\[ y \left (t \right ) = \frac {3 \cos \left (3 t \right )}{130}-\frac {\sin \left (3 t \right )}{65}-\frac {13 \cosh \left (t \right )}{30}+\frac {13 \sinh \left (t \right )}{15}+\frac {16 \,{\mathrm e}^{-2 t}}{39} \]

\[ y(t)\to \frac {1}{780} \left (e^{-2 t} \left (-507 e^t+169 e^{3 t}+320\right )-12 \sin (3 t)+18 \cos (3 t)\right ) \]

5.10 problem 40

5.10.1 Maple step by step solution