problem section 9.3, problem 37

Internal problem ID [1534]
Internal ﬁle name [OUTPUT/1535_Sunday_June_05_2022_02_21_07_AM_21645714/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 37.
ODE order: 4.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-2 y^{\prime \prime }-8 y^{\prime }-8 y={\mathrm e}^{x} \left (8 \cos \left (x \right )+16 \sin \left (x \right )\right )} \] 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 \[ y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-2 y^{\prime \prime }-8 y^{\prime }-8 y = 0 \] The characteristic equation is \[ \lambda ^{4}+2 \lambda ^{3}-2 \lambda ^{2}-8 \lambda -8 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2\\ \lambda _3 &= -1-i\\ \lambda _4 &= -1+i \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{\left (-1-i\right ) x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 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 \[ y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-2 y^{\prime \prime }-8 y^{\prime }-8 y = {\mathrm e}^{x} \left (8 \cos \left (x \right )+16 \sin \left (x \right )\right ) \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x} \left (8 \cos \left (x \right )+16 \sin \left (x \right )\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x} \cos \left (x \right ), {\mathrm e}^{x} \sin \left (x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{\left (-1-i\right ) x}, {\mathrm e}^{\left (-1+i\right ) x}, {\mathrm e}^{-2 x}, {\mathrm e}^{2 x}\} \] Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set. \[ y_p = A_{1} {\mathrm e}^{x} \cos \left (x \right )+A_{2} {\mathrm e}^{x} \sin \left (x \right ) \] The unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coeﬃcients. Substituting the trial solution into the ODE and simplifying gives \[ -24 A_{1} {\mathrm e}^{x} \cos \left (x \right )-24 A_{2} {\mathrm e}^{x} \sin \left (x \right )+8 A_{1} {\mathrm e}^{x} \sin \left (x \right )-8 A_{2} {\mathrm e}^{x} \cos \left (x \right ) = {\mathrm e}^{x} \left (8 \cos \left (x \right )+16 \sin \left (x \right )\right ) \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {1}{10}}, A_{2} = -{\frac {7}{10}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {{\mathrm e}^{x} \cos \left (x \right )}{10}-\frac {7 \,{\mathrm e}^{x} \sin \left (x \right )}{10} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{\left (-1-i\right ) x} c_{4}\right ) + \left (-\frac {{\mathrm e}^{x} \cos \left (x \right )}{10}-\frac {7 \,{\mathrm e}^{x} \sin \left (x \right )}{10}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{\left (-1-i\right ) x} c_{4} -\frac {{\mathrm e}^{x} \cos \left (x \right )}{10}-\frac {7 \,{\mathrm e}^{x} \sin \left (x \right )}{10} \\ \end{align*}

Veriﬁcation of solutions

\[ y = {\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{\left (-1-i\right ) x} c_{4} -\frac {{\mathrm e}^{x} \cos \left (x \right )}{10}-\frac {7 \,{\mathrm e}^{x} \sin \left (x \right )}{10} \] Veriﬁed OK.

19.37.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }+2 \frac {d}{d x}y^{\prime \prime }-2 \frac {d}{d x}y^{\prime }-8 y^{\prime }-8 y={\mathrm e}^{x} \left (8 \cos \left (x \right )+16 \sin \left (x \right )\right ) \\ \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 } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }=8 y+8 \,{\mathrm e}^{x} \cos \left (x \right )+16 \,{\mathrm e}^{x} \sin \left (x \right )-2 \frac {d}{d x}y^{\prime \prime }+2 \frac {d}{d x}y^{\prime }+8 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} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }+2 \frac {d}{d x}y^{\prime \prime }-2 \frac {d}{d x}y^{\prime }-8 y^{\prime }-8 y=8 \,{\mathrm e}^{x} \left (\cos \left (x \right )+2 \sin \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 \\ {} & \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 )=8 \,{\mathrm e}^{x} \cos \left (x \right )+16 \,{\mathrm e}^{x} \sin \left (x \right )-2 y_{4}\left (x \right )+2 y_{3}\left (x \right )+8 y_{2}\left (x \right )+8 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 )=8 \,{\mathrm e}^{x} \cos \left (x \right )+16 \,{\mathrm e}^{x} \sin \left (x \right )-2 y_{4}\left (x \right )+2 y_{3}\left (x \right )+8 y_{2}\left (x \right )+8 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 \\ 8 & 8 & 2 & -2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 8 \,{\mathrm e}^{x} \cos \left (x \right )+16 \,{\mathrm e}^{x} \sin \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 \\ 0 \\ 8 \,{\mathrm e}^{x} \cos \left (x \right )+16 \,{\mathrm e}^{x} \sin \left (x \right ) \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 \\ 8 & 8 & 2 & -2 \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 [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 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 [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\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} -\frac {{\mathrm e}^{-2 x}}{8} & \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 ) \\ \frac {{\mathrm e}^{-2 x}}{4} & \frac {{\mathrm e}^{2 x}}{4} & -\frac {{\mathrm e}^{-x} \sin \left (x \right )}{2} & -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2} \\ -\frac {{\mathrm e}^{-2 x}}{2} & \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}^{-2 x} & {\mathrm e}^{2 x} & {\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 \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}^{-2 x}}{8} & \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 ) \\ \frac {{\mathrm e}^{-2 x}}{4} & \frac {{\mathrm e}^{2 x}}{4} & -\frac {{\mathrm e}^{-x} \sin \left (x \right )}{2} & -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2} \\ -\frac {{\mathrm e}^{-2 x}}{2} & \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}^{-2 x} & {\mathrm e}^{2 x} & {\mathrm e}^{-x} \cos \left (x \right ) & -{\mathrm e}^{-x} \sin \left (x \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -\frac {1}{8} & \frac {1}{8} & \frac {1}{4} & \frac {1}{4} \\ \frac {1}{4} & \frac {1}{4} & 0 & -\frac {1}{2} \\ -\frac {1}{2} & \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 {\left ({\mathrm e}^{4 x}+4 \,{\mathrm e}^{x} \cos \left (x \right )+12 \,{\mathrm e}^{x} \sin \left (x \right )+5\right ) {\mathrm e}^{-2 x}}{10} & -\frac {2 \,{\mathrm e}^{-2 x} \left (-\frac {3 \,{\mathrm e}^{4 x}}{8}-\frac {5}{8}+\left (\cos \left (x \right )-2 \sin \left (x \right )\right ) {\mathrm e}^{x}\right )}{5} & \frac {\left (-\cos \left (x \right )-3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{10}+\frac {{\mathrm e}^{2 x}}{10} & \frac {\left ({\mathrm e}^{4 x}-5+\left (4 \cos \left (x \right )-8 \sin \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{40} \\ \frac {\left ({\mathrm e}^{4 x}-5+\left (4 \cos \left (x \right )-8 \sin \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{5} & \frac {\left (3 \,{\mathrm e}^{4 x}-5+\left (12 \cos \left (x \right )-4 \sin \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{10} & \frac {\left (2 \sin \left (x \right )-\cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{2 x}}{5} & \frac {\left ({\mathrm e}^{4 x}+5+\left (-6 \cos \left (x \right )+2 \sin \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{20} \\ \frac {2 \left (5+{\mathrm e}^{4 x}+2 \left (\sin \left (x \right )-3 \cos \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{5} & \frac {\left (3 \,{\mathrm e}^{4 x}+5+\left (-8 \cos \left (x \right )-4 \sin \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{5} & \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 ({\mathrm e}^{4 x}-5+\left (4 \cos \left (x \right )+2 \sin \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{10} \\ \frac {4 \left (-5+{\mathrm e}^{4 x}+2 \left (2 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{5} & \frac {4 \left (\frac {3 \,{\mathrm e}^{4 x}}{2}-\frac {5}{2}+\left (\cos \left (x \right )+3 \sin \left (x \right )\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{5} & \frac {\left (-4 \cos \left (x \right )-2 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {4 \,{\mathrm e}^{2 x}}{5} & -\frac {{\mathrm e}^{-2 x} \left (-{\mathrm e}^{4 x}-5+\left (\cos \left (x \right )+3 \sin \left (x \right )\right ) {\mathrm e}^{x}\right )}{5} \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 {{\mathrm e}^{-2 x} \left (-\left (7 \sin \left (x \right )+\cos \left (x \right )\right ) {\mathrm e}^{3 x}-3 \,{\mathrm e}^{x} \cos \left (x \right )+{\mathrm e}^{x} \sin \left (x \right )+3 \,{\mathrm e}^{4 x}+1\right )}{10} \\ \frac {{\mathrm e}^{-2 x} \left (\left (-4 \cos \left (x \right )-3 \sin \left (x \right )\right ) {\mathrm e}^{3 x}+2 \,{\mathrm e}^{x} \cos \left (x \right )+{\mathrm e}^{x} \sin \left (x \right )+3 \,{\mathrm e}^{4 x}-1\right )}{5} \\ -\frac {{\mathrm e}^{-2 x} \left (\left (7 \cos \left (x \right )-\sin \left (x \right )\right ) {\mathrm e}^{3 x}+{\mathrm e}^{x} \cos \left (x \right )+3 \,{\mathrm e}^{x} \sin \left (x \right )-6 \,{\mathrm e}^{4 x}-2\right )}{5} \\ -\frac {2 \,{\mathrm e}^{-2 x} \left (\left (3 \cos \left (x \right )-4 \sin \left (x \right )\right ) {\mathrm e}^{3 x}+{\mathrm e}^{x} \cos \left (x \right )-2 \,{\mathrm e}^{x} \sin \left (x \right )-6 \,{\mathrm e}^{4 x}+2\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 {{\mathrm e}^{-2 x} \left (-\left (7 \sin \left (x \right )+\cos \left (x \right )\right ) {\mathrm e}^{3 x}-3 \,{\mathrm e}^{x} \cos \left (x \right )+{\mathrm e}^{x} \sin \left (x \right )+3 \,{\mathrm e}^{4 x}+1\right )}{10} \\ \frac {{\mathrm e}^{-2 x} \left (\left (-4 \cos \left (x \right )-3 \sin \left (x \right )\right ) {\mathrm e}^{3 x}+2 \,{\mathrm e}^{x} \cos \left (x \right )+{\mathrm e}^{x} \sin \left (x \right )+3 \,{\mathrm e}^{4 x}-1\right )}{5} \\ -\frac {{\mathrm e}^{-2 x} \left (\left (7 \cos \left (x \right )-\sin \left (x \right )\right ) {\mathrm e}^{3 x}+{\mathrm e}^{x} \cos \left (x \right )+3 \,{\mathrm e}^{x} \sin \left (x \right )-6 \,{\mathrm e}^{4 x}-2\right )}{5} \\ -\frac {2 \,{\mathrm e}^{-2 x} \left (\left (3 \cos \left (x \right )-4 \sin \left (x \right )\right ) {\mathrm e}^{3 x}+{\mathrm e}^{x} \cos \left (x \right )-2 \,{\mathrm e}^{x} \sin \left (x \right )-6 \,{\mathrm e}^{4 x}+2\right )}{5} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-2 x} \left (\left (7 \sin \left (x \right )+\cos \left (x \right )\right ) {\mathrm e}^{3 x}+\left (-\frac {5 c_{2}}{4}-3\right ) {\mathrm e}^{4 x}+\left (\left (-\frac {5 c_{3}}{2}-\frac {5 c_{4}}{2}+3\right ) \cos \left (x \right )-\frac {5 \sin \left (x \right ) \left (c_{3} -c_{4} +\frac {2}{5}\right )}{2}\right ) {\mathrm e}^{x}+\frac {5 c_{1}}{4}-1\right )}{10} \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`

\[ y \left (x \right ) = -\frac {\left (\left (7 \sin \left (x \right )+\cos \left (x \right )\right ) {\mathrm e}^{3 x}-10 c_{3} \cos \left (x \right ) {\mathrm e}^{x}-10 c_{4} \sin \left (x \right ) {\mathrm e}^{x}-10 c_{2} {\mathrm e}^{4 x}-10 c_{1} \right ) {\mathrm e}^{-2 x}}{10} \]

\[ y(x)\to c_3 e^{-2 x}+c_4 e^{2 x}-\frac {1}{10} e^x (7 \sin (x)+\cos (x))+c_2 e^{-x} \cos (x)+c_1 e^{-x} \sin (x) \]

19.37 problem section 9.3, problem 37

19.37.1 Maple step by step solution