Internal problem ID [14695]
Internal file name [OUTPUT/14375_Wednesday_April_03_2024_02_17_49_PM_84085844/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: 21.
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]]
\[ \boxed {y^{\prime \prime \prime }+3 y^{\prime \prime }-10 y^{\prime }-24 y={\mathrm e}^{-3 t}} \] 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 \[ y^{\prime \prime \prime }+3 y^{\prime \prime }-10 y^{\prime }-24 y = 0 \] The characteristic equation is \[ \lambda ^{3}+3 \lambda ^{2}-10 \lambda -24 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 3\\ \lambda _2 &= -4\\ \lambda _3 &= -2 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-4 t}+c_{3} {\mathrm e}^{3 t} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 t} \\ y_2 &= {\mathrm e}^{-4 t} \\ y_3 &= {\mathrm e}^{3 t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+3 y^{\prime \prime }-10 y^{\prime }-24 y = {\mathrm e}^{-3 t} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{-3 t} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-3 t}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{-4 t}, {\mathrm e}^{-2 t}, {\mathrm e}^{3 t}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{-3 t} \] The unknowns \(\{A_{1}\}\) 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 \[ 6 A_{1} {\mathrm e}^{-3 t} = {\mathrm e}^{-3 t} \] 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 \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{-3 t}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-4 t}+c_{3} {\mathrm e}^{3 t}\right ) + \left (\frac {{\mathrm e}^{-3 t}}{6}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-4 t}+c_{3} {\mathrm e}^{3 t}+\frac {{\mathrm e}^{-3 t}}{6} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-4 t}+c_{3} {\mathrm e}^{3 t}+\frac {{\mathrm e}^{-3 t}}{6} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+3 y^{\prime \prime }-10 y^{\prime }-24 y={\mathrm e}^{-3 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 )={\mathrm e}^{-3 t}-3 y_{3}\left (t \right )+10 y_{2}\left (t \right )+24 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 )={\mathrm e}^{-3 t}-3 y_{3}\left (t \right )+10 y_{2}\left (t \right )+24 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 \\ 24 & 10 & -3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{-3 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}^{-3 t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 24 & 10 & -3 \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 [-4, \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{4} \\ 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 [-4, \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-4 t}\cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{4} \\ 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 [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}}_{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 }{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}^{-4 t}}{16} & \frac {{\mathrm e}^{-2 t}}{4} & \frac {{\mathrm e}^{3 t}}{9} \\ -\frac {{\mathrm e}^{-4 t}}{4} & -\frac {{\mathrm e}^{-2 t}}{2} & \frac {{\mathrm e}^{3 t}}{3} \\ {\mathrm e}^{-4 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} \frac {{\mathrm e}^{-4 t}}{16} & \frac {{\mathrm e}^{-2 t}}{4} & \frac {{\mathrm e}^{3 t}}{9} \\ -\frac {{\mathrm e}^{-4 t}}{4} & -\frac {{\mathrm e}^{-2 t}}{2} & \frac {{\mathrm e}^{3 t}}{3} \\ {\mathrm e}^{-4 t} & {\mathrm e}^{-2 t} & {\mathrm e}^{3 t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{16} & \frac {1}{4} & \frac {1}{9} \\ -\frac {1}{4} & -\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} \frac {\left (8 \,{\mathrm e}^{7 t}+42 \,{\mathrm e}^{2 t}-15\right ) {\mathrm e}^{-4 t}}{35} & \frac {\left (12 \,{\mathrm e}^{7 t}-7 \,{\mathrm e}^{2 t}-5\right ) {\mathrm e}^{-4 t}}{70} & \frac {\left (2 \,{\mathrm e}^{7 t}-7 \,{\mathrm e}^{2 t}+5\right ) {\mathrm e}^{-4 t}}{70} \\ \frac {12 \left (2 \,{\mathrm e}^{7 t}-7 \,{\mathrm e}^{2 t}+5\right ) {\mathrm e}^{-4 t}}{35} & \frac {\left (18 \,{\mathrm e}^{7 t}+7 \,{\mathrm e}^{2 t}+10\right ) {\mathrm e}^{-4 t}}{35} & \frac {\left (3 \,{\mathrm e}^{7 t}+7 \,{\mathrm e}^{2 t}-10\right ) {\mathrm e}^{-4 t}}{35} \\ \frac {24 \left (3 \,{\mathrm e}^{7 t}+7 \,{\mathrm e}^{2 t}-10\right ) {\mathrm e}^{-4 t}}{35} & \frac {2 \left (27 \,{\mathrm e}^{7 t}-7 \,{\mathrm e}^{2 t}-20\right ) {\mathrm e}^{-4 t}}{35} & \frac {\left (9 \,{\mathrm e}^{7 t}-14 \,{\mathrm e}^{2 t}+40\right ) {\mathrm e}^{-4 t}}{35} \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 ({\mathrm e}^{7 t}-21 \,{\mathrm e}^{2 t}+35 \,{\mathrm e}^{t}-15\right ) {\mathrm e}^{-4 t}}{210} \\ \frac {\left ({\mathrm e}^{7 t}+14 \,{\mathrm e}^{2 t}-35 \,{\mathrm e}^{t}+20\right ) {\mathrm e}^{-4 t}}{70} \\ \frac {\left (3 \,{\mathrm e}^{7 t}-28 \,{\mathrm e}^{2 t}+105 \,{\mathrm e}^{t}-80\right ) {\mathrm e}^{-4 t}}{70} \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 ({\mathrm e}^{7 t}-21 \,{\mathrm e}^{2 t}+35 \,{\mathrm e}^{t}-15\right ) {\mathrm e}^{-4 t}}{210} \\ \frac {\left ({\mathrm e}^{7 t}+14 \,{\mathrm e}^{2 t}-35 \,{\mathrm e}^{t}+20\right ) {\mathrm e}^{-4 t}}{70} \\ \frac {\left (3 \,{\mathrm e}^{7 t}-28 \,{\mathrm e}^{2 t}+105 \,{\mathrm e}^{t}-80\right ) {\mathrm e}^{-4 t}}{70} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (560 c_{3} {\mathrm e}^{7 t}+24 \,{\mathrm e}^{7 t}+1260 c_{2} {\mathrm e}^{2 t}-504 \,{\mathrm e}^{2 t}+840 \,{\mathrm e}^{t}+315 c_{1} -360\right ) {\mathrm e}^{-4 t}}{5040} \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: 30
dsolve(diff(y(t),t$3)+3*diff(y(t),t$2)-10*diff(y(t),t)-24*y(t)=exp(-3*t),y(t), singsol=all)
\[ y \left (t \right ) = \frac {\left (6 c_{3} {\mathrm e}^{7 t}+6 c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{t}+6 c_{1} \right ) {\mathrm e}^{-4 t}}{6} \]
✓ Solution by Mathematica
Time used: 0.015 (sec). Leaf size: 40
DSolve[y'''[t]+3*y''[t]-10*y'[t]-24*y[t]==Exp[-3*t],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{6} e^{-4 t} \left (e^t+6 c_2 e^{2 t}+6 c_3 e^{7 t}+6 c_1\right ) \]