problem 39

Internal problem ID [6664]
Internal ﬁle name [OUTPUT/5912_Sunday_June_05_2022_04_01_18_PM_63338079/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: 39.
ODE order: 3.
ODE degree: 1.

\[ \boxed {2 y^{\prime \prime \prime }+3 y^{\prime \prime }-3 y^{\prime }-2 y={\mathrm e}^{-t}} \] 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 \[ 2 s^{3} Y \left (s \right )-2 y^{\prime \prime }\left (0\right )-2 s y^{\prime }\left (0\right )-2 s^{2} y \left (0\right )+3 s^{2} Y \left (s \right )-3 y^{\prime }\left (0\right )-3 s y \left (0\right )-3 s Y \left (s \right )+3 y \left (0\right )-2 Y \left (s \right ) = \frac {1}{s +1}\tag {1} \] 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 \[ 2 s^{3} Y \left (s \right )-2+3 s^{2} Y \left (s \right )-3 s Y \left (s \right )-2 Y \left (s \right ) = \frac {1}{s +1} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {3+2 s}{\left (s +1\right ) \left (2 s^{3}+3 s^{2}-3 s -2\right )} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{9 s +18}-\frac {8}{9 \left (s +\frac {1}{2}\right )}+\frac {5}{18 \left (s -1\right )}+\frac {1}{2 s +2} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{9 s +18}\right ) &= \frac {{\mathrm e}^{-2 t}}{9}\\ \mathcal {L}^{-1}\left (-\frac {8}{9 \left (s +\frac {1}{2}\right )}\right ) &= -\frac {8 \,{\mathrm e}^{-\frac {t}{2}}}{9}\\ \mathcal {L}^{-1}\left (\frac {5}{18 \left (s -1\right )}\right ) &= \frac {5 \,{\mathrm e}^{t}}{18}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s +2}\right ) &= \frac {{\mathrm e}^{-t}}{2} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {7 \cosh \left (t \right )}{9}-\frac {2 \sinh \left (t \right )}{9}+\frac {{\mathrm e}^{-2 t}}{9}-\frac {8 \,{\mathrm e}^{-\frac {t}{2}}}{9} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {7 \cosh \left (t \right )}{9}-\frac {2 \sinh \left (t \right )}{9}+\frac {{\mathrm e}^{-2 t}}{9}-\frac {8 \,{\mathrm e}^{-\frac {t}{2}}}{9} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {7 \cosh \left (t \right )}{9}-\frac {2 \sinh \left (t \right )}{9}+\frac {{\mathrm e}^{-2 t}}{9}-\frac {8 \,{\mathrm e}^{-\frac {t}{2}}}{9} \] Veriﬁed OK.

5.9.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 y^{\prime \prime \prime }+3 y^{\prime \prime }-3 y^{\prime }-2 y={\mathrm e}^{-t}, 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 } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-\frac {3 y^{\prime \prime }}{2}+\frac {3 y^{\prime }}{2}+y+\frac {{\mathrm e}^{-t}}{2} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+\frac {3 y^{\prime \prime }}{2}-\frac {3 y^{\prime }}{2}-y=\frac {{\mathrm e}^{-t}}{2} \\ \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 )=\frac {{\mathrm e}^{-t}}{2}-\frac {3 y_{3}\left (t \right )}{2}+\frac {3 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 )=\frac {{\mathrm e}^{-t}}{2}-\frac {3 y_{3}\left (t \right )}{2}+\frac {3 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 \\ 1 & \frac {3}{2} & -\frac {3}{2} \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ \frac {{\mathrm e}^{-t}}{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ \frac {{\mathrm e}^{-t}}{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & \frac {3}{2} & -\frac {3}{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 [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 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 [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} 4 \\ -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}}_{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} & 4 \,{\mathrm e}^{-\frac {t}{2}} & {\mathrm e}^{t} \\ -\frac {{\mathrm e}^{-2 t}}{2} & -2 \,{\mathrm e}^{-\frac {t}{2}} & {\mathrm e}^{t} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{-\frac {t}{2}} & {\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} & 4 \,{\mathrm e}^{-\frac {t}{2}} & {\mathrm e}^{t} \\ -\frac {{\mathrm e}^{-2 t}}{2} & -2 \,{\mathrm e}^{-\frac {t}{2}} & {\mathrm e}^{t} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{-\frac {t}{2}} & {\mathrm e}^{t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{4} & 4 & 1 \\ -\frac {1}{2} & -2 & 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 (2 \,{\mathrm e}^{3 t}+8 \,{\mathrm e}^{\frac {3 t}{2}}-1\right ) {\mathrm e}^{-2 t}}{9} & \frac {\left (5 \,{\mathrm e}^{3 t}-4 \,{\mathrm e}^{\frac {3 t}{2}}-1\right ) {\mathrm e}^{-2 t}}{9} & \frac {2 \left ({\mathrm e}^{3 t}-2 \,{\mathrm e}^{\frac {3 t}{2}}+1\right ) {\mathrm e}^{-2 t}}{9} \\ \frac {2 \left ({\mathrm e}^{3 t}-2 \,{\mathrm e}^{\frac {3 t}{2}}+1\right ) {\mathrm e}^{-2 t}}{9} & \frac {\left (5 \,{\mathrm e}^{3 t}+2 \,{\mathrm e}^{\frac {3 t}{2}}+2\right ) {\mathrm e}^{-2 t}}{9} & \frac {2 \left ({\mathrm e}^{3 t}+{\mathrm e}^{\frac {3 t}{2}}-2\right ) {\mathrm e}^{-2 t}}{9} \\ \frac {2 \left ({\mathrm e}^{3 t}+{\mathrm e}^{\frac {3 t}{2}}-2\right ) {\mathrm e}^{-2 t}}{9} & \frac {\left (5 \,{\mathrm e}^{3 t}-{\mathrm e}^{\frac {3 t}{2}}-4\right ) {\mathrm e}^{-2 t}}{9} & \frac {\left (2 \,{\mathrm e}^{3 t}-{\mathrm e}^{\frac {3 t}{2}}+8\right ) {\mathrm e}^{-2 t}}{9} \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 ({\mathrm e}^{3 t}-8 \,{\mathrm e}^{\frac {3 t}{2}}+9 \,{\mathrm e}^{t}-2\right ) {\mathrm e}^{-2 t}}{18} \\ -\frac {\left (-{\mathrm e}^{3 t}-4 \,{\mathrm e}^{\frac {3 t}{2}}+9 \,{\mathrm e}^{t}-4\right ) {\mathrm e}^{-2 t}}{18} \\ \frac {\left ({\mathrm e}^{3 t}-2 \,{\mathrm e}^{\frac {3 t}{2}}+9 \,{\mathrm e}^{t}-8\right ) {\mathrm e}^{-2 t}}{18} \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 ({\mathrm e}^{3 t}-8 \,{\mathrm e}^{\frac {3 t}{2}}+9 \,{\mathrm e}^{t}-2\right ) {\mathrm e}^{-2 t}}{18} \\ -\frac {\left (-{\mathrm e}^{3 t}-4 \,{\mathrm e}^{\frac {3 t}{2}}+9 \,{\mathrm e}^{t}-4\right ) {\mathrm e}^{-2 t}}{18} \\ \frac {\left ({\mathrm e}^{3 t}-2 \,{\mathrm e}^{\frac {3 t}{2}}+9 \,{\mathrm e}^{t}-8\right ) {\mathrm e}^{-2 t}}{18} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (36 c_{3} {\mathrm e}^{3 t}+2 \,{\mathrm e}^{3 t}+144 c_{2} {\mathrm e}^{\frac {3 t}{2}}-16 \,{\mathrm e}^{\frac {3 t}{2}}+18 \,{\mathrm e}^{t}+9 c_{1} -4\right ) {\mathrm e}^{-2 t}}{36} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{3} +4 c_{2} +\frac {c_{1}}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (108 c_{3} {\mathrm e}^{3 t}+6 \,{\mathrm e}^{3 t}+216 c_{2} {\mathrm e}^{\frac {3 t}{2}}-24 \,{\mathrm e}^{\frac {3 t}{2}}+18 \,{\mathrm e}^{t}\right ) {\mathrm e}^{-2 t}}{36}-\frac {\left (36 c_{3} {\mathrm e}^{3 t}+2 \,{\mathrm e}^{3 t}+144 c_{2} {\mathrm e}^{\frac {3 t}{2}}-16 \,{\mathrm e}^{\frac {3 t}{2}}+18 \,{\mathrm e}^{t}+9 c_{1} -4\right ) {\mathrm e}^{-2 t}}{18} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{3} -2 c_{2} -\frac {c_{1}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (324 c_{3} {\mathrm e}^{3 t}+18 \,{\mathrm e}^{3 t}+324 c_{2} {\mathrm e}^{\frac {3 t}{2}}-36 \,{\mathrm e}^{\frac {3 t}{2}}+18 \,{\mathrm e}^{t}\right ) {\mathrm e}^{-2 t}}{36}-\frac {\left (108 c_{3} {\mathrm e}^{3 t}+6 \,{\mathrm e}^{3 t}+216 c_{2} {\mathrm e}^{\frac {3 t}{2}}-24 \,{\mathrm e}^{\frac {3 t}{2}}+18 \,{\mathrm e}^{t}\right ) {\mathrm e}^{-2 t}}{9}+\frac {\left (36 c_{3} {\mathrm e}^{3 t}+2 \,{\mathrm e}^{3 t}+144 c_{2} {\mathrm e}^{\frac {3 t}{2}}-16 \,{\mathrm e}^{\frac {3 t}{2}}+18 \,{\mathrm e}^{t}+9 c_{1} -4\right ) {\mathrm e}^{-2 t}}{9} \\ \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 {8}{9}, c_{2} =-\frac {1}{9}, c_{3} =\frac {2}{9}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (5 \,{\mathrm e}^{3 t}-16 \,{\mathrm e}^{\frac {3 t}{2}}+9 \,{\mathrm e}^{t}+2\right ) {\mathrm e}^{-2 t}}{18} \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`

\[ y \left (t \right ) = \frac {7 \cosh \left (t \right )}{9}-\frac {2 \sinh \left (t \right )}{9}+\frac {{\mathrm e}^{-2 t}}{9}-\frac {8 \,{\mathrm e}^{-\frac {t}{2}}}{9} \]

\[ y(t)\to \frac {1}{18} e^{-2 t} \left (9 e^t-16 e^{3 t/2}+5 e^{3 t}+2\right ) \]

5.9 problem 39

5.9.1 Maple step by step solution