14.7 problem 7

Internal problem ID [14681]
Internal file name [OUTPUT/14361_Wednesday_April_03_2024_02_17_39_PM_86988471/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.6, page 187
Problem number: 7.
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, _linear, _nonhomogeneous]]

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

Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-2 t}+{\mathrm e}^{-3 t} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-t} \\ y_2 &= {\mathrm e}^{-2 t} \\ y_3 &= {\mathrm e}^{-3 t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+6 y^{\prime \prime }+11 y^{\prime }+6 y = 2 \,{\mathrm e}^{-3 t}-t \,{\mathrm e}^{-t} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 2 \,{\mathrm e}^{-3 t}-t \,{\mathrm e}^{-t} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-3 t}\}, \{t \,{\mathrm e}^{-t}, {\mathrm e}^{-t}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{-3 t}, {\mathrm e}^{-2 t}, {\mathrm e}^{-t}\} \] Since \({\mathrm e}^{-t}\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{{\mathrm e}^{-3 t}\}, \{t \,{\mathrm e}^{-t}, t^{2} {\mathrm e}^{-t}\}] \] Since \({\mathrm e}^{-3 t}\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{{\mathrm e}^{-3 t} t\}, \{t \,{\mathrm e}^{-t}, t^{2} {\mathrm e}^{-t}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{-3 t} t +A_{2} t \,{\mathrm e}^{-t}+A_{3} t^{2} {\mathrm e}^{-t} \] The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) 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 \[ 2 A_{1} {\mathrm e}^{-3 t}+2 A_{2} {\mathrm e}^{-t}+6 A_{3} {\mathrm e}^{-t}+4 A_{3} t \,{\mathrm e}^{-t} = 2 \,{\mathrm e}^{-3 t}-t \,{\mathrm e}^{-t} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = 1, A_{2} = {\frac {3}{4}}, A_{3} = -{\frac {1}{4}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = {\mathrm e}^{-3 t} t +\frac {3 t \,{\mathrm e}^{-t}}{4}-\frac {t^{2} {\mathrm e}^{-t}}{4} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-2 t}+{\mathrm e}^{-3 t} c_{3}\right ) + \left ({\mathrm e}^{-3 t} t +\frac {3 t \,{\mathrm e}^{-t}}{4}-\frac {t^{2} {\mathrm e}^{-t}}{4}\right ) \\ \end{align*}

Summary

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

Verification of solutions

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

14.7.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+6 y^{\prime \prime }+11 y^{\prime }+6 y=2 \,{\mathrm e}^{-3 t}-t \,{\mathrm e}^{-t} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \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 (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (t \right )=-t \,{\mathrm e}^{-t}+2 \,{\mathrm e}^{-3 t}-6 y_{3}\left (t \right )-11 y_{2}\left (t \right )-6 y_{1}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=y_{1}^{\prime }\left (t \right ), y_{3}\left (t \right )=y_{2}^{\prime }\left (t \right ), y_{3}^{\prime }\left (t \right )=-t \,{\mathrm e}^{-t}+2 \,{\mathrm e}^{-3 t}-6 y_{3}\left (t \right )-11 y_{2}\left (t \right )-6 y_{1}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -11 & -6 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 2 \,{\mathrm e}^{-3 t}-t \,{\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 \\ 2 \,{\mathrm e}^{-3 t}-t \,{\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 }{y}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\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 [-3, \left [\begin {array}{c} \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]\right ], \left [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ]\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 }{y}}_{1}={\mathrm e}^{-3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ -\frac {1}{3} \\ 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 }{y}}_{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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 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 }{y}}_{p}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{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} \frac {{\mathrm e}^{-3 t}}{9} & \frac {{\mathrm e}^{-2 t}}{4} & {\mathrm e}^{-t} \\ -\frac {{\mathrm e}^{-3 t}}{3} & -\frac {{\mathrm e}^{-2 t}}{2} & -{\mathrm e}^{-t} \\ {\mathrm e}^{-3 t} & {\mathrm e}^{-2 t} & {\mathrm e}^{-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} \frac {{\mathrm e}^{-3 t}}{9} & \frac {{\mathrm e}^{-2 t}}{4} & {\mathrm e}^{-t} \\ -\frac {{\mathrm e}^{-3 t}}{3} & -\frac {{\mathrm e}^{-2 t}}{2} & -{\mathrm e}^{-t} \\ {\mathrm e}^{-3 t} & {\mathrm e}^{-2 t} & {\mathrm e}^{-t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{9} & \frac {1}{4} & 1 \\ -\frac {1}{3} & -\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}{ccc} {\mathrm e}^{-3 t}-3 \,{\mathrm e}^{-2 t}+3 \,{\mathrm e}^{-t} & \frac {3 \,{\mathrm e}^{-3 t}}{2}-4 \,{\mathrm e}^{-2 t}+\frac {5 \,{\mathrm e}^{-t}}{2} & \frac {{\mathrm e}^{-t}}{2}+\frac {{\mathrm e}^{-3 t}}{2}-{\mathrm e}^{-2 t} \\ -3 \,{\mathrm e}^{-3 t}+6 \,{\mathrm e}^{-2 t}-3 \,{\mathrm e}^{-t} & -\frac {9 \,{\mathrm e}^{-3 t}}{2}+8 \,{\mathrm e}^{-2 t}-\frac {5 \,{\mathrm e}^{-t}}{2} & -\frac {3 \,{\mathrm e}^{-3 t}}{2}+2 \,{\mathrm e}^{-2 t}-\frac {{\mathrm e}^{-t}}{2} \\ 9 \,{\mathrm e}^{-3 t}-12 \,{\mathrm e}^{-2 t}+3 \,{\mathrm e}^{-t} & \frac {27 \,{\mathrm e}^{-3 t}}{2}-16 \,{\mathrm e}^{-2 t}+\frac {5 \,{\mathrm e}^{-t}}{2} & \frac {9 \,{\mathrm e}^{-3 t}}{2}-4 \,{\mathrm e}^{-2 t}+\frac {{\mathrm e}^{-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 }{y}}_{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 }{y}}_{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 }{y}}_{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 }{y}}_{p}\left (t \right )=\left [\begin {array}{c} \frac {\left (-2 t^{2}+6 t -3\right ) {\mathrm e}^{-t}}{8}+\frac {\left (8 t +11\right ) {\mathrm e}^{-3 t}}{8}-{\mathrm e}^{-2 t} \\ \frac {\left (2 t^{2}-10 t +9\right ) {\mathrm e}^{-t}}{8}+\frac {\left (-24 t -25\right ) {\mathrm e}^{-3 t}}{8}+2 \,{\mathrm e}^{-2 t} \\ \frac {\left (-2 t^{2}+14 t -19\right ) {\mathrm e}^{-t}}{8}+\frac {\left (72 t +51\right ) {\mathrm e}^{-3 t}}{8}-4 \,{\mathrm e}^{-2 t} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} \frac {\left (-2 t^{2}+6 t -3\right ) {\mathrm e}^{-t}}{8}+\frac {\left (8 t +11\right ) {\mathrm e}^{-3 t}}{8}-{\mathrm e}^{-2 t} \\ \frac {\left (2 t^{2}-10 t +9\right ) {\mathrm e}^{-t}}{8}+\frac {\left (-24 t -25\right ) {\mathrm e}^{-3 t}}{8}+2 \,{\mathrm e}^{-2 t} \\ \frac {\left (-2 t^{2}+14 t -19\right ) {\mathrm e}^{-t}}{8}+\frac {\left (72 t +51\right ) {\mathrm e}^{-3 t}}{8}-4 \,{\mathrm e}^{-2 t} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (-2 t^{2}+8 c_{3} +6 t -3\right ) {\mathrm e}^{-t}}{8}+\frac {\left (72 t +8 c_{1} +99\right ) {\mathrm e}^{-3 t}}{72}+\frac {{\mathrm e}^{-2 t} \left (c_{2} -4\right )}{4} \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: 0.0 (sec). Leaf size: 44

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

\[ y \left (t \right ) = \frac {\left (-2 t^{2}+8 c_{3} +6 t -7\right ) {\mathrm e}^{-t}}{8}+\frac {\left (3+2 t +2 c_{1} \right ) {\mathrm e}^{-3 t}}{2}+c_{2} {\mathrm e}^{-2 t} \]

✓ Solution by Mathematica

Time used: 0.06 (sec). Leaf size: 50

DSolve[y'''[t]+6*y''[t]+11*y'[t]+6*y[t]==2*Exp[-3*t]-t*Exp[-t],y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{8} e^{-3 t} \left (e^{2 t} \left (-2 t^2+6 t-7+8 c_3\right )+8 t+8 c_2 e^t+12+8 c_1\right ) \]