problem 1(b)

Internal problem ID [11501]
Internal ﬁle name [OUTPUT/10484_Thursday_May_18_2023_04_20_48_AM_44934501/index.tex]

Book: A First Course in Diﬀerential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section: Chapter 2, Second order linear equations. Section 2.5 Higher order equations. Exercises page 130
Problem number: 1(b).
ODE order: 3.
ODE degree: 1.

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

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

Therefore the homogeneous solution is \[ x_h(t)=c_{1} +{\mathrm e}^{i t} c_{2} +{\mathrm e}^{-i t} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} x_1 &= 1 \\ x_2 &= {\mathrm e}^{i t} \\ x_3 &= {\mathrm e}^{-i t} \\ \end{align*} Now the particular solution to the given ODE is found \[ x^{\prime \prime \prime }+x^{\prime } = 1 \] 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 \[ \{1, {\mathrm e}^{i t}, {\mathrm e}^{-i t}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t\}] \] 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. \[ x_p = A_{1} t \] 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 \[ A_{1} = 1 \] Solving for the unknowns by comparing coeﬃcients results in \[ [A_{1} = 1] \] Substituting the above back in the above trial solution \(x_p\), gives the particular solution \[ x_p = t \] Therefore the general solution is \begin{align*} x &= x_h + x_p \\ &= \left (c_{1} +{\mathrm e}^{i t} c_{2} +{\mathrm e}^{-i t} c_{3}\right ) + \left (t\right ) \\ \end{align*}

Summary

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

Veriﬁcation of solutions

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

14.2.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime \prime \prime }+x^{\prime }=1 \\ \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 )=1-x_{2}\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 )=1-x_{2}\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 \\ 0 & -1 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{x}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [\mathrm {-I}, \left [\begin {array}{c} -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I}, \left [\begin {array}{c} -1 \\ \mathrm {-I} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I}, \left [\begin {array}{c} -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} t}\cdot \left [\begin {array}{c} -1 \\ \mathrm {I} \\ 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 \\ {} & {} & \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right )\cdot \left [\begin {array}{c} -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\cos \left (t \right )+\mathrm {I} \sin \left (t \right ) \\ \mathrm {I} \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \cos \left (t \right )-\mathrm {I} \sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{x}}_{2}\left (t \right )=\left [\begin {array}{c} -\cos \left (t \right ) \\ \sin \left (t \right ) \\ \cos \left (t \right ) \end {array}\right ], {\moverset {\rightarrow }{x}}_{3}\left (t \right )=\left [\begin {array}{c} \sin \left (t \right ) \\ \cos \left (t \right ) \\ -\sin \left (t \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 }{x}}_{p}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=c_{1} {\moverset {\rightarrow }{x}}_{1}+c_{2} {\moverset {\rightarrow }{x}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{x}}_{3}\left (t \right )+{\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} 1 & -\cos \left (t \right ) & \sin \left (t \right ) \\ 0 & \sin \left (t \right ) & \cos \left (t \right ) \\ 0 & \cos \left (t \right ) & -\sin \left (t \right ) \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} 1 & -\cos \left (t \right ) & \sin \left (t \right ) \\ 0 & \sin \left (t \right ) & \cos \left (t \right ) \\ 0 & \cos \left (t \right ) & -\sin \left (t \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} 1 & \sin \left (t \right ) & -\cos \left (t \right )+1 \\ 0 & \cos \left (t \right ) & \sin \left (t \right ) \\ 0 & -\sin \left (t \right ) & \cos \left (t \right ) \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} t -\sin \left (t \right ) \\ -\cos \left (t \right )+1 \\ \sin \left (t \right ) \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}\left (t \right )+c_{3} {\moverset {\rightarrow }{x}}_{3}\left (t \right )+\left [\begin {array}{c} t -\sin \left (t \right ) \\ -\cos \left (t \right )+1 \\ \sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & x=\left (c_{3} -1\right ) \sin \left (t \right )-c_{2} \cos \left (t \right )+t +c_{1} \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] 
-> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = -_b(_a)+1, _b(_a)`   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying high order exact linear fully integrable 
   trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
   trying a double symmetry of the form [xi=0, eta=F(x)] 
   -> Try solving first the homogeneous part of the ODE 
      checking if the LODE has constant coefficients 
      <- constant coefficients successful 
   <- solving first the homogeneous part of the ODE successful 
<- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`

\[ x \left (t \right ) = c_{1} \sin \left (t \right )-c_{2} \cos \left (t \right )+t +c_{3} \]

14.2 problem 1(b)

14.2.1 Maple step by step solution