19.19 problem section 9.3, problem 19
Internal
problem
ID
[2166]
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
19
Date
solved
:
Thursday, October 17, 2024 at 02:21:08 AM
CAS
classification
:
[[_high_order, _linear, _nonhomogeneous]]
Solve
\begin{align*} 2 y^{\prime \prime \prime \prime }+5 y^{\prime \prime \prime }-5 y^{\prime }-2 y&=18 \,{\mathrm e}^{x} \left (5+2 x \right ) \end{align*}
19.19.1 Solved as higher order constant coeff ode
Time used: 0.114 (sec)
The characteristic equation is
\[ 2 \lambda ^{4}+5 \lambda ^{3}-5 \lambda -2 = 0 \]
The roots of the above equation are
\begin{align*} \lambda _1 &= 1\\ \lambda _2 &= -{\frac {1}{2}}\\ \lambda _3 &= -2\\ \lambda _4 &= -1 \end{align*}
Therefore the homogeneous solution is
\[ y_h(x)={\mathrm e}^{-x} c_1 +{\mathrm e}^{-2 x} c_2 +{\mathrm e}^{x} c_3 +{\mathrm e}^{-\frac {x}{2}} c_4 \]
The fundamental set of solutions for the
homogeneous solution are the following
\begin{align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{-2 x}\\ y_3 &= {\mathrm e}^{x}\\ y_4 &= {\mathrm e}^{-\frac {x}{2}} \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
\[ 2 y^{\prime \prime \prime \prime }+5 y^{\prime \prime \prime }-5 y^{\prime }-2 y = 0 \]
Now the particular solution to the given ODE is found
\[
2 y^{\prime \prime \prime \prime }+5 y^{\prime \prime \prime }-5 y^{\prime }-2 y = -\left (-90-36 x \right ) {\mathrm e}^{x}
\]
The particular solution
is now found using the method of undetermined coefficients.
Looking at the RHS of the ode, which is
\[ -\left (-90-36 x \right ) {\mathrm e}^{x} \]
Shows that the corresponding undetermined set of
the basis functions (UC_set) for the trial solution is
\[ [\{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \]
While the set of the basis
functions for the homogeneous solution found earlier is
\[ \left \{{\mathrm e}^{x}, {\mathrm e}^{-2 x}, {\mathrm e}^{-x}, {\mathrm e}^{-\frac {x}{2}}\right \} \]
Since \({\mathrm e}^{x}\) is duplicated in
the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes
\[ [\{x^{2} {\mathrm e}^{x}, {\mathrm e}^{x} x\}] \]
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.
\[
y_p = A_{1} x^{2} {\mathrm e}^{x}+A_{2} {\mathrm e}^{x} x
\]
The unknowns \(\{A_{1}, A_{2}\}\) 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
\[
54 A_{1} {\mathrm e}^{x}+36 A_{1} x \,{\mathrm e}^{x}+18 A_{2} {\mathrm e}^{x} = 18 \,{\mathrm e}^{x} \left (5+2 x \right )
\]
Solving for the
unknowns by comparing coefficients results in
\[ [A_{1} = 1, A_{2} = 2] \]
Substituting the above back in the
above trial solution \(y_p\), gives the particular solution
\[
y_p = x^{2} {\mathrm e}^{x}+2 \,{\mathrm e}^{x} x
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left ({\mathrm e}^{-x} c_1 +{\mathrm e}^{-2 x} c_2 +{\mathrm e}^{x} c_3 +{\mathrm e}^{-\frac {x}{2}} c_4\right ) + \left (x^{2} {\mathrm e}^{x}+2 \,{\mathrm e}^{x} x\right ) \\
\end{align*}
19.19.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 \frac {d^{4}}{d x^{4}}y \left (x \right )+5 \frac {d^{3}}{d x^{3}}y \left (x \right )-5 \frac {d}{d x}y \left (x \right )-2 y \left (x \right )=18 \,{\mathrm e}^{x} \left (5+2 x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right )=y \left (x \right )-\frac {5 \frac {d^{3}}{d x^{3}}y \left (x \right )}{2}+\frac {5 \frac {d}{d x}y \left (x \right )}{2}+18 \,{\mathrm e}^{x} x +45 \,{\mathrm e}^{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right )+\frac {5 \frac {d^{3}}{d x^{3}}y \left (x \right )}{2}-\frac {5 \frac {d}{d x}y \left (x \right )}{2}-y \left (x \right )=9 \,{\mathrm e}^{x} \left (5+2 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 {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=\frac {d^{3}}{d x^{3}}y \left (x \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d x}y_{4}\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y_{4}\left (x \right )=18 \,{\mathrm e}^{x} x +45 \,{\mathrm e}^{x}-\frac {5 y_{4}\left (x \right )}{2}+\frac {5 y_{2}\left (x \right )}{2}+y_{1}\left (x \right ) \\ & {} & \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 ), y_{4}\left (x \right )=\frac {d}{d x}y_{3}\left (x \right ), \frac {d}{d x}y_{4}\left (x \right )=18 \,{\mathrm e}^{x} x +45 \,{\mathrm e}^{x}-\frac {5 y_{4}\left (x \right )}{2}+\frac {5 y_{2}\left (x \right )}{2}+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 ) \\ y_{4}\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}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & \frac {5}{2} & 0 & -\frac {5}{2} \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 18 \,{\mathrm e}^{x} x +45 \,{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 18 \,{\mathrm e}^{x} x +45 \,{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & \frac {5}{2} & 0 & -\frac {5}{2} \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 [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \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 x}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \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 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 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 (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}+\mathit {C4} {\moverset {\rightarrow }{y}}_{4}+{\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}{cccc} -\frac {{\mathrm e}^{-2 x}}{8} & -{\mathrm e}^{-x} & -8 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{x} \\ \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{-x} & 4 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{x} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -{\mathrm e}^{-x} & -2 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{x} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-x} & {\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{x} \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}{cccc} -\frac {{\mathrm e}^{-2 x}}{8} & -{\mathrm e}^{-x} & -8 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{x} \\ \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{-x} & 4 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{x} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -{\mathrm e}^{-x} & -2 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{x} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-x} & {\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{x} \end {array}\right ]\cdot \left [\begin {array}{cccc} -\frac {1}{8} & -1 & -8 & 1 \\ \frac {1}{4} & 1 & 4 & 1 \\ -\frac {1}{2} & -1 & -2 & 1 \\ 1 & 1 & 1 & 1 \end {array}\right ]^{-1} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{cccc} -\frac {\left (-{\mathrm e}^{3 x}-16 \,{\mathrm e}^{\frac {3 x}{2}}+9 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{9} & -\frac {\left (-7 \,{\mathrm e}^{3 x}-16 \,{\mathrm e}^{\frac {3 x}{2}}+27 \,{\mathrm e}^{x}-4\right ) {\mathrm e}^{-2 x}}{18} & \frac {\left (7 \,{\mathrm e}^{3 x}-32 \,{\mathrm e}^{\frac {3 x}{2}}+27 \,{\mathrm e}^{x}-2\right ) {\mathrm e}^{-2 x}}{18} & \frac {\left ({\mathrm e}^{3 x}-8 \,{\mathrm e}^{\frac {3 x}{2}}+9 \,{\mathrm e}^{x}-2\right ) {\mathrm e}^{-2 x}}{9} \\ \frac {\left ({\mathrm e}^{3 x}-8 \,{\mathrm e}^{\frac {3 x}{2}}+9 \,{\mathrm e}^{x}-2\right ) {\mathrm e}^{-2 x}}{9} & \frac {\left (7 \,{\mathrm e}^{3 x}-8 \,{\mathrm e}^{\frac {3 x}{2}}+27 \,{\mathrm e}^{x}-8\right ) {\mathrm e}^{-2 x}}{18} & -\frac {\left (-7 \,{\mathrm e}^{3 x}-16 \,{\mathrm e}^{\frac {3 x}{2}}+27 \,{\mathrm e}^{x}-4\right ) {\mathrm e}^{-2 x}}{18} & -\frac {\left (-{\mathrm e}^{3 x}-4 \,{\mathrm e}^{\frac {3 x}{2}}+9 \,{\mathrm e}^{x}-4\right ) {\mathrm e}^{-2 x}}{9} \\ -\frac {\left (-{\mathrm e}^{3 x}-4 \,{\mathrm e}^{\frac {3 x}{2}}+9 \,{\mathrm e}^{x}-4\right ) {\mathrm e}^{-2 x}}{9} & -\frac {\left (-7 \,{\mathrm e}^{3 x}-4 \,{\mathrm e}^{\frac {3 x}{2}}+27 \,{\mathrm e}^{x}-16\right ) {\mathrm e}^{-2 x}}{18} & \frac {\left (7 \,{\mathrm e}^{3 x}-8 \,{\mathrm e}^{\frac {3 x}{2}}+27 \,{\mathrm e}^{x}-8\right ) {\mathrm e}^{-2 x}}{18} & \frac {\left ({\mathrm e}^{3 x}-2 \,{\mathrm e}^{\frac {3 x}{2}}+9 \,{\mathrm e}^{x}-8\right ) {\mathrm e}^{-2 x}}{9} \\ \frac {\left ({\mathrm e}^{3 x}-2 \,{\mathrm e}^{\frac {3 x}{2}}+9 \,{\mathrm e}^{x}-8\right ) {\mathrm e}^{-2 x}}{9} & \frac {\left (7 \,{\mathrm e}^{3 x}-2 \,{\mathrm e}^{\frac {3 x}{2}}+27 \,{\mathrm e}^{x}-32\right ) {\mathrm e}^{-2 x}}{18} & -\frac {\left (-7 \,{\mathrm e}^{3 x}-4 \,{\mathrm e}^{\frac {3 x}{2}}+27 \,{\mathrm e}^{x}-16\right ) {\mathrm e}^{-2 x}}{18} & -\frac {\left (-{\mathrm e}^{3 x}-{\mathrm e}^{\frac {3 x}{2}}+9 \,{\mathrm e}^{x}-16\right ) {\mathrm e}^{-2 x}}{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 (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 (-9 x^{2} {\mathrm e}^{3 x}-18 x \,{\mathrm e}^{3 x}+40 \,{\mathrm e}^{3 x}-176 \,{\mathrm e}^{\frac {3 x}{2}}+162 \,{\mathrm e}^{x}-26\right ) {\mathrm e}^{-2 x}}{9} \\ \frac {\left (9 x^{2} {\mathrm e}^{3 x}+36 x \,{\mathrm e}^{3 x}-22 \,{\mathrm e}^{3 x}-88 \,{\mathrm e}^{\frac {3 x}{2}}+162 \,{\mathrm e}^{x}-52\right ) {\mathrm e}^{-2 x}}{9} \\ -\frac {\left (-9 x^{2} {\mathrm e}^{3 x}-54 x \,{\mathrm e}^{3 x}-14 \,{\mathrm e}^{3 x}-44 \,{\mathrm e}^{\frac {3 x}{2}}+162 \,{\mathrm e}^{x}-104\right ) {\mathrm e}^{-2 x}}{9} \\ \frac {\left (9 x^{2} {\mathrm e}^{3 x}+72 x \,{\mathrm e}^{3 x}+68 \,{\mathrm e}^{3 x}-22 \,{\mathrm e}^{\frac {3 x}{2}}+162 \,{\mathrm e}^{x}-208\right ) {\mathrm e}^{-2 x}}{9} \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}+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}+\mathit {C4} {\moverset {\rightarrow }{y}}_{4}+\left [\begin {array}{c} -\frac {\left (-9 x^{2} {\mathrm e}^{3 x}-18 x \,{\mathrm e}^{3 x}+40 \,{\mathrm e}^{3 x}-176 \,{\mathrm e}^{\frac {3 x}{2}}+162 \,{\mathrm e}^{x}-26\right ) {\mathrm e}^{-2 x}}{9} \\ \frac {\left (9 x^{2} {\mathrm e}^{3 x}+36 x \,{\mathrm e}^{3 x}-22 \,{\mathrm e}^{3 x}-88 \,{\mathrm e}^{\frac {3 x}{2}}+162 \,{\mathrm e}^{x}-52\right ) {\mathrm e}^{-2 x}}{9} \\ -\frac {\left (-9 x^{2} {\mathrm e}^{3 x}-54 x \,{\mathrm e}^{3 x}-14 \,{\mathrm e}^{3 x}-44 \,{\mathrm e}^{\frac {3 x}{2}}+162 \,{\mathrm e}^{x}-104\right ) {\mathrm e}^{-2 x}}{9} \\ \frac {\left (9 x^{2} {\mathrm e}^{3 x}+72 x \,{\mathrm e}^{3 x}+68 \,{\mathrm e}^{3 x}-22 \,{\mathrm e}^{\frac {3 x}{2}}+162 \,{\mathrm e}^{x}-208\right ) {\mathrm e}^{-2 x}}{9} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )={\mathrm e}^{-2 x} \left (\left (-8 \mathit {C3} +\frac {176}{9}\right ) {\mathrm e}^{\frac {3 x}{2}}+\left (\mathit {C4} +x^{2}+2 x -\frac {40}{9}\right ) {\mathrm e}^{3 x}+\left (-\mathit {C2} -18\right ) {\mathrm e}^{x}-\frac {\mathit {C1}}{8}+\frac {26}{9}\right ) \end {array} \]
19.19.3 Maple trace
`Methods for high order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
trying differential order: 4; linear nonhomogeneous with symmetry [0,1]
trying high order linear exact nonhomogeneous
trying differential order: 4; missing the dependent variable
checking if the LODE has constant coefficients
<- constant coefficients successful`
19.19.4 Maple dsolve solution
Solving time : 0.009
(sec)
Leaf size : 34
dsolve(2*diff(diff(diff(diff(y(x),x),x),x),x)+5*diff(diff(diff(y(x),x),x),x)-5*diff(y(x),x)-2*y(x) = 18*exp(x)*(5+2*x),
y(x),singsol=all)
\[
y = \left ({\mathrm e}^{\frac {3 x}{2}} c_4 +\left (x^{2}+c_1 +2 x \right ) {\mathrm e}^{3 x}+c_3 \,{\mathrm e}^{x}+c_2 \right ) {\mathrm e}^{-2 x}
\]
19.19.5 Mathematica DSolve solution
Solving time : 0.072
(sec)
Leaf size : 48
DSolve[{2*D[y[x],{x,4}]+5*D[y[x],{x,3}]+0*D[y[x],{x,2}]-5*D[y[x],x]-2*y[x]==18*Exp[x]*(5+2*x),{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to e^{-2 x} \left (e^{3 x} \left (x^2+2 x-\frac {40}{9}+c_4\right )+c_1 e^{3 x/2}+c_3 e^x+c_2\right )
\]