Internal problem ID [1512]
Internal file name [OUTPUT/1513_Sunday_June_05_2022_02_20_12_AM_13644452/index.tex]
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 15.
ODE order: 4.
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
[[_high_order, _missing_y]]
\[ \boxed {y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime }+24 y^{\prime \prime }+32 y^{\prime }=-16 \,{\mathrm e}^{-2 x} \left (-x^{3}+x^{2}+x +1\right )} \] 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 \[ y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime }+24 y^{\prime \prime }+32 y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{4}+8 \lambda ^{3}+24 \lambda ^{2}+32 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= -4\\ \lambda _3 &= -2-2 i\\ \lambda _4 &= -2+2 i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} +{\mathrm e}^{-4 x} c_{2} +{\mathrm e}^{\left (-2+2 i\right ) x} c_{3} +{\mathrm e}^{\left (-2-2 i\right ) x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{-4 x} \\ y_3 &= {\mathrm e}^{\left (-2+2 i\right ) x} \\ y_4 &= {\mathrm e}^{\left (-2-2 i\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime }+24 y^{\prime \prime }+32 y^{\prime } = -16 \,{\mathrm e}^{-2 x} \left (-x^{3}+x^{2}+x +1\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ -16 \,{\mathrm e}^{-2 x} \left (-x^{3}+x^{2}+x +1\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x^{2} {\mathrm e}^{-2 x}, x^{3} {\mathrm e}^{-2 x}, {\mathrm e}^{-2 x} x, {\mathrm e}^{-2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{\left (-2-2 i\right ) x}, {\mathrm e}^{\left (-2+2 i\right ) x}, {\mathrm e}^{-4 x}\} \] 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. \[ y_p = A_{1} x^{2} {\mathrm e}^{-2 x}+A_{2} x^{3} {\mathrm e}^{-2 x}+A_{3} {\mathrm e}^{-2 x} x +A_{4} {\mathrm e}^{-2 x} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}\}\) 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 \[ -16 A_{4} {\mathrm e}^{-2 x}-16 A_{1} x^{2} {\mathrm e}^{-2 x}-16 A_{2} x^{3} {\mathrm e}^{-2 x}-16 A_{3} {\mathrm e}^{-2 x} x = -16 \,{\mathrm e}^{-2 x} \left (-x^{3}+x^{2}+x +1\right ) \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 1, A_{2} = -1, A_{3} = 1, A_{4} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = x^{2} {\mathrm e}^{-2 x}-x^{3} {\mathrm e}^{-2 x}+{\mathrm e}^{-2 x} x +{\mathrm e}^{-2 x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +{\mathrm e}^{-4 x} c_{2} +{\mathrm e}^{\left (-2+2 i\right ) x} c_{3} +{\mathrm e}^{\left (-2-2 i\right ) x} c_{4}\right ) + \left (x^{2} {\mathrm e}^{-2 x}-x^{3} {\mathrm e}^{-2 x}+{\mathrm e}^{-2 x} x +{\mathrm e}^{-2 x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +{\mathrm e}^{-4 x} c_{2} +{\mathrm e}^{\left (-2+2 i\right ) x} c_{3} +{\mathrm e}^{\left (-2-2 i\right ) x} c_{4} +x^{2} {\mathrm e}^{-2 x}-x^{3} {\mathrm e}^{-2 x}+{\mathrm e}^{-2 x} x +{\mathrm e}^{-2 x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} +{\mathrm e}^{-4 x} c_{2} +{\mathrm e}^{\left (-2+2 i\right ) x} c_{3} +{\mathrm e}^{\left (-2-2 i\right ) x} c_{4} +x^{2} {\mathrm e}^{-2 x}-x^{3} {\mathrm e}^{-2 x}+{\mathrm e}^{-2 x} x +{\mathrm e}^{-2 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime }+24 y^{\prime \prime }+32 y^{\prime }=-16 \,{\mathrm e}^{-2 x} \left (-x^{3}+x^{2}+x +1\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=16 x^{3} {\mathrm e}^{-2 x}-16 x^{2} {\mathrm e}^{-2 x}-16 \,{\mathrm e}^{-2 x} x -16 \,{\mathrm e}^{-2 x}-8 y^{\prime \prime \prime }-24 y^{\prime \prime }-32 y^{\prime } \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime }+24 y^{\prime \prime }+32 y^{\prime }=16 \,{\mathrm e}^{-2 x} \left (x^{3}-x^{2}-x -1\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 \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=16 x^{3} {\mathrm e}^{-2 x}-16 x^{2} {\mathrm e}^{-2 x}-16 \,{\mathrm e}^{-2 x} x -8 y_{4}\left (x \right )-24 y_{3}\left (x \right )-32 y_{2}\left (x \right )-16 \,{\mathrm e}^{-2 x} \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{4}^{\prime }\left (x \right )=16 x^{3} {\mathrm e}^{-2 x}-16 x^{2} {\mathrm e}^{-2 x}-16 \,{\mathrm e}^{-2 x} x -8 y_{4}\left (x \right )-24 y_{3}\left (x \right )-32 y_{2}\left (x \right )-16 \,{\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} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & -32 & -24 & -8 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 16 x^{3} {\mathrm e}^{-2 x}-16 x^{2} {\mathrm e}^{-2 x}-16 \,{\mathrm e}^{-2 x} x -16 \,{\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 \\ 16 x^{3} {\mathrm e}^{-2 x}-16 x^{2} {\mathrm e}^{-2 x}-16 \,{\mathrm e}^{-2 x} x -16 \,{\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 \\ 0 & -32 & -24 & -8 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\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 [-4, \left [\begin {array}{c} -\frac {1}{64} \\ \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [-2-2 \,\mathrm {I}, \left [\begin {array}{c} \frac {1}{32}+\frac {\mathrm {I}}{32} \\ -\frac {\mathrm {I}}{8} \\ -\frac {1}{4}+\frac {\mathrm {I}}{4} \\ 1 \end {array}\right ]\right ], \left [-2+2 \,\mathrm {I}, \left [\begin {array}{c} \frac {1}{32}-\frac {\mathrm {I}}{32} \\ \frac {\mathrm {I}}{8} \\ -\frac {1}{4}-\frac {\mathrm {I}}{4} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-4, \left [\begin {array}{c} -\frac {1}{64} \\ \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-4 x}\cdot \left [\begin {array}{c} -\frac {1}{64} \\ \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2-2 \,\mathrm {I}, \left [\begin {array}{c} \frac {1}{32}+\frac {\mathrm {I}}{32} \\ -\frac {\mathrm {I}}{8} \\ -\frac {1}{4}+\frac {\mathrm {I}}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-2-2 \,\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{32}+\frac {\mathrm {I}}{32} \\ -\frac {\mathrm {I}}{8} \\ -\frac {1}{4}+\frac {\mathrm {I}}{4} \\ 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}^{-2 x}\cdot \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right )\cdot \left [\begin {array}{c} \frac {1}{32}+\frac {\mathrm {I}}{32} \\ -\frac {\mathrm {I}}{8} \\ -\frac {1}{4}+\frac {\mathrm {I}}{4} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \left (\frac {1}{32}+\frac {\mathrm {I}}{32}\right ) \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ -\frac {\mathrm {I}}{8} \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \left (-\frac {1}{4}+\frac {\mathrm {I}}{4}\right ) \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \cos \left (2 x \right )-\mathrm {I} \sin \left (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}}_{3}\left (x \right )={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {\cos \left (2 x \right )}{32}+\frac {\sin \left (2 x \right )}{32} \\ -\frac {\sin \left (2 x \right )}{8} \\ -\frac {\cos \left (2 x \right )}{4}+\frac {\sin \left (2 x \right )}{4} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {\cos \left (2 x \right )}{32}-\frac {\sin \left (2 x \right )}{32} \\ -\frac {\cos \left (2 x \right )}{8} \\ \frac {\sin \left (2 x \right )}{4}+\frac {\cos \left (2 x \right )}{4} \\ -\sin \left (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 )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\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}{cccc} -\frac {{\mathrm e}^{-4 x}}{64} & 1 & {\mathrm e}^{-2 x} \left (\frac {\cos \left (2 x \right )}{32}+\frac {\sin \left (2 x \right )}{32}\right ) & {\mathrm e}^{-2 x} \left (\frac {\cos \left (2 x \right )}{32}-\frac {\sin \left (2 x \right )}{32}\right ) \\ \frac {{\mathrm e}^{-4 x}}{16} & 0 & -\frac {{\mathrm e}^{-2 x} \sin \left (2 x \right )}{8} & -\frac {{\mathrm e}^{-2 x} \cos \left (2 x \right )}{8} \\ -\frac {{\mathrm e}^{-4 x}}{4} & 0 & {\mathrm e}^{-2 x} \left (-\frac {\cos \left (2 x \right )}{4}+\frac {\sin \left (2 x \right )}{4}\right ) & {\mathrm e}^{-2 x} \left (\frac {\sin \left (2 x \right )}{4}+\frac {\cos \left (2 x \right )}{4}\right ) \\ {\mathrm e}^{-4 x} & 0 & {\mathrm e}^{-2 x} \cos \left (2 x \right ) & -{\mathrm e}^{-2 x} \sin \left (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 \frac {1}{\phi \left (0\right )} \\ {} & \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}^{-4 x}}{64} & 1 & {\mathrm e}^{-2 x} \left (\frac {\cos \left (2 x \right )}{32}+\frac {\sin \left (2 x \right )}{32}\right ) & {\mathrm e}^{-2 x} \left (\frac {\cos \left (2 x \right )}{32}-\frac {\sin \left (2 x \right )}{32}\right ) \\ \frac {{\mathrm e}^{-4 x}}{16} & 0 & -\frac {{\mathrm e}^{-2 x} \sin \left (2 x \right )}{8} & -\frac {{\mathrm e}^{-2 x} \cos \left (2 x \right )}{8} \\ -\frac {{\mathrm e}^{-4 x}}{4} & 0 & {\mathrm e}^{-2 x} \left (-\frac {\cos \left (2 x \right )}{4}+\frac {\sin \left (2 x \right )}{4}\right ) & {\mathrm e}^{-2 x} \left (\frac {\sin \left (2 x \right )}{4}+\frac {\cos \left (2 x \right )}{4}\right ) \\ {\mathrm e}^{-4 x} & 0 & {\mathrm e}^{-2 x} \cos \left (2 x \right ) & -{\mathrm e}^{-2 x} \sin \left (2 x \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -\frac {1}{64} & 1 & \frac {1}{32} & \frac {1}{32} \\ \frac {1}{16} & 0 & 0 & -\frac {1}{8} \\ -\frac {1}{4} & 0 & -\frac {1}{4} & \frac {1}{4} \\ 1 & 0 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{cccc} 1 & \frac {\left (-2 \cos \left (2 x \right )-2 \sin \left (2 x \right )\right ) {\mathrm e}^{-2 x}}{4}-\frac {{\mathrm e}^{-4 x}}{4}+\frac {3}{4} & \frac {\left (-\cos \left (2 x \right )-3 \sin \left (2 x \right )\right ) {\mathrm e}^{-2 x}}{8}-\frac {{\mathrm e}^{-4 x}}{8}+\frac {1}{4} & -\frac {{\mathrm e}^{-2 x} \sin \left (2 x \right )}{16}-\frac {{\mathrm e}^{-4 x}}{32}+\frac {1}{32} \\ 0 & {\mathrm e}^{-4 x}+2 \,{\mathrm e}^{-2 x} \sin \left (2 x \right ) & \frac {\left (2 \sin \left (2 x \right )-\cos \left (2 x \right )\right ) {\mathrm e}^{-2 x}}{2}+\frac {{\mathrm e}^{-4 x}}{2} & \frac {{\mathrm e}^{-2 x} \left (-\cos \left (2 x \right )+\sin \left (2 x \right )\right )}{8}+\frac {{\mathrm e}^{-4 x}}{8} \\ 0 & -4 \,{\mathrm e}^{-4 x}+\left (4 \cos \left (2 x \right )-4 \sin \left (2 x \right )\right ) {\mathrm e}^{-2 x} & -2 \,{\mathrm e}^{-4 x}+\left (3 \cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-2 x} & \frac {{\mathrm e}^{-2 x} \cos \left (2 x \right )}{2}-\frac {{\mathrm e}^{-4 x}}{2} \\ 0 & 16 \,{\mathrm e}^{-4 x}-16 \,{\mathrm e}^{-2 x} \cos \left (2 x \right ) & 8 \,{\mathrm e}^{-4 x}+\left (-8 \cos \left (2 x \right )-4 \sin \left (2 x \right )\right ) {\mathrm e}^{-2 x} & 2 \,{\mathrm e}^{-4 x}+\left (-\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-2 x} \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} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \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 {\moverset {\rightarrow }{v}}^{\prime }\left (x \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 {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\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}\frac {1}{\Phi \left (s \right )}\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}\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 (x \right )=\left [\begin {array}{c} -\frac {5}{16}+\frac {\left (-8 x^{3}+8 x^{2}+8 x -2 \cos \left (2 x \right )-5 \sin \left (2 x \right )+8\right ) {\mathrm e}^{-2 x}}{8}-\frac {7 \,{\mathrm e}^{-4 x}}{16} \\ \frac {\left (8 x^{3}-20 x^{2}-3 \cos \left (2 x \right )+7 \sin \left (2 x \right )-4\right ) {\mathrm e}^{-2 x}}{4}+\frac {7 \,{\mathrm e}^{-4 x}}{4} \\ \left (-4 x^{3}+16 x^{2}-10 x +5 \cos \left (2 x \right )-2 \sin \left (2 x \right )+2\right ) {\mathrm e}^{-2 x}-7 \,{\mathrm e}^{-4 x} \\ \left (8 x^{3}-44 x^{2}+52 x -14 \cos \left (2 x \right )-6 \sin \left (2 x \right )-14\right ) {\mathrm e}^{-2 x}+28 \,{\mathrm e}^{-4 x} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right )+\left [\begin {array}{c} -\frac {5}{16}+\frac {\left (-8 x^{3}+8 x^{2}+8 x -2 \cos \left (2 x \right )-5 \sin \left (2 x \right )+8\right ) {\mathrm e}^{-2 x}}{8}-\frac {7 \,{\mathrm e}^{-4 x}}{16} \\ \frac {\left (8 x^{3}-20 x^{2}-3 \cos \left (2 x \right )+7 \sin \left (2 x \right )-4\right ) {\mathrm e}^{-2 x}}{4}+\frac {7 \,{\mathrm e}^{-4 x}}{4} \\ \left (-4 x^{3}+16 x^{2}-10 x +5 \cos \left (2 x \right )-2 \sin \left (2 x \right )+2\right ) {\mathrm e}^{-2 x}-7 \,{\mathrm e}^{-4 x} \\ \left (8 x^{3}-44 x^{2}+52 x -14 \cos \left (2 x \right )-6 \sin \left (2 x \right )-14\right ) {\mathrm e}^{-2 x}+28 \,{\mathrm e}^{-4 x} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {5}{16}+\frac {\left (32+\left (-8+c_{3} +c_{4} \right ) \cos \left (2 x \right )+\left (-20+c_{3} -c_{4} \right ) \sin \left (2 x \right )-32 x^{3}+32 x^{2}+32 x \right ) {\mathrm e}^{-2 x}}{32}+\frac {\left (-c_{1} -28\right ) {\mathrm e}^{-4 x}}{64}+c_{2} \end {array} \]
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] -> Calling odsolve with the ODE`, diff(diff(diff(_b(_a), _a), _a), _a) = 16*exp(-2*_a)*_a^3-16*exp(-2*_a)*_a^2-16*exp(-2*_a)*_a-16*e 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 <- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 56
dsolve(diff(y(x),x$4)+8*diff(y(x),x$3)+24*diff(y(x),x$2)+32*diff(y(x),x)=-16*exp(-2*x)*(1+x+x^2-x^3),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (-c_{2} -c_{3} \right ) \cos \left (2 x \right )+\left (c_{2} -c_{3} \right ) \sin \left (2 x \right )-4 x^{3}+4 x^{2}+4 x +4\right ) {\mathrm e}^{-2 x}}{4}-\frac {{\mathrm e}^{-4 x} c_{1}}{4}+c_{4} \]
✓ Solution by Mathematica
Time used: 0.712 (sec). Leaf size: 64
DSolve[y''''[x]+8*y'''[x]+24*y''[x]+32*y'[x]==-16*Exp[-2*x]*(1+x+x^2-x^3),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} e^{-2 x} \left (-4 x^3+4 x^2+4 x-c_3 e^{-2 x}-(c_1+c_2) \cos (2 x)+(c_2-c_1) \sin (2 x)+4\right )+c_4 \]