problem 16.1 (iv)

Internal problem ID [12046]
Internal ﬁle name [OUTPUT/10699_Sunday_September_03_2023_12_36_41_PM_2817466/index.tex]

Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section: Chapter 16, Higher order linear equations with constant coeﬃcients. Exercises page 153
Problem number: 16.1 (iv).
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {x^{\prime \prime \prime \prime }-5 x^{\prime \prime }+4 x={\mathrm e}^{t}} \] This is higher order nonhomogeneous ODE. Let the solution be \[ x = x_h + x_p \] Where \(x_h\) is the solution to the homogeneous ODE And \(x_p\) is a particular solution to the nonhomogeneous ODE. \(x_h\) is the solution to \[ x^{\prime \prime \prime \prime }-5 x^{\prime \prime }+4 x = 0 \] The characteristic equation is \[ \lambda ^{4}-5 \lambda ^{2}+4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2\\ \lambda _3 &= 1\\ \lambda _4 &= -1 \end {align*}

Therefore the homogeneous solution is \[ x_h(t)=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-2 t}+{\mathrm e}^{t} c_{3} +{\mathrm e}^{2 t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} x_1 &= {\mathrm e}^{-t} \\ x_2 &= {\mathrm e}^{-2 t} \\ x_3 &= {\mathrm e}^{t} \\ x_4 &= {\mathrm e}^{2 t} \\ \end{align*} Now the particular solution to the given ODE is found \[ x^{\prime \prime \prime \prime }-5 x^{\prime \prime }+4 x = {\mathrm e}^{t} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{t} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{t}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{t}, {\mathrm e}^{-2 t}, {\mathrm e}^{-t}, {\mathrm e}^{2 t}\} \] Since \({\mathrm e}^{t}\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t \,{\mathrm e}^{t}\}] \] 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. \[ x_p = A_{1} t \,{\mathrm e}^{t} \] The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(x_p\) into the ODE and comparing coeﬃcients. Substituting the trial solution into the ODE and simplifying gives \[ -6 A_{1} {\mathrm e}^{t} = {\mathrm e}^{t} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(x_p\), gives the particular solution \[ x_p = -\frac {t \,{\mathrm e}^{t}}{6} \] Therefore the general solution is \begin{align*} x &= x_h + x_p \\ &= \left (c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-2 t}+{\mathrm e}^{t} c_{3} +{\mathrm e}^{2 t} c_{4}\right ) + \left (-\frac {t \,{\mathrm e}^{t}}{6}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-2 t}+{\mathrm e}^{t} c_{3} +{\mathrm e}^{2 t} c_{4} -\frac {t \,{\mathrm e}^{t}}{6} \\ \end{align*}

Veriﬁcation of solutions

\[ x = c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-2 t}+{\mathrm e}^{t} c_{3} +{\mathrm e}^{2 t} c_{4} -\frac {t \,{\mathrm e}^{t}}{6} \] Veriﬁed OK.

9.4.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime \prime \prime \prime }-5 x^{\prime \prime }+4 x={\mathrm e}^{t} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & x^{\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} x_{1}\left (t \right ) \\ {} & {} & x_{1}\left (t \right )=x \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} x_{2}\left (t \right ) \\ {} & {} & x_{2}\left (t \right )=x^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} x_{3}\left (t \right ) \\ {} & {} & x_{3}\left (t \right )=x^{\prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} x_{4}\left (t \right ) \\ {} & {} & x_{4}\left (t \right )=x^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} x_{4}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & x_{4}^{\prime }\left (t \right )={\mathrm e}^{t}+5 x_{3}\left (t \right )-4 x_{1}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [x_{2}\left (t \right )=x_{1}^{\prime }\left (t \right ), x_{3}\left (t \right )=x_{2}^{\prime }\left (t \right ), x_{4}\left (t \right )=x_{3}^{\prime }\left (t \right ), x_{4}^{\prime }\left (t \right )={\mathrm e}^{t}+5 x_{3}\left (t \right )-4 x_{1}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=\left [\begin {array}{c} x_{1}\left (t \right ) \\ x_{2}\left (t \right ) \\ x_{3}\left (t \right ) \\ x_{4}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}^{\prime }\left (t \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -4 & 0 & 5 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{x}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ {\mathrm e}^{t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ {\mathrm e}^{t} \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 & 5 & 0 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{x}}\left (t \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 [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \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 ]\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 }{x}}_{1}={\mathrm e}^{-2 t}\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 [-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 }{x}}_{2}={\mathrm e}^{-t}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\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 }{x}}_{3}={\mathrm e}^{t}\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 }{x}}_{4}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{x}}_{p}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=c_{1} {\moverset {\rightarrow }{x}}_{1}+c_{2} {\moverset {\rightarrow }{x}}_{2}+c_{3} {\moverset {\rightarrow }{x}}_{3}+c_{4} {\moverset {\rightarrow }{x}}_{4}+{\moverset {\rightarrow }{x}}_{p}\left (t \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (t \right )=\left [\begin {array}{cccc} -\frac {{\mathrm e}^{-2 t}}{8} & -{\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{8} \\ \frac {{\mathrm e}^{-2 t}}{4} & {\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{4} \\ -\frac {{\mathrm e}^{-2 t}}{2} & -{\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{2} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{-t} & {\mathrm e}^{t} & {\mathrm e}^{2 t} \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (t \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (t \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 (t \right )=\phi \left (t \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{cccc} -\frac {{\mathrm e}^{-2 t}}{8} & -{\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{8} \\ \frac {{\mathrm e}^{-2 t}}{4} & {\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{4} \\ -\frac {{\mathrm e}^{-2 t}}{2} & -{\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{2} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{-t} & {\mathrm e}^{t} & {\mathrm e}^{2 t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -\frac {1}{8} & -1 & 1 & \frac {1}{8} \\ \frac {1}{4} & 1 & 1 & \frac {1}{4} \\ -\frac {1}{2} & -1 & 1 & \frac {1}{2} \\ 1 & 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{cccc} -\frac {\left ({\mathrm e}^{4 t}-4 \,{\mathrm e}^{3 t}-4 \,{\mathrm e}^{t}+1\right ) {\mathrm e}^{-2 t}}{6} & -\frac {\left ({\mathrm e}^{4 t}-8 \,{\mathrm e}^{3 t}+8 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{12} & -\frac {\left (-{\mathrm e}^{4 t}+{\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{6} & \frac {\left ({\mathrm e}^{4 t}-2 \,{\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{12} \\ -\frac {\left ({\mathrm e}^{4 t}-2 \,{\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{3} & -\frac {\left ({\mathrm e}^{4 t}-4 \,{\mathrm e}^{3 t}-4 \,{\mathrm e}^{t}+1\right ) {\mathrm e}^{-2 t}}{6} & \frac {\left (2 \,{\mathrm e}^{4 t}-{\mathrm e}^{3 t}+{\mathrm e}^{t}-2\right ) {\mathrm e}^{-2 t}}{6} & -\frac {\left (-{\mathrm e}^{4 t}+{\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{6} \\ \frac {2 \left (-{\mathrm e}^{4 t}+{\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{3} & -\frac {\left ({\mathrm e}^{4 t}-2 \,{\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{3} & -\frac {\left (-4 \,{\mathrm e}^{4 t}+{\mathrm e}^{3 t}+{\mathrm e}^{t}-4\right ) {\mathrm e}^{-2 t}}{6} & \frac {\left (2 \,{\mathrm e}^{4 t}-{\mathrm e}^{3 t}+{\mathrm e}^{t}-2\right ) {\mathrm e}^{-2 t}}{6} \\ -\frac {2 \left (2 \,{\mathrm e}^{4 t}-{\mathrm e}^{3 t}+{\mathrm e}^{t}-2\right ) {\mathrm e}^{-2 t}}{3} & \frac {2 \left (-{\mathrm e}^{4 t}+{\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{3} & \frac {\left (8 \,{\mathrm e}^{4 t}-{\mathrm e}^{3 t}+{\mathrm e}^{t}-8\right ) {\mathrm e}^{-2 t}}{6} & -\frac {\left (-4 \,{\mathrm e}^{4 t}+{\mathrm e}^{3 t}+{\mathrm e}^{t}-4\right ) {\mathrm e}^{-2 t}}{6} \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 (t \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{x}}_{p}\left (t \right )=\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{p}^{\prime }\left (t \right )=\Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \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 (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )={\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=\frac {1}{\Phi \left (t \right )}\cdot {\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (t \right )=\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{p}\left (t \right )=\Phi \left (t \right )\cdot \left (\int _{0}^{t}\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 }{x}}_{p}\left (t \right )=\left [\begin {array}{c} -\frac {\left (-3 \,{\mathrm e}^{4 t}+6 t \,{\mathrm e}^{3 t}+{\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{36} \\ \frac {\left (6 \,{\mathrm e}^{4 t}-6 t \,{\mathrm e}^{3 t}-7 \,{\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-2\right ) {\mathrm e}^{-2 t}}{36} \\ \frac {\left (12 \,{\mathrm e}^{4 t}-6 t \,{\mathrm e}^{3 t}-13 \,{\mathrm e}^{3 t}-3 \,{\mathrm e}^{t}+4\right ) {\mathrm e}^{-2 t}}{36} \\ \frac {\left (24 \,{\mathrm e}^{4 t}-6 t \,{\mathrm e}^{3 t}-19 \,{\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-8\right ) {\mathrm e}^{-2 t}}{36} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=c_{1} {\moverset {\rightarrow }{x}}_{1}+c_{2} {\moverset {\rightarrow }{x}}_{2}+c_{3} {\moverset {\rightarrow }{x}}_{3}+c_{4} {\moverset {\rightarrow }{x}}_{4}+\left [\begin {array}{c} -\frac {\left (-3 \,{\mathrm e}^{4 t}+6 t \,{\mathrm e}^{3 t}+{\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-2 t}}{36} \\ \frac {\left (6 \,{\mathrm e}^{4 t}-6 t \,{\mathrm e}^{3 t}-7 \,{\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-2\right ) {\mathrm e}^{-2 t}}{36} \\ \frac {\left (12 \,{\mathrm e}^{4 t}-6 t \,{\mathrm e}^{3 t}-13 \,{\mathrm e}^{3 t}-3 \,{\mathrm e}^{t}+4\right ) {\mathrm e}^{-2 t}}{36} \\ \frac {\left (24 \,{\mathrm e}^{4 t}-6 t \,{\mathrm e}^{3 t}-19 \,{\mathrm e}^{3 t}+3 \,{\mathrm e}^{t}-8\right ) {\mathrm e}^{-2 t}}{36} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & x=-\frac {\left (t -6 c_{3} +\frac {1}{6}\right ) {\mathrm e}^{-2 t} {\mathrm e}^{3 t}}{6}-\frac {\left (72 c_{2} {\mathrm e}^{t}-9 \,{\mathrm e}^{4 t} c_{4} +9 c_{1} +6 \,{\mathrm e}^{t}-6 \,{\mathrm e}^{4 t}-2\right ) {\mathrm e}^{-2 t}}{72} \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`

\[ x \left (t \right ) = -\frac {{\mathrm e}^{-2 t} \left (\left (t -6 c_{1} \right ) {\mathrm e}^{3 t}-6 c_{3} {\mathrm e}^{t}-6 c_{4} {\mathrm e}^{4 t}-6 c_{2} \right )}{6} \]

\[ x(t)\to e^{-2 t} \left (c_2 e^t+e^{3 t} \left (-\frac {t}{6}-\frac {1}{36}+c_3\right )+c_4 e^{4 t}+c_1\right ) \]

9.4 problem 16.1 (iv)

9.4.1 Maple step by step solution