Internal problem ID [5211]
Internal file name [OUTPUT/4704_Sunday_June_05_2022_03_03_43_PM_92798513/index.tex]
Book: Schaums Outline Differential Equations, 4th edition. Bronson and Costa. McGraw Hill
2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems.
page 248
Problem number: Problem 24.35.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_laplace"
Maple gives the following as the ode type
[[_3rd_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime }-y=5} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\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 ) \end {align*}
The given ode becomes an algebraic equation in the Laplace domain \[ s^{3} Y \left (s \right )-y^{\prime \prime }\left (0\right )-s y^{\prime }\left (0\right )-s^{2} y \left (0\right )-Y \left (s \right ) = \frac {5}{s}\tag {1} \] 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 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{3} Y \left (s \right )-Y \left (s \right ) = \frac {5}{s} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {5}{s \left (s^{3}-1\right )} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {5}{3 \left (s -1\right )}+\frac {5}{3 \left (s +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}+\frac {5}{3 \left (s +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}-\frac {5}{s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {5}{3 \left (s -1\right )}\right ) &= \frac {5 \,{\mathrm e}^{x}}{3}\\ \mathcal {L}^{-1}\left (\frac {5}{3 \left (s +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}\right ) &= \frac {5 \,{\mathrm e}^{-\frac {\left (1-i \sqrt {3}\right ) x}{2}}}{3}\\ \mathcal {L}^{-1}\left (\frac {5}{3 \left (s +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}\right ) &= \frac {5 \,{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}}}{3}\\ \mathcal {L}^{-1}\left (-\frac {5}{s}\right ) &= -5 \end {align*}
Adding the above results and simplifying gives \[ y=-5+\frac {10 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {5 \,{\mathrm e}^{x}}{3} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -5+\frac {10 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {5 \,{\mathrm e}^{x}}{3} \\ \end{align*}
Verification of solutions
\[ y = -5+\frac {10 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {5 \,{\mathrm e}^{x}}{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-y=5, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\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 (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=5+y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=5+y_{1}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 5 \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 5 \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \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 [-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{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}^{x}\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 [-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{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 \\ {} & {} & {\mathrm e}^{-\frac {x}{2}}\cdot \left (\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )\right )\cdot \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ \cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\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 (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -\frac {\sqrt {3}\, \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ \frac {\sqrt {3}\, \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ -\sin \left (\frac {\sqrt {3}\, x}{2}\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 (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\sqrt {3}\, \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) & {\mathrm e}^{-\frac {x}{2}} \left (\frac {\sqrt {3}\, \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) & -{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \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 (x \right )=\phi \left (x \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\sqrt {3}\, \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) & {\mathrm e}^{-\frac {x}{2}} \left (\frac {\sqrt {3}\, \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) & -{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & -\frac {1}{2} & -\frac {\sqrt {3}}{2} \\ 1 & -\frac {1}{2} & \frac {\sqrt {3}}{2} \\ 1 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{x}}{3}+\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} & \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {{\mathrm e}^{-\frac {x}{2}} \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{3} & \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{3} \\ \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{3} & \frac {{\mathrm e}^{x}}{3}+\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} & \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {{\mathrm e}^{-\frac {x}{2}} \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{3} \\ \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {{\mathrm e}^{-\frac {x}{2}} \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{3} & \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{3} & \frac {{\mathrm e}^{x}}{3}+\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\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 (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \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 (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\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 (x \right )=\left [\begin {array}{c} -5+\frac {10 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {5 \,{\mathrm e}^{x}}{3} \\ -\frac {5 \,{\mathrm e}^{-\frac {x}{2}} \left (\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )+\cos \left (\frac {\sqrt {3}\, x}{2}\right )-{\mathrm e}^{\frac {3 x}{2}}\right )}{3} \\ \frac {5 \,{\mathrm e}^{-\frac {x}{2}} \left ({\mathrm e}^{\frac {3 x}{2}}+\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )-\cos \left (\frac {\sqrt {3}\, x}{2}\right )\right )}{3} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+\left [\begin {array}{c} -5+\frac {10 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {5 \,{\mathrm e}^{x}}{3} \\ -\frac {5 \,{\mathrm e}^{-\frac {x}{2}} \left (\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )+\cos \left (\frac {\sqrt {3}\, x}{2}\right )-{\mathrm e}^{\frac {3 x}{2}}\right )}{3} \\ \frac {5 \,{\mathrm e}^{-\frac {x}{2}} \left ({\mathrm e}^{\frac {3 x}{2}}+\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )-\cos \left (\frac {\sqrt {3}\, x}{2}\right )\right )}{3} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-5-\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} -\frac {20}{3}\right ) \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{2} \sqrt {3}-c_{3} \right ) \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\left (5+3 c_{1} \right ) {\mathrm e}^{x}}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-\frac {c_{3} \sqrt {3}}{2}-\frac {c_{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} -\frac {20}{3}\right ) \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} -\frac {20}{3}\right ) \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{2} \sqrt {3}-c_{3} \right ) \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{4}-\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{2} \sqrt {3}-c_{3} \right ) \sqrt {3}\, \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{4}+\frac {\left (5+3 c_{1} \right ) {\mathrm e}^{x}}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {c_{3} \sqrt {3}}{4}+\frac {c_{2}}{4}-\frac {\left (c_{2} \sqrt {3}-c_{3} \right ) \sqrt {3}}{4}+c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} -\frac {20}{3}\right ) \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{4}-\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} -\frac {20}{3}\right ) \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{2} \sqrt {3}-c_{3} \right ) \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{2} \sqrt {3}-c_{3} \right ) \sqrt {3}\, \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{4}+\frac {\left (5+3 c_{1} \right ) {\mathrm e}^{x}}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {c_{3} \sqrt {3}}{4}+\frac {c_{2}}{4}+\frac {\left (c_{2} \sqrt {3}-c_{3} \right ) \sqrt {3}}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =0, c_{2} =0, c_{3} =0\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-5+\frac {10 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {5 \,{\mathrm e}^{x}}{3} \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`
✓ Solution by Maple
Time used: 0.422 (sec). Leaf size: 23
dsolve([diff(y(x),x$3)-y(x)=5,y(0) = 0, D(y)(0) = 0, (D@@2)(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = -5+\frac {5 \,{\mathrm e}^{x}}{3}+\frac {10 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 34
DSolve[{y'''[x]-y[x]==5,{y[0]==0,y'[0]==0,y''[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {5}{3} \left (e^x+2 e^{-x/2} \cos \left (\frac {\sqrt {3} x}{2}\right )-3\right ) \]