problem Problem 11.7

Internal problem ID [5169]
Internal ﬁle name [OUTPUT/4662_Sunday_June_05_2022_03_02_55_PM_22257302/index.tex]

Book: Schaums Outline Diﬀerential Equations, 4th edition. Bronson and Costa. McGraw Hill 2014
Section: Chapter 11. THE METHOD OF UNDETERMINED COEFFICIENTS. page 95
Problem number: Problem 11.7.
ODE order: 3.
ODE degree: 1.

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

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

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{3 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{2 x} \\ y_3 &= {\mathrm e}^{3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-6 y^{\prime \prime }+11 y^{\prime }-6 y = 2 x \,{\mathrm e}^{-x} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ 2 x \,{\mathrm e}^{-x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{-x}, {\mathrm e}^{-x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{2 x}, {\mathrm e}^{3 x}\} \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{-x}+A_{2} {\mathrm e}^{-x} \] The unknowns \(\{A_{1}, A_{2}\}\) 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 \[ 26 A_{1} {\mathrm e}^{-x}-24 A_{1} x \,{\mathrm e}^{-x}-24 A_{2} {\mathrm e}^{-x} = 2 x \,{\mathrm e}^{-x} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {1}{12}}, A_{2} = -{\frac {13}{144}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {x \,{\mathrm e}^{-x}}{12}-\frac {13 \,{\mathrm e}^{-x}}{144} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{3 x} c_{3}\right ) + \left (-\frac {x \,{\mathrm e}^{-x}}{12}-\frac {13 \,{\mathrm e}^{-x}}{144}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{3 x} c_{3} -\frac {x \,{\mathrm e}^{-x}}{12}-\frac {13 \,{\mathrm e}^{-x}}{144} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{3 x} c_{3} -\frac {x \,{\mathrm e}^{-x}}{12}-\frac {13 \,{\mathrm e}^{-x}}{144} \] Veriﬁed OK.

1.7.1 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 x \,{\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 )=2 x \,{\mathrm e}^{-x}+6 y_{3}\left (x \right )-11 y_{2}\left (x \right )+6 y_{1}\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 )=2 x \,{\mathrm e}^{-x}+6 y_{3}\left (x \right )-11 y_{2}\left (x \right )+6 y_{1}\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 \\ 6 & -11 & 6 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 2 x \,{\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 \\ 2 x \,{\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 \\ 6 & -11 & 6 \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 [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 }{y}}_{1}={\mathrm e}^{x}\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 }{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 [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 x}\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 (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} {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{4} & \frac {{\mathrm e}^{3 x}}{9} \\ {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{2} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{x} & {\mathrm e}^{2 x} & {\mathrm e}^{3 x} \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} {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{4} & \frac {{\mathrm e}^{3 x}}{9} \\ {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{2} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{x} & {\mathrm e}^{2 x} & {\mathrm e}^{3 x} \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 (x \right )=\left [\begin {array}{ccc} 3 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{2 x}+{\mathrm e}^{3 x} & -\frac {5 \,{\mathrm e}^{x}}{2}+4 \,{\mathrm e}^{2 x}-\frac {3 \,{\mathrm e}^{3 x}}{2} & \frac {{\mathrm e}^{x}}{2}-{\mathrm e}^{2 x}+\frac {{\mathrm e}^{3 x}}{2} \\ 3 \,{\mathrm e}^{x}-6 \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{3 x} & -\frac {5 \,{\mathrm e}^{x}}{2}+8 \,{\mathrm e}^{2 x}-\frac {9 \,{\mathrm e}^{3 x}}{2} & \frac {{\mathrm e}^{x}}{2}-2 \,{\mathrm e}^{2 x}+\frac {3 \,{\mathrm e}^{3 x}}{2} \\ 3 \,{\mathrm e}^{x}-12 \,{\mathrm e}^{2 x}+9 \,{\mathrm e}^{3 x} & -\frac {5 \,{\mathrm e}^{x}}{2}+16 \,{\mathrm e}^{2 x}-\frac {27 \,{\mathrm e}^{3 x}}{2} & \frac {{\mathrm e}^{x}}{2}-4 \,{\mathrm e}^{2 x}+\frac {9 \,{\mathrm e}^{3 x}}{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 (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} \frac {\left (-12 x -13\right ) {\mathrm e}^{-x}}{144}+\frac {{\mathrm e}^{x}}{4}-\frac {2 \,{\mathrm e}^{2 x}}{9}+\frac {{\mathrm e}^{3 x}}{16} \\ \frac {\left (12 x +1\right ) {\mathrm e}^{-x}}{144}+\frac {{\mathrm e}^{x}}{4}-\frac {4 \,{\mathrm e}^{2 x}}{9}+\frac {3 \,{\mathrm e}^{3 x}}{16} \\ \frac {\left (-12 x +11\right ) {\mathrm e}^{-x}}{144}+\frac {{\mathrm e}^{x}}{4}-\frac {8 \,{\mathrm e}^{2 x}}{9}+\frac {9 \,{\mathrm e}^{3 x}}{16} \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} \frac {\left (-12 x -13\right ) {\mathrm e}^{-x}}{144}+\frac {{\mathrm e}^{x}}{4}-\frac {2 \,{\mathrm e}^{2 x}}{9}+\frac {{\mathrm e}^{3 x}}{16} \\ \frac {\left (12 x +1\right ) {\mathrm e}^{-x}}{144}+\frac {{\mathrm e}^{x}}{4}-\frac {4 \,{\mathrm e}^{2 x}}{9}+\frac {3 \,{\mathrm e}^{3 x}}{16} \\ \frac {\left (-12 x +11\right ) {\mathrm e}^{-x}}{144}+\frac {{\mathrm e}^{x}}{4}-\frac {8 \,{\mathrm e}^{2 x}}{9}+\frac {9 \,{\mathrm e}^{3 x}}{16} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (-12 x -13\right ) {\mathrm e}^{-x}}{144}+\frac {\left (-8+9 c_{2} \right ) {\mathrm e}^{2 x}}{36}+\frac {\left (16 c_{3} +9\right ) {\mathrm e}^{3 x}}{144}+\frac {{\mathrm e}^{x} \left (1+4 c_{1} \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`

\[ y \left (x \right ) = \frac {\left (-12 x -13\right ) {\mathrm e}^{-x}}{144}+{\mathrm e}^{x} c_{1} +c_{2} {\mathrm e}^{2 x}+c_{3} {\mathrm e}^{3 x} \]

1.7 problem Problem 11.7

1.7.1 Maple step by step solution