19.31 problem section 9.3, problem 31
Internal
problem
ID
[2178]
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
31
Date
solved
:
Thursday, October 17, 2024 at 02:21:14 AM
CAS
classification
:
[[_3rd_order, _linear, _nonhomogeneous]]
Solve
\begin{align*} y^{\prime \prime \prime }-y^{\prime \prime }+2 y^{\prime }-2 y&={\mathrm e}^{2 x} \left (\left (-x^{2}+5 x +27\right ) \cos \left (x \right )+\left (9 x^{2}+13 x +2\right ) \sin \left (x \right )\right ) \end{align*}
19.31.1 Solved as higher order constant coeff ode
Time used: 0.239 (sec)
The characteristic equation is
\[ \lambda ^{3}-\lambda ^{2}+2 \lambda -2 = 0 \]
The roots of the above equation are
\begin{align*} \lambda _1 &= 1\\ \lambda _2 &= i \sqrt {2}\\ \lambda _3 &= -i \sqrt {2} \end{align*}
Therefore the homogeneous solution is
\[ y_h(x)={\mathrm e}^{x} c_1 +{\mathrm e}^{i \sqrt {2}\, x} c_2 +{\mathrm e}^{-i \sqrt {2}\, x} c_3 \]
The fundamental set of solutions for the
homogeneous solution are the following
\begin{align*} y_1 &= {\mathrm e}^{x}\\ y_2 &= {\mathrm e}^{i \sqrt {2}\, x}\\ y_3 &= {\mathrm e}^{-i \sqrt {2}\, 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
\[ y^{\prime \prime \prime }-y^{\prime \prime }+2 y^{\prime }-2 y = 0 \]
Now the particular solution to the given ODE is found
\[
y^{\prime \prime \prime }-y^{\prime \prime }+2 y^{\prime }-2 y = -\left (\left (x^{2}-5 x -27\right ) \cos \left (x \right )-9 \sin \left (x \right ) x^{2}-13 \sin \left (x \right ) x -2 \sin \left (x \right )\right ) {\mathrm e}^{2 x}
\]
The particular solution
is now found using the method of undetermined coefficients.
Looking at the RHS of the ode, which is
\[ -\left (\left (x^{2}-5 x -27\right ) \cos \left (x \right )-9 \sin \left (x \right ) x^{2}-13 \sin \left (x \right ) x -2 \sin \left (x \right )\right ) {\mathrm e}^{2 x} \]
Shows that the corresponding undetermined set of
the basis functions (UC_set) for the trial solution is
\[ [\{{\mathrm e}^{2 x} \cos \left (x \right ), {\mathrm e}^{2 x} \sin \left (x \right ), {\mathrm e}^{2 x} \cos \left (x \right ) x, {\mathrm e}^{2 x} \cos \left (x \right ) x^{2}, {\mathrm e}^{2 x} \sin \left (x \right ) x, {\mathrm e}^{2 x} \sin \left (x \right ) x^{2}\}] \]
While the set of the basis functions for
the homogeneous solution found earlier is
\[ \left \{{\mathrm e}^{x}, {\mathrm e}^{i \sqrt {2}\, x}, {\mathrm e}^{-i \sqrt {2}\, x}\right \} \]
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.
\[
y_p = A_{1} {\mathrm e}^{2 x} \cos \left (x \right )+A_{2} {\mathrm e}^{2 x} \sin \left (x \right )+A_{3} {\mathrm e}^{2 x} \cos \left (x \right ) x +A_{4} {\mathrm e}^{2 x} \cos \left (x \right ) x^{2}+A_{5} {\mathrm e}^{2 x} \sin \left (x \right ) x +A_{6} {\mathrm e}^{2 x} \sin \left (x \right ) x^{2}
\]
The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) 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
\[
A_{3} {\mathrm e}^{2 x} \cos \left (x \right ) x +10 A_{6} {\mathrm e}^{2 x} \sin \left (x \right )+9 A_{5} {\mathrm e}^{2 x} \cos \left (x \right ) x +7 A_{5} {\mathrm e}^{2 x} \sin \left (x \right )+9 A_{6} {\mathrm e}^{2 x} \cos \left (x \right ) x^{2}-10 A_{3} {\mathrm e}^{2 x} \sin \left (x \right )+10 A_{4} {\mathrm e}^{2 x} \cos \left (x \right )+10 A_{5} {\mathrm e}^{2 x} \cos \left (x \right )+7 A_{3} {\mathrm e}^{2 x} \cos \left (x \right )-9 A_{3} {\mathrm e}^{2 x} \sin \left (x \right ) x -9 A_{4} {\mathrm e}^{2 x} \sin \left (x \right ) x^{2}+A_{4} {\mathrm e}^{2 x} \cos \left (x \right ) x^{2}+A_{5} {\mathrm e}^{2 x} \sin \left (x \right ) x +A_{6} {\mathrm e}^{2 x} \sin \left (x \right ) x^{2}-20 A_{4} {\mathrm e}^{2 x} \sin \left (x \right ) x +20 A_{6} {\mathrm e}^{2 x} \cos \left (x \right ) x +14 A_{4} {\mathrm e}^{2 x} \cos \left (x \right ) x +14 A_{6} {\mathrm e}^{2 x} \sin \left (x \right ) x -6 A_{4} {\mathrm e}^{2 x} \sin \left (x \right )-9 A_{1} {\mathrm e}^{2 x} \sin \left (x \right )+9 A_{2} {\mathrm e}^{2 x} \cos \left (x \right )+6 A_{6} {\mathrm e}^{2 x} \cos \left (x \right )+A_{1} {\mathrm e}^{2 x} \cos \left (x \right )+A_{2} {\mathrm e}^{2 x} \sin \left (x \right ) = {\mathrm e}^{2 x} \left (\left (-x^{2}+5 x +27\right ) \cos \left (x \right )+\left (9 x^{2}+13 x +2\right ) \sin \left (x \right )\right )
\]
Solving for the
unknowns by comparing coefficients results in
\[ [A_{1} = 1, A_{2} = 1, A_{3} = 1, A_{4} = -1, A_{5} = 2, A_{6} = 0] \]
Substituting the above back in the
above trial solution \(y_p\), gives the particular solution
\[
y_p = {\mathrm e}^{2 x} \cos \left (x \right )+{\mathrm e}^{2 x} \sin \left (x \right )+{\mathrm e}^{2 x} \cos \left (x \right ) x -{\mathrm e}^{2 x} \cos \left (x \right ) x^{2}+2 \,{\mathrm e}^{2 x} \sin \left (x \right ) x
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left ({\mathrm e}^{x} c_1 +{\mathrm e}^{i \sqrt {2}\, x} c_2 +{\mathrm e}^{-i \sqrt {2}\, x} c_3\right ) + \left ({\mathrm e}^{2 x} \cos \left (x \right )+{\mathrm e}^{2 x} \sin \left (x \right )+{\mathrm e}^{2 x} \cos \left (x \right ) x -{\mathrm e}^{2 x} \cos \left (x \right ) x^{2}+2 \,{\mathrm e}^{2 x} \sin \left (x \right ) x\right ) \\
\end{align*}
19.31.2 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 \frac {d}{d x}y \left (x \right )-2 y \left (x \right )={\mathrm e}^{2 x} \left (\left (-x^{2}+5 x +27\right ) \cos \left (x \right )+\left (9 x^{2}+13 x +2\right ) \sin \left (x \right )\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 ) \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )=2 y \left (x \right )+9 x^{2} \sin \left (x \right ) {\mathrm e}^{2 x}-x^{2} \cos \left (x \right ) {\mathrm e}^{2 x}+13 \sin \left (x \right ) x \,{\mathrm e}^{2 x}+5 x \cos \left (x \right ) {\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x} \sin \left (x \right )+27 \,{\mathrm e}^{2 x} \cos \left (x \right )+\frac {d^{2}}{d x^{2}}y \left (x \right )-2 \frac {d}{d x}y \left (x \right ) \\ \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^{3}}{d x^{3}}y \left (x \right )-\frac {d^{2}}{d x^{2}}y \left (x \right )+2 \frac {d}{d x}y \left (x \right )-2 y \left (x \right )={\mathrm e}^{2 x} \left (9 x^{2} \sin \left (x \right )-x^{2} \cos \left (x \right )+13 \sin \left (x \right ) x +5 x \cos \left (x \right )+2 \sin \left (x \right )+27 \cos \left (x \right )\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 )=9 x^{2} \sin \left (x \right ) {\mathrm e}^{2 x}-x^{2} \cos \left (x \right ) {\mathrm e}^{2 x}+13 \sin \left (x \right ) x \,{\mathrm e}^{2 x}+5 x \cos \left (x \right ) {\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x} \sin \left (x \right )+27 \,{\mathrm e}^{2 x} \cos \left (x \right )+y_{3}\left (x \right )-2 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 ), \frac {d}{d x}y_{3}\left (x \right )=9 x^{2} \sin \left (x \right ) {\mathrm e}^{2 x}-x^{2} \cos \left (x \right ) {\mathrm e}^{2 x}+13 \sin \left (x \right ) x \,{\mathrm e}^{2 x}+5 x \cos \left (x \right ) {\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x} \sin \left (x \right )+27 \,{\mathrm e}^{2 x} \cos \left (x \right )+y_{3}\left (x \right )-2 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 ) \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 & -2 & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 9 x^{2} \sin \left (x \right ) {\mathrm e}^{2 x}-x^{2} \cos \left (x \right ) {\mathrm e}^{2 x}+13 \sin \left (x \right ) x \,{\mathrm e}^{2 x}+5 x \cos \left (x \right ) {\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x} \sin \left (x \right )+27 \,{\mathrm e}^{2 x} \cos \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 9 x^{2} \sin \left (x \right ) {\mathrm e}^{2 x}-x^{2} \cos \left (x \right ) {\mathrm e}^{2 x}+13 \sin \left (x \right ) x \,{\mathrm e}^{2 x}+5 x \cos \left (x \right ) {\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x} \sin \left (x \right )+27 \,{\mathrm e}^{2 x} \cos \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & -2 & 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 [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ -\frac {\mathrm {I}}{2} \sqrt {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 [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} \sqrt {2}\, x}\cdot \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {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 \\ {} & {} & \left (\cos \left (\sqrt {2}\, x \right )-\mathrm {I} \sin \left (\sqrt {2}\, x \right )\right )\cdot \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\cos \left (\sqrt {2}\, x \right )}{2}+\frac {\mathrm {I} \sin \left (\sqrt {2}\, x \right )}{2} \\ \frac {\mathrm {I}}{2} \left (\cos \left (\sqrt {2}\, x \right )-\mathrm {I} \sin \left (\sqrt {2}\, x \right )\right ) \sqrt {2} \\ \cos \left (\sqrt {2}\, x \right )-\mathrm {I} \sin \left (\sqrt {2}\, 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 )=\left [\begin {array}{c} -\frac {\cos \left (\sqrt {2}\, x \right )}{2} \\ \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} \\ \cos \left (\sqrt {2}\, x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} \frac {\sin \left (\sqrt {2}\, x \right )}{2} \\ \frac {\sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{2} \\ -\sin \left (\sqrt {2}\, 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 {\cos \left (\sqrt {2}\, x \right )}{2} & \frac {\sin \left (\sqrt {2}\, x \right )}{2} \\ {\mathrm e}^{x} & \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} & \frac {\sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{2} \\ {\mathrm e}^{x} & \cos \left (\sqrt {2}\, x \right ) & -\sin \left (\sqrt {2}\, 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 {\cos \left (\sqrt {2}\, x \right )}{2} & \frac {\sin \left (\sqrt {2}\, x \right )}{2} \\ {\mathrm e}^{x} & \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} & \frac {\sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{2} \\ {\mathrm e}^{x} & \cos \left (\sqrt {2}\, x \right ) & -\sin \left (\sqrt {2}\, x \right ) \end {array}\right ]\cdot \left [\begin {array}{ccc} 1 & -\frac {1}{2} & 0 \\ 1 & 0 & \frac {\sqrt {2}}{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 {2 \,{\mathrm e}^{x}}{3}+\frac {\cos \left (\sqrt {2}\, x \right )}{3}-\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{3} & \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} & \frac {{\mathrm e}^{x}}{3}-\frac {\cos \left (\sqrt {2}\, x \right )}{3}-\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{6} \\ \frac {2 \,{\mathrm e}^{x}}{3}-\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{3}-\frac {2 \cos \left (\sqrt {2}\, x \right )}{3} & \cos \left (\sqrt {2}\, x \right ) & \frac {{\mathrm e}^{x}}{3}+\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{3}-\frac {\cos \left (\sqrt {2}\, x \right )}{3} \\ \frac {2 \,{\mathrm e}^{x}}{3}-\frac {2 \cos \left (\sqrt {2}\, x \right )}{3}+\frac {2 \sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{3} & -\sqrt {2}\, \sin \left (\sqrt {2}\, x \right ) & \frac {{\mathrm e}^{x}}{3}+\frac {2 \cos \left (\sqrt {2}\, x \right )}{3}+\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \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} \\ {} & {} & \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} 4 \cos \left (\sqrt {2}\, x \right )+\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2}+\left (\left (-x^{2}+x +1\right ) \cos \left (x \right )+\left (2 x +1\right ) \sin \left (x \right )\right ) {\mathrm e}^{2 x}-5 \,{\mathrm e}^{x} \\ \cos \left (\sqrt {2}\, x \right )-4 \sqrt {2}\, \sin \left (\sqrt {2}\, x \right )+\left (\left (-2 x^{2}+2 x +4\right ) \cos \left (x \right )+\left (x^{2}+3 x +3\right ) \sin \left (x \right )\right ) {\mathrm e}^{2 x}-5 \,{\mathrm e}^{x} \\ -8 \cos \left (\sqrt {2}\, x \right )-\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )+\left (\left (-3 x^{2}+3 x +13\right ) \cos \left (x \right )+\left (4 x^{2}+6 x +5\right ) \sin \left (x \right )\right ) {\mathrm e}^{2 x}-5 \,{\mathrm e}^{x} \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} 4 \cos \left (\sqrt {2}\, x \right )+\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2}+\left (\left (-x^{2}+x +1\right ) \cos \left (x \right )+\left (2 x +1\right ) \sin \left (x \right )\right ) {\mathrm e}^{2 x}-5 \,{\mathrm e}^{x} \\ \cos \left (\sqrt {2}\, x \right )-4 \sqrt {2}\, \sin \left (\sqrt {2}\, x \right )+\left (\left (-2 x^{2}+2 x +4\right ) \cos \left (x \right )+\left (x^{2}+3 x +3\right ) \sin \left (x \right )\right ) {\mathrm e}^{2 x}-5 \,{\mathrm e}^{x} \\ -8 \cos \left (\sqrt {2}\, x \right )-\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )+\left (\left (-3 x^{2}+3 x +13\right ) \cos \left (x \right )+\left (4 x^{2}+6 x +5\right ) \sin \left (x \right )\right ) {\mathrm e}^{2 x}-5 \,{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\left (-\mathit {C2} +8\right ) \cos \left (\sqrt {2}\, x \right )}{2}+\frac {\left (\mathit {C3} +\sqrt {2}\right ) \sin \left (\sqrt {2}\, x \right )}{2}+\left (\left (-x^{2}+x +1\right ) \cos \left (x \right )+\left (2 x +1\right ) \sin \left (x \right )\right ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \left (\mathit {C1} -5\right ) \end {array} \]
19.31.3 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`
19.31.4 Maple dsolve solution
Solving time : 0.141
(sec)
Leaf size : 178
dsolve(diff(diff(diff(y(x),x),x),x)-diff(diff(y(x),x),x)+2*diff(y(x),x)-2*y(x) = exp(2*x)*((-x^2+5*x+27)*cos(x)+(9*x^2+13*x+2)*sin(x)),
y(x),singsol=all)
\[
y = \frac {\sqrt {2}\, \left (\int \left (\left (-9 x^{2}-13 x -2\right ) \sin \left (x \right )+\left (x^{2}-5 x -27\right ) \cos \left (x \right )\right ) \left (\sqrt {2}\, \cos \left (\sqrt {2}\, x \right )-\sin \left (\sqrt {2}\, x \right )\right ) {\mathrm e}^{2 x}d x \right ) \cos \left (\sqrt {2}\, x \right )}{6}+\frac {\sqrt {2}\, \left (\int \left (\left (-9 x^{2}-13 x -2\right ) \sin \left (x \right )+\left (x^{2}-5 x -27\right ) \cos \left (x \right )\right ) \left (\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )+\cos \left (\sqrt {2}\, x \right )\right ) {\mathrm e}^{2 x}d x \right ) \sin \left (\sqrt {2}\, x \right )}{6}+c_2 \cos \left (\sqrt {2}\, x \right )+c_3 \sin \left (\sqrt {2}\, x \right )+\frac {\left (5 \left (-x^{2}+x +3\right ) \cos \left (x \right )+4 \sin \left (x \right ) \left (x^{2}+\frac {5}{2} x +\frac {7}{4}\right )\right ) {\mathrm e}^{2 x}}{3}+{\mathrm e}^{x} c_1
\]
19.31.5 Mathematica DSolve solution
Solving time : 0.019
(sec)
Leaf size : 60
DSolve[{1*D[y[x],{x,3}]-1*D[y[x],{x,2}]+2*D[y[x],x]-2*y[x]==Exp[2*x]*((27+5*x-x^2)*Cos[1*x]+(2+13*x+9*x^2)*Sin[1*x]),{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to e^{2 x} \left (\left (-x^2+x+1\right ) \cos (x)+(2 x+1) \sin (x)\right )+c_3 e^x+c_1 \cos \left (\sqrt {2} x\right )+c_2 \sin \left (\sqrt {2} x\right )
\]