Internal problem ID [2744]
Internal file name [OUTPUT/2236_Sunday_June_05_2022_02_55_49_AM_28455852/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 8, Linear differential equations of order n. Section 8.1, General Theory for
Linear Differential Equations. page 502
Problem number: Problem 42.
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, _missing_y]]
\[ \boxed {y^{\prime \prime \prime }+5 y^{\prime \prime }+6 y^{\prime }=6 \,{\mathrm e}^{-x}} \] 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 \[ y^{\prime \prime \prime }+5 y^{\prime \prime }+6 y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{3}+5 \lambda ^{2}+6 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= -2\\ \lambda _3 &= -3 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} +c_{2} {\mathrm e}^{-2 x}+{\mathrm e}^{-3 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{-2 x} \\ y_3 &= {\mathrm e}^{-3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+5 y^{\prime \prime }+6 y^{\prime } = 6 \,{\mathrm e}^{-x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 6 \,{\mathrm e}^{-x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{-3 x}, {\mathrm e}^{-2 x}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{-x} \] 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 \[ -2 A_{1} {\mathrm e}^{-x} = 6 \,{\mathrm e}^{-x} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = -3] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -3 \,{\mathrm e}^{-x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +c_{2} {\mathrm e}^{-2 x}+{\mathrm e}^{-3 x} c_{3}\right ) + \left (-3 \,{\mathrm e}^{-x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +c_{2} {\mathrm e}^{-2 x}+{\mathrm e}^{-3 x} c_{3} -3 \,{\mathrm e}^{-x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} +c_{2} {\mathrm e}^{-2 x}+{\mathrm e}^{-3 x} c_{3} -3 \,{\mathrm e}^{-x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+5 y^{\prime \prime }+6 y^{\prime }=6 \,{\mathrm e}^{-x} \\ \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 (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=6 \,{\mathrm e}^{-x}-5 y_{3}\left (x \right )-6 y_{2}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=6 \,{\mathrm e}^{-x}-5 y_{3}\left (x \right )-6 y_{2}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -6 & -5 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 6 \,{\mathrm e}^{-x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 6 \,{\mathrm e}^{-x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -6 & -5 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \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 x}\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 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\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 }{y}}_{3}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \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 (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{-3 x}}{9} & \frac {{\mathrm e}^{-2 x}}{4} & 1 \\ -\frac {{\mathrm e}^{-3 x}}{3} & -\frac {{\mathrm e}^{-2 x}}{2} & 0 \\ {\mathrm e}^{-3 x} & {\mathrm e}^{-2 x} & 0 \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \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 (x \right )=\phi \left (x \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{-3 x}}{9} & \frac {{\mathrm e}^{-2 x}}{4} & 1 \\ -\frac {{\mathrm e}^{-3 x}}{3} & -\frac {{\mathrm e}^{-2 x}}{2} & 0 \\ {\mathrm e}^{-3 x} & {\mathrm e}^{-2 x} & 0 \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{9} & \frac {1}{4} & 1 \\ -\frac {1}{3} & -\frac {1}{2} & 0 \\ 1 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} 1 & \frac {2 \,{\mathrm e}^{-3 x}}{3}-\frac {3 \,{\mathrm e}^{-2 x}}{2}+\frac {5}{6} & \frac {{\mathrm e}^{-3 x}}{3}-\frac {{\mathrm e}^{-2 x}}{2}+\frac {1}{6} \\ 0 & -2 \,{\mathrm e}^{-3 x}+3 \,{\mathrm e}^{-2 x} & -{\mathrm e}^{-3 x}+{\mathrm e}^{-2 x} \\ 0 & 6 \,{\mathrm e}^{-3 x}-6 \,{\mathrm e}^{-2 x} & 3 \,{\mathrm e}^{-3 x}-2 \,{\mathrm e}^{-2 x} \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 (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \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 (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\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 (x \right )=\left [\begin {array}{c} -3 \,{\mathrm e}^{-x}+1+3 \,{\mathrm e}^{-2 x}-{\mathrm e}^{-3 x} \\ 3 \,{\mathrm e}^{-x}-6 \,{\mathrm e}^{-2 x}+3 \,{\mathrm e}^{-3 x} \\ -3 \,{\mathrm e}^{-x}+12 \,{\mathrm e}^{-2 x}-9 \,{\mathrm e}^{-3 x} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} -3 \,{\mathrm e}^{-x}+1+3 \,{\mathrm e}^{-2 x}-{\mathrm e}^{-3 x} \\ 3 \,{\mathrm e}^{-x}-6 \,{\mathrm e}^{-2 x}+3 \,{\mathrm e}^{-3 x} \\ -3 \,{\mathrm e}^{-x}+12 \,{\mathrm e}^{-2 x}-9 \,{\mathrm e}^{-3 x} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=1+\frac {\left (-9+c_{1} \right ) {\mathrm e}^{-3 x}}{9}+\frac {\left (12+c_{2} \right ) {\mathrm e}^{-2 x}}{4}+c_{3} -3 \,{\mathrm e}^{-x} \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) = -5*(diff(_b(_a), _a))-6*_b(_a)+6*exp(-_a), _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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 26
dsolve(diff(y(x),x$3)+5*diff(y(x),x$2)+6*diff(y(x),x)=6*exp(-x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {c_{1} {\mathrm e}^{-3 x}}{3}-\frac {{\mathrm e}^{-2 x} c_{2}}{2}-3 \,{\mathrm e}^{-x}+c_{3} \]
✓ Solution by Mathematica
Time used: 0.05 (sec). Leaf size: 37
DSolve[y'''[x]+5*y''[x]+6*y'[x]==6*Exp[-x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -3 e^{-x}-\frac {1}{3} c_1 e^{-3 x}-\frac {1}{2} c_2 e^{-2 x}+c_3 \]