19.61 problem section 9.3, problem 61
Internal
problem
ID
[2208]
Book
:
Elementary
differential
equations
with
boundary
value
problems.
William
F.
Trench.
Brooks/Cole
2001
Section
:
Chapter
9
Introduction
to
Linear
Higher
Order
Equations.
Section
9.3.
Undetermined
Coefficients
for
Higher
Order
Equations.
Page
495
Problem
number
:
section
9.3,
problem
61
Date
solved
:
Thursday, October 17, 2024 at 02:21:43 AM
CAS
classification
:
[[_3rd_order, _linear, _nonhomogeneous]]
Solve
\begin{align*} y^{\prime \prime \prime }+y^{\prime \prime }-2 y&=-{\mathrm e}^{3 x} \left (17 x^{2}+67 x +9\right ) \end{align*}
19.61.1 Solved as higher order constant coeff ode
Time used: 0.115 (sec)
The characteristic equation is
\[ \lambda ^{3}+\lambda ^{2}-2 = 0 \]
The roots of the above equation are
\begin{align*} \lambda _1 &= 1\\ \lambda _2 &= -1-i\\ \lambda _3 &= -1+i \end{align*}
Therefore the homogeneous solution is
\[ y_h(x)={\mathrm e}^{x} c_1 +{\mathrm e}^{\left (-1-i\right ) x} c_2 +{\mathrm e}^{\left (-1+i\right ) x} c_3 \]
The fundamental set of solutions for the
homogeneous solution are the following
\begin{align*} y_1 &= {\mathrm e}^{x}\\ y_2 &= {\mathrm e}^{\left (-1-i\right ) x}\\ y_3 &= {\mathrm e}^{\left (-1+i\right ) x} \end{align*}
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
\[ y^{\prime \prime \prime }+y^{\prime \prime }-2 y = 0 \]
Now the particular solution to the given ODE is found
\[
y^{\prime \prime \prime }+y^{\prime \prime }-2 y = -{\mathrm e}^{3 x} \left (17 x^{2}+67 x +9\right )
\]
The particular solution
is now found using the method of undetermined coefficients.
Looking at the RHS of the ode, which is
\[ -{\mathrm e}^{3 x} \left (17 x^{2}+67 x +9\right ) \]
Shows that the corresponding undetermined set of
the basis functions (UC_set) for the trial solution is
\[ [\{{\mathrm e}^{3 x} x, {\mathrm e}^{3 x} x^{2}, {\mathrm e}^{3 x}\}] \]
While the set of the basis functions for
the homogeneous solution found earlier is
\[ \{{\mathrm e}^{x}, {\mathrm e}^{\left (-1-i\right ) x}, {\mathrm e}^{\left (-1+i\right ) x}\} \]
Since there is no duplication between the basis
function in the UC_set and the basis functions of the homogeneous solution, the trial
solution is a linear combination of all the basis in the UC_set.
\[
y_p = A_{1} {\mathrm e}^{3 x} x +A_{2} {\mathrm e}^{3 x} x^{2}+A_{3} {\mathrm e}^{3 x}
\]
The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) 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
\[
34 A_{1} {\mathrm e}^{3 x} x +33 A_{1} {\mathrm e}^{3 x}+34 A_{2} {\mathrm e}^{3 x} x^{2}+66 A_{2} {\mathrm e}^{3 x} x +20 A_{2} {\mathrm e}^{3 x}+34 A_{3} {\mathrm e}^{3 x} = -{\mathrm e}^{3 x} \left (17 x^{2}+67 x +9\right )
\]
Solving for the
unknowns by comparing coefficients results in
\[ \left [A_{1} = -1, A_{2} = -{\frac {1}{2}}, A_{3} = 1\right ] \]
Substituting the above back in the
above trial solution \(y_p\), gives the particular solution
\[
y_p = -{\mathrm e}^{3 x} x -\frac {{\mathrm e}^{3 x} x^{2}}{2}+{\mathrm e}^{3 x}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left ({\mathrm e}^{x} c_1 +{\mathrm e}^{\left (-1-i\right ) x} c_2 +{\mathrm e}^{\left (-1+i\right ) x} c_3\right ) + \left (-{\mathrm e}^{3 x} x -\frac {{\mathrm e}^{3 x} x^{2}}{2}+{\mathrm e}^{3 x}\right ) \\
\end{align*}
19.61.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )+\frac {d^{2}}{d x^{2}}y \left (x \right )-2 y \left (x \right )=-{\mathrm e}^{3 x} \left (17 x^{2}+67 x +9\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right ) \\ \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 \left (x \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=\frac {d}{d x}y \left (x \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d x}y_{3}\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y_{3}\left (x \right )=-17 x^{2} {\mathrm e}^{3 x}-67 x \,{\mathrm e}^{3 x}-y_{3}\left (x \right )+2 y_{1}\left (x \right )-9 \,{\mathrm e}^{3 x} \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=\frac {d}{d x}y_{1}\left (x \right ), y_{3}\left (x \right )=\frac {d}{d x}y_{2}\left (x \right ), \frac {d}{d x}y_{3}\left (x \right )=-17 x^{2} {\mathrm e}^{3 x}-67 x \,{\mathrm e}^{3 x}-y_{3}\left (x \right )+2 y_{1}\left (x \right )-9 \,{\mathrm e}^{3 x}\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} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 0 & -1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ -17 x^{2} {\mathrm e}^{3 x}-67 x \,{\mathrm e}^{3 x}-9 \,{\mathrm e}^{3 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ -17 x^{2} {\mathrm e}^{3 x}-67 x \,{\mathrm e}^{3 x}-9 \,{\mathrm e}^{3 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 0 & -1 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}\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 [-1-\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [-1+\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{2} \\ -\frac {1}{2}-\frac {\mathrm {I}}{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 [-1-\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-1-\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\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 \\ {} & {} & {\mathrm e}^{-x}\cdot \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \left (-\frac {1}{2}+\frac {\mathrm {I}}{2}\right ) \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \cos \left (x \right )-\mathrm {I} \sin \left (x \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}^{-x}\cdot \left [\begin {array}{c} -\frac {\sin \left (x \right )}{2} \\ -\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {\cos \left (x \right )}{2} \\ \frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ -\sin \left (x \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 )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+\mathit {C3} {\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} & -\frac {{\mathrm e}^{-x} \sin \left (x \right )}{2} & -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2} \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \left (-\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) & {\mathrm e}^{-x} \left (\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \cos \left (x \right ) & -{\mathrm e}^{-x} \sin \left (x \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 \phi \left (0\right )^{-1} \\ {} & \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} & -\frac {{\mathrm e}^{-x} \sin \left (x \right )}{2} & -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2} \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \left (-\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) & {\mathrm e}^{-x} \left (\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \cos \left (x \right ) & -{\mathrm e}^{-x} \sin \left (x \right ) \end {array}\right ]\cdot \left [\begin {array}{ccc} 1 & 0 & -\frac {1}{2} \\ 1 & -\frac {1}{2} & \frac {1}{2} \\ 1 & 1 & 0 \end {array}\right ]^{-1} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {\left (3 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (-2 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (-\cos \left (x \right )-2 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{x}}{5} \\ \frac {\left (-4 \sin \left (x \right )-2 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (3 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (-\cos \left (x \right )+3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{x}}{5} \\ \frac {\left (-2 \cos \left (x \right )+6 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (-4 \sin \left (x \right )-2 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (4 \cos \left (x \right )-2 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{x}}{5} \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} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left (\frac {d}{d x}\Phi \left (x \right )\right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}\Phi \left (x \right )\right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\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 \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\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 \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )=\Phi \left (x \right )^{-1}\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}\Phi \left (s \right )^{-1}\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}\Phi \left (s \right )^{-1}\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} \frac {\left (-x^{2}-2 x +2\right ) {\mathrm e}^{3 x}}{2}+\frac {\left (3 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}-\frac {8 \,{\mathrm e}^{x}}{5} \\ \frac {\left (-3 x^{2}-8 x +4\right ) {\mathrm e}^{3 x}}{2}+\frac {2 \left (-\cos \left (x \right )-2 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}-\frac {8 \,{\mathrm e}^{x}}{5} \\ \frac {\left (-9 x^{2}-30 x +4\right ) {\mathrm e}^{3 x}}{2}+\frac {2 \left (-\cos \left (x \right )+3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}-\frac {8 \,{\mathrm e}^{x}}{5} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+\left [\begin {array}{c} \frac {\left (-x^{2}-2 x +2\right ) {\mathrm e}^{3 x}}{2}+\frac {\left (3 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}-\frac {8 \,{\mathrm e}^{x}}{5} \\ \frac {\left (-3 x^{2}-8 x +4\right ) {\mathrm e}^{3 x}}{2}+\frac {2 \left (-\cos \left (x \right )-2 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}-\frac {8 \,{\mathrm e}^{x}}{5} \\ \frac {\left (-9 x^{2}-30 x +4\right ) {\mathrm e}^{3 x}}{2}+\frac {2 \left (-\cos \left (x \right )+3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}-\frac {8 \,{\mathrm e}^{x}}{5} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\left (\left (-5 \mathit {C3} +6\right ) \cos \left (x \right )+\left (-5 \mathit {C2} +2\right ) \sin \left (x \right )\right ) {\mathrm e}^{-x}}{10}+\frac {\left (-x^{2}-2 x +2\right ) {\mathrm e}^{3 x}}{2}+\frac {{\mathrm e}^{x} \left (5 \mathit {C1} -8\right )}{5} \end {array} \]
19.61.3 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`
19.61.4 Maple dsolve solution
Solving time : 0.007
(sec)
Leaf size : 41
dsolve(diff(diff(diff(y(x),x),x),x)+diff(diff(y(x),x),x)-2*y(x) = -exp(3*x)*(17*x^2+67*x+9),
y(x),singsol=all)
\[
y = -\frac {\left (\left (x^{2}+2 x -2\right ) {\mathrm e}^{4 x}-2 \,{\mathrm e}^{2 x} c_1 -2 c_2 \cos \left (x \right )-2 c_3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{2}
\]
19.61.5 Mathematica DSolve solution
Solving time : 0.007
(sec)
Leaf size : 49
DSolve[{D[y[x],{x,3}]+1*D[y[x],{x,2}]-0*D[y[x],x]-2*y[x]==-Exp[3*x]*(9+67*x+17*x^2),{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to -\frac {1}{2} e^{3 x} \left (x^2+2 x-2\right )+c_3 e^x+c_2 e^{-x} \cos (x)+c_1 e^{-x} \sin (x)
\]