Internal problem ID [14854]
Internal file name [OUTPUT/14534_Monday_April_08_2024_06_26_40_AM_28449704/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Chapter 4 review exercises, page 219
Problem number: 34.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_3rd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime \prime }+3 y^{\prime \prime }-9 y^{\prime }+5 y={\mathrm e}^{t}} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime }+3 y^{\prime \prime }-9 y^{\prime }+5 y = 0 \] The characteristic equation is \[ \lambda ^{3}+3 \lambda ^{2}-9 \lambda +5 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -5\\ \lambda _2 &= 1\\ \lambda _3 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{t} t +{\mathrm e}^{-5 t} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{t} \\ y_2 &= t \,{\mathrm e}^{t} \\ y_3 &= {\mathrm e}^{-5 t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+3 y^{\prime \prime }-9 y^{\prime }+5 y = {\mathrm e}^{t} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{t} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{t}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{t \,{\mathrm e}^{t}, {\mathrm e}^{t}, {\mathrm e}^{-5 t}\} \] Since \({\mathrm e}^{t}\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t \,{\mathrm e}^{t}\}] \] Since \(t \,{\mathrm e}^{t}\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{{\mathrm e}^{t} t^{2}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{1} {\mathrm e}^{t} t^{2} \] The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives \[ 12 A_{1} {\mathrm e}^{t} = {\mathrm e}^{t} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{12}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{t} t^{2}}{12} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{t} t +{\mathrm e}^{-5 t} c_{3}\right ) + \left (\frac {{\mathrm e}^{t} t^{2}}{12}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{t} t +{\mathrm e}^{-5 t} c_{3} +\frac {{\mathrm e}^{t} t^{2}}{12} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{t} t +{\mathrm e}^{-5 t} c_{3} +\frac {{\mathrm e}^{t} t^{2}}{12} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+3 y^{\prime \prime }-9 y^{\prime }+5 y={\mathrm e}^{t} \\ \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 )={\mathrm e}^{t}-3 y_{3}\left (t \right )+9 y_{2}\left (t \right )-5 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 )={\mathrm e}^{t}-3 y_{3}\left (t \right )+9 y_{2}\left (t \right )-5 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 \\ -5 & 9 & -3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ {\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 \\ {\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 \\ -5 & 9 & -3 \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 [-5, \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-5, \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-5 t}\cdot \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} 1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (t \right )={\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =1\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{\lambda t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} t \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (t \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda t} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda t} A \right )\cdot \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda t} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda t} \left (\lambda t {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (t \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} 1 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & 9 & -3 \end {array}\right ]-1\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} 1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{t}\cdot \left (t \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \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} \frac {{\mathrm e}^{-5 t}}{25} & {\mathrm e}^{t} & {\mathrm e}^{t} \left (t -1\right ) \\ -\frac {{\mathrm e}^{-5 t}}{5} & {\mathrm e}^{t} & t \,{\mathrm e}^{t} \\ {\mathrm e}^{-5 t} & {\mathrm e}^{t} & 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}^{-5 t}}{25} & {\mathrm e}^{t} & {\mathrm e}^{t} \left (t -1\right ) \\ -\frac {{\mathrm e}^{-5 t}}{5} & {\mathrm e}^{t} & t \,{\mathrm e}^{t} \\ {\mathrm e}^{-5 t} & {\mathrm e}^{t} & t \,{\mathrm e}^{t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{25} & 1 & -1 \\ -\frac {1}{5} & 1 & 0 \\ 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}{ccc} -{\mathrm e}^{t} \left (t -1\right ) & \frac {\left (24 \,{\mathrm e}^{6 t} t +{\mathrm e}^{6 t}-1\right ) {\mathrm e}^{-5 t}}{30} & \frac {{\mathrm e}^{-5 t} \left (-1+6 t \right ) {\mathrm e}^{6 t}}{30}+\frac {{\mathrm e}^{-5 t}}{30} \\ -t \,{\mathrm e}^{t} & \frac {\left (24 t +25\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{30}+\frac {{\mathrm e}^{-5 t}}{6} & \frac {\left (6 t +5\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{30}-\frac {{\mathrm e}^{-5 t}}{6} \\ -t \,{\mathrm e}^{t} & \frac {\left (24 t +25\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{30}-\frac {5 \,{\mathrm e}^{-5 t}}{6} & \frac {\left (6 t +5\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{30}+\frac {5 \,{\mathrm e}^{-5 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 (18 t^{2}-6 t +1\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{180}-\frac {{\mathrm e}^{-5 t}}{180} \\ \frac {\left (18 t^{2}+30 t -5\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{180}+\frac {{\mathrm e}^{-5 t}}{36} \\ \frac {\left (18 t^{2}+30 t +25\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{180}-\frac {5 \,{\mathrm e}^{-5 t}}{36} \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 {\left (18 t^{2}-6 t +1\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{180}-\frac {{\mathrm e}^{-5 t}}{180} \\ \frac {\left (18 t^{2}+30 t -5\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{180}+\frac {{\mathrm e}^{-5 t}}{36} \\ \frac {\left (18 t^{2}+30 t +25\right ) {\mathrm e}^{-5 t} {\mathrm e}^{6 t}}{180}-\frac {5 \,{\mathrm e}^{-5 t}}{36} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-5 t} \left (\left (t^{2}+\left (10 c_{3} -\frac {1}{3}\right ) t +10 c_{2} -10 c_{3} +\frac {1}{18}\right ) {\mathrm e}^{6 t}+\frac {2 c_{1}}{5}-\frac {1}{18}\right )}{10} \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: 0.0 (sec). Leaf size: 30
dsolve(diff(y(t),t$3)+3*diff(y(t),t$2)-9*diff(y(t),t)+5*y(t)=exp(t),y(t), singsol=all)
\[ y \left (t \right ) = \frac {\left (\left (12 c_{3} t +t^{2}+12 c_{1} \right ) {\mathrm e}^{6 t}+12 c_{2} \right ) {\mathrm e}^{-5 t}}{12} \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 39
DSolve[y'''[t]+3*y''[t]-9*y'[t]+5*y[t]==Exp[t],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^t \left (\frac {t^2}{12}+\left (-\frac {1}{36}+c_3\right ) t+\frac {1}{216}+c_2\right )+c_1 e^{-5 t} \]