Internal problem ID [1517]
Internal file name [OUTPUT/1518_Sunday_June_05_2022_02_20_23_AM_71425407/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 20.
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, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }-2 y^{\prime \prime }-6 y^{\prime }-4 y=-{\mathrm e}^{2 x} \left (15 x^{2}+28 x +4\right )} \] 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 \[ y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }-2 y^{\prime \prime }-6 y^{\prime }-4 y = 0 \] The characteristic equation is \[ \lambda ^{4}+\lambda ^{3}-2 \lambda ^{2}-6 \lambda -4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -1\\ \lambda _3 &= -1-i\\ \lambda _4 &= -1+i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-1-i\right ) x} c_{3} +{\mathrm e}^{\left (-1+i\right ) x} c_{4} \] 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}^{\left (-1-i\right ) x} \\ y_4 &= {\mathrm e}^{\left (-1+i\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }-2 y^{\prime \prime }-6 y^{\prime }-4 y = -{\mathrm e}^{2 x} \left (15 x^{2}+28 x +4\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ -{\mathrm e}^{2 x} \left (15 x^{2}+28 x +4\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{2 x}, x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{\left (-1-i\right ) x}, {\mathrm e}^{\left (-1+i\right ) x}, {\mathrm e}^{-x}, {\mathrm e}^{2 x}\} \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{2 x}, x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x^{3}\}] \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{2 x}+A_{2} x^{2} {\mathrm e}^{2 x}+A_{3} {\mathrm e}^{2 x} x^{3} \] 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 \[ 168 A_{3} {\mathrm e}^{2 x} x +60 A_{2} x \,{\mathrm e}^{2 x}+90 A_{3} {\mathrm e}^{2 x} x^{2}+30 A_{1} {\mathrm e}^{2 x}+56 A_{2} {\mathrm e}^{2 x}+54 A_{3} {\mathrm e}^{2 x} = -{\mathrm e}^{2 x} \left (15 x^{2}+28 x +4\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{6}}, A_{2} = 0, A_{3} = -{\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x \,{\mathrm e}^{2 x}}{6}-\frac {{\mathrm e}^{2 x} x^{3}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-1-i\right ) x} c_{3} +{\mathrm e}^{\left (-1+i\right ) x} c_{4}\right ) + \left (\frac {x \,{\mathrm e}^{2 x}}{6}-\frac {{\mathrm e}^{2 x} x^{3}}{6}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-1-i\right ) x} c_{3} +{\mathrm e}^{\left (-1+i\right ) x} c_{4} +\frac {x \,{\mathrm e}^{2 x}}{6}-\frac {{\mathrm e}^{2 x} x^{3}}{6} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{\left (-1-i\right ) x} c_{3} +{\mathrm e}^{\left (-1+i\right ) x} c_{4} +\frac {x \,{\mathrm e}^{2 x}}{6}-\frac {{\mathrm e}^{2 x} x^{3}}{6} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }-2 y^{\prime \prime }-6 y^{\prime }-4 y=-{\mathrm e}^{2 x} \left (15 x^{2}+28 x +4\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \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 )=-15 x^{2} {\mathrm e}^{2 x}-28 x \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{2 x}-y_{4}\left (x \right )+2 y_{3}\left (x \right )+6 y_{2}\left (x \right )+4 y_{1}\left (x \right ) \\ & {} & \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 )=-15 x^{2} {\mathrm e}^{2 x}-28 x \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{2 x}-y_{4}\left (x \right )+2 y_{3}\left (x \right )+6 y_{2}\left (x \right )+4 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} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 4 & 6 & 2 & -1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ -15 x^{2} {\mathrm e}^{2 x}-28 x \,{\mathrm e}^{2 x}-4 \,{\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 \\ -15 x^{2} {\mathrm e}^{2 x}-28 x \,{\mathrm e}^{2 x}-4 \,{\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 \\ 4 & 6 & 2 & -1 \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 [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1-\mathrm {I}, \left [\begin {array}{c} \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\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 {1}{4}-\frac {\mathrm {I}}{4} \\ \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 \\ 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 \\ 1 \end {array}\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}}_{2}={\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 complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-1-\mathrm {I}, \left [\begin {array}{c} \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\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 {1}{4}+\frac {\mathrm {I}}{4} \\ -\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 {1}{4}+\frac {\mathrm {I}}{4} \\ -\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} \left (\frac {1}{4}+\frac {\mathrm {I}}{4}\right ) \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ -\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}}_{3}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {\cos \left (x \right )}{4}+\frac {\sin \left (x \right )}{4} \\ -\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}}_{4}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \\ -\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 )=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} -{\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{8} & {\mathrm e}^{-x} \left (\frac {\cos \left (x \right )}{4}+\frac {\sin \left (x \right )}{4}\right ) & {\mathrm e}^{-x} \left (\frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4}\right ) \\ {\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{4} & -\frac {\sin \left (x \right ) {\mathrm e}^{-x}}{2} & -\frac {\cos \left (x \right ) {\mathrm e}^{-x}}{2} \\ -{\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{2} & {\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}^{2 x} & \cos \left (x \right ) {\mathrm e}^{-x} & -\sin \left (x \right ) {\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 \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} -{\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{8} & {\mathrm e}^{-x} \left (\frac {\cos \left (x \right )}{4}+\frac {\sin \left (x \right )}{4}\right ) & {\mathrm e}^{-x} \left (\frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4}\right ) \\ {\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{4} & -\frac {\sin \left (x \right ) {\mathrm e}^{-x}}{2} & -\frac {\cos \left (x \right ) {\mathrm e}^{-x}}{2} \\ -{\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{2} & {\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}^{2 x} & \cos \left (x \right ) {\mathrm e}^{-x} & -\sin \left (x \right ) {\mathrm e}^{-x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -1 & \frac {1}{8} & \frac {1}{4} & \frac {1}{4} \\ 1 & \frac {1}{4} & 0 & -\frac {1}{2} \\ -1 & \frac {1}{2} & -\frac {1}{2} & \frac {1}{2} \\ 1 & 1 & 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} \frac {2 \left (10-3 \cos \left (x \right )+6 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {{\mathrm e}^{2 x}}{15} & \frac {\left (10-12 \cos \left (x \right )+9 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {2 \,{\mathrm e}^{2 x}}{15} & \frac {\left (-\cos \left (x \right )-3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{10}+\frac {{\mathrm e}^{2 x}}{10} & \frac {\left (-10+9 \cos \left (x \right )-3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{30}+\frac {{\mathrm e}^{2 x}}{30} \\ \frac {2 \left (-10+9 \cos \left (x \right )-3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {2 \,{\mathrm e}^{2 x}}{15} & \frac {\left (-10+3 \sin \left (x \right )+21 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {4 \,{\mathrm e}^{2 x}}{15} & \frac {\left (2 \sin \left (x \right )-\cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{2 x}}{5} & \frac {\left (5-3 \sin \left (x \right )-6 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {{\mathrm e}^{2 x}}{15} \\ \frac {4 \left (5-3 \sin \left (x \right )-6 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {4 \,{\mathrm e}^{2 x}}{15} & \frac {2 \left (5-9 \cos \left (x \right )-12 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {8 \,{\mathrm e}^{2 x}}{15} & \frac {\left (3 \cos \left (x \right )-\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{2 x}}{5} & \frac {\left (-5+3 \cos \left (x \right )+9 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {2 \,{\mathrm e}^{2 x}}{15} \\ \frac {4 \left (-5+3 \cos \left (x \right )+9 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {8 \,{\mathrm e}^{2 x}}{15} & \frac {2 \left (-5-3 \cos \left (x \right )+21 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {16 \,{\mathrm e}^{2 x}}{15} & \frac {\left (-2 \sin \left (x \right )-4 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {4 \,{\mathrm e}^{2 x}}{5} & \frac {\left (5+6 \cos \left (x \right )-12 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{15}+\frac {4 \,{\mathrm e}^{2 x}}{15} \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 {\left (-9 \cos \left (x \right )+18 \sin \left (x \right )+20\right ) {\mathrm e}^{-x}}{90}-\frac {\left (x^{3}-x +\frac {11}{15}\right ) {\mathrm e}^{2 x}}{6} \\ \frac {\left (-30 x^{3}-45 x^{2}+30 x -7\right ) {\mathrm e}^{2 x}}{90}+\frac {3 \left (\cos \left (x \right )-\frac {\sin \left (x \right )}{3}-\frac {20}{27}\right ) {\mathrm e}^{-x}}{10} \\ \frac {\left (-30 x^{3}-90 x^{2}-15 x +8\right ) {\mathrm e}^{2 x}}{45}-\frac {2 \,{\mathrm e}^{-x} \left (\cos \left (x \right )+\frac {\sin \left (x \right )}{2}-\frac {5}{9}\right )}{5} \\ \frac {\left (-60 x^{3}-270 x^{2}-210 x +1\right ) {\mathrm e}^{2 x}}{45}+\frac {{\mathrm e}^{-x} \left (\cos \left (x \right )+3 \sin \left (x \right )-\frac {10}{9}\right )}{5} \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 {\left (-9 \cos \left (x \right )+18 \sin \left (x \right )+20\right ) {\mathrm e}^{-x}}{90}-\frac {\left (x^{3}-x +\frac {11}{15}\right ) {\mathrm e}^{2 x}}{6} \\ \frac {\left (-30 x^{3}-45 x^{2}+30 x -7\right ) {\mathrm e}^{2 x}}{90}+\frac {3 \left (\cos \left (x \right )-\frac {\sin \left (x \right )}{3}-\frac {20}{27}\right ) {\mathrm e}^{-x}}{10} \\ \frac {\left (-30 x^{3}-90 x^{2}-15 x +8\right ) {\mathrm e}^{2 x}}{45}-\frac {2 \,{\mathrm e}^{-x} \left (\cos \left (x \right )+\frac {\sin \left (x \right )}{2}-\frac {5}{9}\right )}{5} \\ \frac {\left (-60 x^{3}-270 x^{2}-210 x +1\right ) {\mathrm e}^{2 x}}{45}+\frac {{\mathrm e}^{-x} \left (\cos \left (x \right )+3 \sin \left (x \right )-\frac {10}{9}\right )}{5} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (90 c_{3} +90 c_{4} -36\right ) \cos \left (x \right )+\left (90 c_{3} -90 c_{4} +72\right ) \sin \left (x \right )-360 c_{1} +80\right ) {\mathrm e}^{-x}}{360}-\frac {{\mathrm e}^{2 x} \left (x^{3}-x -\frac {3}{4} c_{2} +\frac {11}{15}\right )}{6} \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] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 36
dsolve(1*diff(y(x),x$4)+1*diff(y(x),x$3)-2*diff(y(x),x$2)-6*diff(y(x),x)-4*y(x)=-exp(2*x)*(4+28*x+15*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \left (\cos \left (x \right ) c_{3} +c_{4} \sin \left (x \right )+c_{1} \right ) {\mathrm e}^{-x}-\frac {{\mathrm e}^{2 x} \left (x^{3}-6 c_{2} -x \right )}{6} \]
✓ Solution by Mathematica
Time used: 0.193 (sec). Leaf size: 65
DSolve[1*y''''[x]+1*y'''[x]-2*y''[x]-6*y'[x]-4*y[x]==-Exp[2*x]*(4+28*x+15*x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{90} e^{-x} \left (-15 e^{3 x} x^3+15 e^{3 x} x-11 e^{3 x}+90 c_4 e^{3 x}+90 c_2 \cos (x)+90 c_1 \sin (x)+90 c_3\right ) \]