Internal problem ID [9859]
Internal file name [OUTPUT/8804_Monday_June_06_2022_05_31_11_AM_39533865/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1535.
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_x]]
\[ \boxed {y^{\prime \prime \prime \prime }+4 y=f} \] 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 \[ y^{\prime \prime \prime \prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{4}+4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1-i\\ \lambda _2 &= 1+i\\ \lambda _3 &= -1-i\\ \lambda _4 &= -1+i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{\left (1-i\right ) x} c_{1} +{\mathrm e}^{\left (1+i\right ) x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{\left (-1-i\right ) x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{\left (1-i\right ) x} \\ y_2 &= {\mathrm e}^{\left (1+i\right ) 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 \[ y^{\prime \prime \prime \prime }+4 y = f \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{\left (-1-i\right ) x}, {\mathrm e}^{\left (-1+i\right ) x}, {\mathrm e}^{\left (1-i\right ) x}, {\mathrm e}^{\left (1+i\right ) x}\} \] 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. \[ y_p = A_{1} \] The unknowns \(\{A_{1}\}\) 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 \[ 4 A_{1} = f \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = \frac {f}{4}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {f}{4} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{\left (1-i\right ) x} c_{1} +{\mathrm e}^{\left (1+i\right ) x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{\left (-1-i\right ) x} c_{4}\right ) + \left (\frac {f}{4}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\left (1-i\right ) x} c_{1} +{\mathrm e}^{\left (1+i\right ) x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{\left (-1-i\right ) x} c_{4} +\frac {f}{4} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{\left (1-i\right ) x} c_{1} +{\mathrm e}^{\left (1+i\right ) x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{\left (-1-i\right ) x} c_{4} +\frac {f}{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }+4 y=f \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\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 )=\frac {d}{d x}y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=\frac {d}{d x}y^{\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 )=f -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 )=f -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 & 0 & 0 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ f \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ f \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 & 0 & 0 & 0 \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-\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 ], \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 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}}_{1}\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}}_{2}\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 {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2} \\ -\sin \left (x \right ) \end {array}\right ]\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 {\sin \left (x \right )}{2}+\frac {\cos \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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+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} \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} \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 ) \\ -\frac {{\mathrm e}^{-x} \sin \left (x \right )}{2} & -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2} & \frac {{\mathrm e}^{x} \sin \left (x \right )}{2} & \frac {{\mathrm e}^{x} \cos \left (x \right )}{2} \\ {\mathrm e}^{-x} \left (-\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) & {\mathrm e}^{-x} \left (\frac {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2}\right ) & {\mathrm e}^{x} \left (\frac {\sin \left (x \right )}{2}+\frac {\cos \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} \cos \left (x \right ) & -{\mathrm e}^{-x} \sin \left (x \right ) & {\mathrm e}^{x} \cos \left (x \right ) & -{\mathrm e}^{x} \sin \left (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}{cccc} {\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} \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 ) \\ -\frac {{\mathrm e}^{-x} \sin \left (x \right )}{2} & -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2} & \frac {{\mathrm e}^{x} \sin \left (x \right )}{2} & \frac {{\mathrm e}^{x} \cos \left (x \right )}{2} \\ {\mathrm e}^{-x} \left (-\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) & {\mathrm e}^{-x} \left (\frac {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2}\right ) & {\mathrm e}^{x} \left (\frac {\sin \left (x \right )}{2}+\frac {\cos \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} \cos \left (x \right ) & -{\mathrm e}^{-x} \sin \left (x \right ) & {\mathrm e}^{x} \cos \left (x \right ) & -{\mathrm e}^{x} \sin \left (x \right ) \end {array}\right ]\cdot \left [\begin {array}{cccc} \frac {1}{4} & \frac {1}{4} & -\frac {1}{4} & \frac {1}{4} \\ 0 & -\frac {1}{2} & 0 & \frac {1}{2} \\ -\frac {1}{2} & \frac {1}{2} & \frac {1}{2} & \frac {1}{2} \\ 1 & 0 & 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}{cccc} \frac {\cos \left (x \right ) \left ({\mathrm e}^{-x}+{\mathrm e}^{x}\right )}{2} & \frac {{\mathrm e}^{-x} \left (-\cos \left (x \right )+\sin \left (x \right )\right )}{4}+\frac {{\mathrm e}^{x} \left (\sin \left (x \right )+\cos \left (x \right )\right )}{4} & -\frac {\sin \left (x \right ) \left ({\mathrm e}^{-x}-{\mathrm e}^{x}\right )}{4} & \frac {{\mathrm e}^{-x} \left (\sin \left (x \right )+\cos \left (x \right )\right )}{8}-\frac {{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right )}{8} \\ \frac {\left (-\sin \left (x \right )-\cos \left (x \right )\right ) {\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right )}{2} & \frac {\cos \left (x \right ) \left ({\mathrm e}^{-x}+{\mathrm e}^{x}\right )}{2} & \frac {{\mathrm e}^{-x} \left (-\cos \left (x \right )+\sin \left (x \right )\right )}{4}+\frac {{\mathrm e}^{x} \left (\sin \left (x \right )+\cos \left (x \right )\right )}{4} & -\frac {\sin \left (x \right ) \left ({\mathrm e}^{-x}-{\mathrm e}^{x}\right )}{4} \\ \sin \left (x \right ) \left ({\mathrm e}^{-x}-{\mathrm e}^{x}\right ) & \frac {\left (-\sin \left (x \right )-\cos \left (x \right )\right ) {\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right )}{2} & \frac {\cos \left (x \right ) \left ({\mathrm e}^{-x}+{\mathrm e}^{x}\right )}{2} & \frac {{\mathrm e}^{-x} \left (-\cos \left (x \right )+\sin \left (x \right )\right )}{4}+\frac {{\mathrm e}^{x} \left (\sin \left (x \right )+\cos \left (x \right )\right )}{4} \\ {\mathrm e}^{-x} \left (\cos \left (x \right )-\sin \left (x \right )\right )-{\mathrm e}^{x} \left (\sin \left (x \right )+\cos \left (x \right )\right ) & \sin \left (x \right ) \left ({\mathrm e}^{-x}-{\mathrm e}^{x}\right ) & \frac {\left (-\sin \left (x \right )-\cos \left (x \right )\right ) {\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right )}{2} & \frac {\cos \left (x \right ) \left ({\mathrm e}^{-x}+{\mathrm e}^{x}\right )}{2} \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 )=\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 {f \left ({\mathrm e}^{-x} \cos \left (x \right )-2+{\mathrm e}^{x} \cos \left (x \right )\right )}{8} \\ -\frac {\left (\left (-\sin \left (x \right )-\cos \left (x \right )\right ) {\mathrm e}^{-x}+{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right )\right ) f}{8} \\ -\frac {f \sin \left (x \right ) \left ({\mathrm e}^{-x}-{\mathrm e}^{x}\right )}{4} \\ \frac {\left ({\mathrm e}^{-x} \left (-\cos \left (x \right )+\sin \left (x \right )\right )+{\mathrm e}^{x} \left (\sin \left (x \right )+\cos \left (x \right )\right )\right ) f}{4} \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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right )+\left [\begin {array}{c} -\frac {f \left ({\mathrm e}^{-x} \cos \left (x \right )-2+{\mathrm e}^{x} \cos \left (x \right )\right )}{8} \\ -\frac {\left (\left (-\sin \left (x \right )-\cos \left (x \right )\right ) {\mathrm e}^{-x}+{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right )\right ) f}{8} \\ -\frac {f \sin \left (x \right ) \left ({\mathrm e}^{-x}-{\mathrm e}^{x}\right )}{4} \\ \frac {\left ({\mathrm e}^{-x} \left (-\cos \left (x \right )+\sin \left (x \right )\right )+{\mathrm e}^{x} \left (\sin \left (x \right )+\cos \left (x \right )\right )\right ) f}{4} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (-f +2 c_{1} +2 c_{2} \right ) \cos \left (x \right )+2 \sin \left (x \right ) \left (c_{1} -c_{2} \right )\right ) {\mathrm e}^{-x}}{8}-\frac {{\mathrm e}^{x} \left (f +2 c_{3} -2 c_{4} \right ) \cos \left (x \right )}{8}+\frac {\sin \left (x \right ) \left (c_{3} +c_{4} \right ) {\mathrm e}^{x}}{4}+\frac {f}{4} \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: 0.015 (sec). Leaf size: 36
dsolve(diff(diff(diff(diff(y(x),x),x),x),x)+4*y(x)-f=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {f}{4}+\cos \left (x \right ) c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{x} \sin \left (x \right )+c_{3} {\mathrm e}^{-x} \cos \left (x \right )+c_{4} \sin \left (x \right ) {\mathrm e}^{-x} \]
✓ Solution by Mathematica
Time used: 0.308 (sec). Leaf size: 172
DSolve[-f[x] + 4*y[x] + Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x} \left (\cos (x) \int _1^x\frac {1}{8} e^{K[1]} f(K[1]) (\cos (K[1])-\sin (K[1]))dK[1]+e^{2 x} \cos (x) \int _1^x-\frac {1}{8} e^{-K[4]} f(K[4]) (\cos (K[4])+\sin (K[4]))dK[4]+\sin (x) \int _1^x\frac {1}{8} e^{K[2]} f(K[2]) (\cos (K[2])+\sin (K[2]))dK[2]+e^{2 x} \sin (x) \int _1^x\frac {1}{8} e^{-K[3]} f(K[3]) (\cos (K[3])-\sin (K[3]))dK[3]+c_1 \cos (x)+c_4 e^{2 x} \cos (x)+c_2 \sin (x)+c_3 e^{2 x} \sin (x)\right ) \]