19.14 problem section 9.3, problem 14
Internal
problem
ID
[2161]
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
14
Date
solved
:
Thursday, October 17, 2024 at 02:21:06 AM
CAS
classification
:
[[_high_order, _linear, _nonhomogeneous]]
Solve
\begin{align*} y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime }+y^{\prime \prime }-3 y^{\prime }-2 y&=-3 \,{\mathrm e}^{2 x} \left (11+12 x \right ) \end{align*}
19.14.1 Solved as higher order constant coeff ode
Time used: 0.115 (sec)
The characteristic equation is
\[ \lambda ^{4}+3 \lambda ^{3}+\lambda ^{2}-3 \lambda -2 = 0 \]
The roots of the above equation are
\begin{align*} \lambda _1 &= -2\\ \lambda _2 &= 1\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end{align*}
Therefore the homogeneous solution is
\[ y_h(x)={\mathrm e}^{-x} c_1 +x \,{\mathrm e}^{-x} c_2 +{\mathrm e}^{-2 x} c_3 +{\mathrm e}^{x} c_4 \]
The fundamental set of solutions for the
homogeneous solution are the following
\begin{align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= x \,{\mathrm e}^{-x}\\ y_3 &= {\mathrm e}^{-2 x}\\ y_4 &= {\mathrm e}^{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
\[ y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime }+y^{\prime \prime }-3 y^{\prime }-2 y = 0 \]
Now the particular solution to the given ODE is found
\[
y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime }+y^{\prime \prime }-3 y^{\prime }-2 y = -\left (33+36 x \right ) {\mathrm e}^{2 x}
\]
The particular solution
is now found using the method of undetermined coefficients.
Looking at the RHS of the ode, which is
\[ -\left (33+36 x \right ) {\mathrm e}^{2 x} \]
Shows that the corresponding undetermined set of
the basis functions (UC_set) for the trial solution is
\[ [\{{\mathrm e}^{2 x} x, {\mathrm e}^{2 x}\}] \]
While the set of the basis functions for
the homogeneous solution found earlier is
\[ \{x \,{\mathrm e}^{-x}, {\mathrm e}^{x}, {\mathrm e}^{-2 x}, {\mathrm e}^{-x}\} \]
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.
\[
y_p = A_{1} {\mathrm e}^{2 x} x +A_{2} {\mathrm e}^{2 x}
\]
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
\[
36 A_{1} {\mathrm e}^{2 x} x +69 A_{1} {\mathrm e}^{2 x}+36 A_{2} {\mathrm e}^{2 x} = -3 \,{\mathrm e}^{2 x} \left (11+12 x \right )
\]
Solving for the
unknowns by comparing coefficients results in
\[ [A_{1} = -1, A_{2} = 1] \]
Substituting the above back in the
above trial solution \(y_p\), gives the particular solution
\[
y_p = -{\mathrm e}^{2 x} x +{\mathrm e}^{2 x}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left ({\mathrm e}^{-x} c_1 +x \,{\mathrm e}^{-x} c_2 +{\mathrm e}^{-2 x} c_3 +{\mathrm e}^{x} c_4\right ) + \left (-{\mathrm e}^{2 x} x +{\mathrm e}^{2 x}\right ) \\
\end{align*}
19.14.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right )+3 \frac {d^{3}}{d x^{3}}y \left (x \right )+\frac {d^{2}}{d x^{2}}y \left (x \right )-3 \frac {d}{d x}y \left (x \right )-2 y \left (x \right )=-3 \,{\mathrm e}^{2 x} \left (11+12 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 ) \\ \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 )=-36 x \,{\mathrm e}^{2 x}-3 y_{4}\left (x \right )-y_{3}\left (x \right )+3 y_{2}\left (x \right )+2 y_{1}\left (x \right )-33 \,{\mathrm e}^{2 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 ), y_{4}\left (x \right )=\frac {d}{d x}y_{3}\left (x \right ), \frac {d}{d x}y_{4}\left (x \right )=-36 x \,{\mathrm e}^{2 x}-3 y_{4}\left (x \right )-y_{3}\left (x \right )+3 y_{2}\left (x \right )+2 y_{1}\left (x \right )-33 \,{\mathrm e}^{2 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 ) \\ 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 \\ 2 & 3 & -1 & -3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ -36 x \,{\mathrm e}^{2 x}-33 \,{\mathrm e}^{2 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 \\ -36 x \,{\mathrm e}^{2 x}-33 \,{\mathrm e}^{2 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 \\ 2 & 3 & -1 & -3 \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 [-1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \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, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -1 \\ 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 (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \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 (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\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 x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\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 (x \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}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 2 & 3 & -1 & -3 \end {array}\right ]-\left (-1\right )\cdot \left [\begin {array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 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 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-x}\cdot \left (x \cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\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}\left (x \right )+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+\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} & {\mathrm e}^{-x} \left (-x -1\right ) & {\mathrm e}^{x} \\ \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & {\mathrm e}^{x} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -{\mathrm e}^{-x} & -x \,{\mathrm e}^{-x} & {\mathrm e}^{x} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & {\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} & {\mathrm e}^{-x} \left (-x -1\right ) & {\mathrm e}^{x} \\ \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & {\mathrm e}^{x} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -{\mathrm e}^{-x} & -x \,{\mathrm e}^{-x} & {\mathrm e}^{x} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & {\mathrm e}^{x} \end {array}\right ]\cdot \left [\begin {array}{cccc} -\frac {1}{8} & -1 & -1 & 1 \\ \frac {1}{4} & 1 & 0 & 1 \\ -\frac {1}{2} & -1 & 0 & 1 \\ 1 & 1 & 0 & 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} \left (x +1\right ) {\mathrm e}^{-x} & \frac {\left (2 \,{\mathrm e}^{3 x}+1+\left (3 x -3\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{6} & -\frac {{\mathrm e}^{-x}}{2}-x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{x}}{2} & \frac {\left ({\mathrm e}^{3 x}-3 \,{\mathrm e}^{x} x -1\right ) {\mathrm e}^{-2 x}}{6} \\ -x \,{\mathrm e}^{-x} & \frac {\left (2 \,{\mathrm e}^{3 x}-2+\left (6-3 x \right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{6} & -\frac {{\mathrm e}^{-x}}{2}+x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{x}}{2} & \frac {\left ({\mathrm e}^{3 x}+2+\left (3 x -3\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{6} \\ x \,{\mathrm e}^{-x} & \frac {\left (2 \,{\mathrm e}^{3 x}+4+\left (3 x -6\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{6} & \frac {{\mathrm e}^{-x}}{2}-x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{x}}{2} & \frac {\left ({\mathrm e}^{3 x}-4+\left (-3 x +3\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{6} \\ -x \,{\mathrm e}^{-x} & \frac {\left (2 \,{\mathrm e}^{3 x}-8+\left (6-3 x \right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{6} & -\frac {{\mathrm e}^{-x}}{2}+x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{x}}{2} & \frac {\left ({\mathrm e}^{3 x}+8+\left (3 x -3\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{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 (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 (5 \,{\mathrm e}^{4 x} x -4 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+7 \,{\mathrm e}^{x} x +{\mathrm e}^{x}+2\right ) {\mathrm e}^{-2 x}}{2} \\ -\frac {\left (10 \,{\mathrm e}^{4 x} x -3 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}-7 \,{\mathrm e}^{x} x +6 \,{\mathrm e}^{x}-4\right ) {\mathrm e}^{-2 x}}{2} \\ -\frac {\left (8 \,{\mathrm e}^{4 x} x -3 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+7 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}+8\right ) {\mathrm e}^{-2 x}}{2} \\ -\frac {\left (28 \,{\mathrm e}^{4 x} x +9 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}-7 \,{\mathrm e}^{x} x +6 \,{\mathrm e}^{x}-16\right ) {\mathrm e}^{-2 x}}{2} \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 )+\mathit {C4} {\moverset {\rightarrow }{y}}_{4}+\left [\begin {array}{c} -\frac {\left (5 \,{\mathrm e}^{4 x} x -4 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+7 \,{\mathrm e}^{x} x +{\mathrm e}^{x}+2\right ) {\mathrm e}^{-2 x}}{2} \\ -\frac {\left (10 \,{\mathrm e}^{4 x} x -3 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}-7 \,{\mathrm e}^{x} x +6 \,{\mathrm e}^{x}-4\right ) {\mathrm e}^{-2 x}}{2} \\ -\frac {\left (8 \,{\mathrm e}^{4 x} x -3 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+7 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}+8\right ) {\mathrm e}^{-2 x}}{2} \\ -\frac {\left (28 \,{\mathrm e}^{4 x} x +9 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}-7 \,{\mathrm e}^{x} x +6 \,{\mathrm e}^{x}-16\right ) {\mathrm e}^{-2 x}}{2} \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 (-\mathit {C4} +\frac {1}{2}\right ) {\mathrm e}^{3 x}+\left (\frac {5 x}{2}-2\right ) {\mathrm e}^{4 x}+\left (\left (\mathit {C3} +\frac {7}{2}\right ) x +\mathit {C2} +\mathit {C3} +\frac {1}{2}\right ) {\mathrm e}^{x}+\frac {\mathit {C1}}{8}+1\right ) \end {array} \]
19.14.3 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`
19.14.4 Maple dsolve solution
Solving time : 0.010
(sec)
Leaf size : 37
dsolve(diff(diff(diff(diff(y(x),x),x),x),x)+3*diff(diff(diff(y(x),x),x),x)+diff(diff(y(x),x),x)-3*diff(y(x),x)-2*y(x) = -3*exp(2*x)*(11+12*x),
y(x),singsol=all)
\[
y = \left (-{\mathrm e}^{4 x} x +{\mathrm e}^{4 x}+c_1 \,{\mathrm e}^{3 x}+c_4 x \,{\mathrm e}^{x}+c_3 \,{\mathrm e}^{x}+c_2 \right ) {\mathrm e}^{-2 x}
\]
19.14.5 Mathematica DSolve solution
Solving time : 0.006
(sec)
Leaf size : 43
DSolve[{D[y[x],{x,4}]+3*D[y[x],{x,3}]+D[y[x],{x,2}]-3*D[y[x],x]-2*y[x]==-3*Exp[2*x]*(11+12*x),{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to e^{-2 x} \left (-e^{4 x} (x-1)+e^x (c_3 x+c_2)+c_4 e^{3 x}+c_1\right )
\]