9.1 problem 16.1 (i)

Internal problem ID [12043]
Internal file name [OUTPUT/10696_Sunday_September_03_2023_12_36_40_PM_92005170/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 (i).
ODE order: 3.
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

[[_3rd_order, _with_linear_symmetries]]

\[ \boxed {x^{\prime \prime \prime }-6 x^{\prime \prime }+11 x^{\prime }-6 x={\mathrm e}^{-t}} \] 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 \[ x^{\prime \prime \prime }-6 x^{\prime \prime }+11 x^{\prime }-6 x = 0 \] The characteristic equation is \[ \lambda ^{3}-6 \lambda ^{2}+11 \lambda -6 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 2\\ \lambda _3 &= 3 \end {align*}

Therefore the homogeneous solution is \[ x_h(t)={\mathrm e}^{t} c_{1} +c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{3 t} c_{3} \] 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}^{3 t} \\ \end{align*} Now the particular solution to the given ODE is found \[ x^{\prime \prime \prime }-6 x^{\prime \prime }+11 x^{\prime }-6 x = {\mathrm e}^{-t} \] 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 \[ \{{\mathrm e}^{t}, {\mathrm e}^{2 t}, {\mathrm e}^{3 t}\} \] 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. \[ x_p = A_{1} {\mathrm e}^{-t} \] 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 \[ -24 A_{1} {\mathrm e}^{-t} = {\mathrm e}^{-t} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{24}}\right ] \] Substituting the above back in the above trial solution \(x_p\), gives the particular solution \[ x_p = -\frac {{\mathrm e}^{-t}}{24} \] Therefore the general solution is \begin{align*} x &= x_h + x_p \\ &= \left ({\mathrm e}^{t} c_{1} +c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{3 t} c_{3}\right ) + \left (-\frac {{\mathrm e}^{-t}}{24}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= {\mathrm e}^{t} c_{1} +c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{3 t} c_{3} -\frac {{\mathrm e}^{-t}}{24} \\ \end{align*}

Verification of solutions

\[ x = {\mathrm e}^{t} c_{1} +c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{3 t} c_{3} -\frac {{\mathrm e}^{-t}}{24} \] Verified OK.

9.1.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime \prime \prime }-6 x^{\prime \prime }+11 x^{\prime }-6 x={\mathrm e}^{-t} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & x^{\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 {Isolate for}\hspace {3pt} x_{3}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & x_{3}^{\prime }\left (t \right )={\mathrm e}^{-t}+6 x_{3}\left (t \right )-11 x_{2}\left (t \right )+6 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_{3}^{\prime }\left (t \right )={\mathrm e}^{-t}+6 x_{3}\left (t \right )-11 x_{2}\left (t \right )+6 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 ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end {array}\right ]\cdot {\moverset {\rightarrow }{x}}\left (t \right )+\left [\begin {array}{c} 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 \\ {\mathrm e}^{-t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \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 [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{1}={\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{2}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{3}={\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 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}+{\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}{ccc} {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{4} & \frac {{\mathrm e}^{3 t}}{9} \\ {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{2} & \frac {{\mathrm e}^{3 t}}{3} \\ {\mathrm e}^{t} & {\mathrm e}^{2 t} & {\mathrm e}^{3 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}{ccc} {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{4} & \frac {{\mathrm e}^{3 t}}{9} \\ {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{2} & \frac {{\mathrm e}^{3 t}}{3} \\ {\mathrm e}^{t} & {\mathrm e}^{2 t} & {\mathrm e}^{3 t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & \frac {1}{4} & \frac {1}{9} \\ 1 & \frac {1}{2} & \frac {1}{3} \\ 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}{ccc} 3 \,{\mathrm e}^{t}-3 \,{\mathrm e}^{2 t}+{\mathrm e}^{3 t} & -\frac {5 \,{\mathrm e}^{t}}{2}+4 \,{\mathrm e}^{2 t}-\frac {3 \,{\mathrm e}^{3 t}}{2} & \frac {{\mathrm e}^{t}}{2}-{\mathrm e}^{2 t}+\frac {{\mathrm e}^{3 t}}{2} \\ 3 \,{\mathrm e}^{t}-6 \,{\mathrm e}^{2 t}+3 \,{\mathrm e}^{3 t} & -\frac {5 \,{\mathrm e}^{t}}{2}+8 \,{\mathrm e}^{2 t}-\frac {9 \,{\mathrm e}^{3 t}}{2} & \frac {{\mathrm e}^{t}}{2}-2 \,{\mathrm e}^{2 t}+\frac {3 \,{\mathrm e}^{3 t}}{2} \\ 3 \,{\mathrm e}^{t}-12 \,{\mathrm e}^{2 t}+9 \,{\mathrm e}^{3 t} & -\frac {5 \,{\mathrm e}^{t}}{2}+16 \,{\mathrm e}^{2 t}-\frac {27 \,{\mathrm e}^{3 t}}{2} & \frac {{\mathrm e}^{t}}{2}-4 \,{\mathrm e}^{2 t}+\frac {9 \,{\mathrm e}^{3 t}}{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 (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 {{\mathrm e}^{t}}{4}-\frac {{\mathrm e}^{-t}}{24}-\frac {{\mathrm e}^{2 t}}{3}+\frac {{\mathrm e}^{3 t}}{8} \\ \frac {{\mathrm e}^{t}}{4}+\frac {{\mathrm e}^{-t}}{24}-\frac {2 \,{\mathrm e}^{2 t}}{3}+\frac {3 \,{\mathrm e}^{3 t}}{8} \\ \frac {{\mathrm e}^{t}}{4}-\frac {{\mathrm e}^{-t}}{24}-\frac {4 \,{\mathrm e}^{2 t}}{3}+\frac {9 \,{\mathrm e}^{3 t}}{8} \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}+\left [\begin {array}{c} \frac {{\mathrm e}^{t}}{4}-\frac {{\mathrm e}^{-t}}{24}-\frac {{\mathrm e}^{2 t}}{3}+\frac {{\mathrm e}^{3 t}}{8} \\ \frac {{\mathrm e}^{t}}{4}+\frac {{\mathrm e}^{-t}}{24}-\frac {2 \,{\mathrm e}^{2 t}}{3}+\frac {3 \,{\mathrm e}^{3 t}}{8} \\ \frac {{\mathrm e}^{t}}{4}-\frac {{\mathrm e}^{-t}}{24}-\frac {4 \,{\mathrm e}^{2 t}}{3}+\frac {9 \,{\mathrm e}^{3 t}}{8} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & x=\frac {\left (18 c_{2} -24\right ) {\mathrm e}^{2 t}}{72}+\frac {\left (8 c_{3} +9\right ) {\mathrm e}^{3 t}}{72}-\frac {{\mathrm e}^{-t}}{24}+\frac {\left (72 c_{1} +18\right ) {\mathrm e}^{t}}{72} \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 3; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 3; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 27

dsolve(diff(x(t),t$3)-6*diff(x(t),t$2)+11*diff(x(t),t)-6*x(t)=exp(-t),x(t), singsol=all)
 

\[ x \left (t \right ) = -\frac {{\mathrm e}^{-t}}{24}+c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{2 t}+c_{3} {\mathrm e}^{3 t} \]

✓ Solution by Mathematica

Time used: 0.029 (sec). Leaf size: 37

DSolve[x'''[t]-6*x''[t]+11*x'[t]-6*x[t]==Exp[-t],x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to -\frac {e^{-t}}{24}+c_1 e^t+c_2 e^{2 t}+c_3 e^{3 t} \]