problem 9

Internal problem ID [11782]
Internal ﬁle name [OUTPUT/11792_Thursday_April_11_2024_08_49_42_PM_60476993/index.tex]

Book: Diﬀerential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi. 2004.
Section: Chapter 4, Section 4.3. The method of undetermined coeﬃcients. Exercises page 151
Problem number: 9.
ODE order: 3.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime }+4 y^{\prime \prime }+y^{\prime }-6 y=-18 x^{2}+1} \] 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 \[ y^{\prime \prime \prime }+4 y^{\prime \prime }+y^{\prime }-6 y = 0 \] The characteristic equation is \[ \lambda ^{3}+4 \lambda ^{2}+\lambda -6 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -3\\ \lambda _3 &= -2 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 x} \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{-3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+4 y^{\prime \prime }+y^{\prime }-6 y = -18 x^{2}+1 \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ x^{2}+1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{-3 x}, {\mathrm e}^{-2 x}\} \] 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. \[ y_p = A_{3} x^{2}+A_{2} x +A_{1} \] 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 \[ -6 A_{3} x^{2}-6 A_{2} x +2 x A_{3}-6 A_{1}+A_{2}+8 A_{3} = -18 x^{2}+1 \] Solving for the unknowns by comparing coeﬃcients results in \[ [A_{1} = 4, A_{2} = 1, A_{3} = 3] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = 3 x^{2}+x +4 \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{3}\right ) + \left (3 x^{2}+x +4\right ) \\ \end{align*}

Summary

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

Veriﬁcation of solutions

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

11.9.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+4 y^{\prime \prime }+y^{\prime }-6 y=-18 x^{2}+1 \\ \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 )=-18 x^{2}-4 y_{3}\left (x \right )-y_{2}\left (x \right )+6 y_{1}\left (x \right )+1 \\ & {} & \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 )=-18 x^{2}-4 y_{3}\left (x \right )-y_{2}\left (x \right )+6 y_{1}\left (x \right )+1\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 & -1 & -4 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ -18 x^{2}+1 \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ -18 x^{2}+1 \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -1 & -4 \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 [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 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 [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}^{x}\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 (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} & {\mathrm e}^{x} \\ -\frac {{\mathrm e}^{-3 x}}{3} & -\frac {{\mathrm e}^{-2 x}}{2} & {\mathrm e}^{x} \\ {\mathrm e}^{-3 x} & {\mathrm e}^{-2 x} & {\mathrm e}^{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} \frac {{\mathrm e}^{-3 x}}{9} & \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{x} \\ -\frac {{\mathrm e}^{-3 x}}{3} & -\frac {{\mathrm e}^{-2 x}}{2} & {\mathrm e}^{x} \\ {\mathrm e}^{-3 x} & {\mathrm e}^{-2 x} & {\mathrm e}^{x} \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 (x \right )=\left [\begin {array}{ccc} \frac {\left ({\mathrm e}^{4 x}+2 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-3 x}}{2} & \frac {\left (5 \,{\mathrm e}^{4 x}-8 \,{\mathrm e}^{x}+3\right ) {\mathrm e}^{-3 x}}{12} & \frac {\left ({\mathrm e}^{4 x}-4 \,{\mathrm e}^{x}+3\right ) {\mathrm e}^{-3 x}}{12} \\ \frac {\left ({\mathrm e}^{4 x}-4 \,{\mathrm e}^{x}+3\right ) {\mathrm e}^{-3 x}}{2} & \frac {\left (5 \,{\mathrm e}^{4 x}+16 \,{\mathrm e}^{x}-9\right ) {\mathrm e}^{-3 x}}{12} & \frac {\left ({\mathrm e}^{4 x}+8 \,{\mathrm e}^{x}-9\right ) {\mathrm e}^{-3 x}}{12} \\ \frac {\left ({\mathrm e}^{4 x}+8 \,{\mathrm e}^{x}-9\right ) {\mathrm e}^{-3 x}}{2} & \frac {\left (5 \,{\mathrm e}^{4 x}-32 \,{\mathrm e}^{x}+27\right ) {\mathrm e}^{-3 x}}{12} & \frac {\left ({\mathrm e}^{4 x}-16 \,{\mathrm e}^{x}+27\right ) {\mathrm e}^{-3 x}}{12} \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} {\mathrm e}^{-3 x} \left (\left (3 x^{2}+x +4\right ) {\mathrm e}^{3 x}-\frac {4 \,{\mathrm e}^{x}}{3}-\frac {35 \,{\mathrm e}^{4 x}}{12}+\frac {1}{4}\right ) \\ -\frac {\left (35 \,{\mathrm e}^{4 x}-72 x \,{\mathrm e}^{3 x}-12 \,{\mathrm e}^{3 x}-32 \,{\mathrm e}^{x}+9\right ) {\mathrm e}^{-3 x}}{12} \\ -\frac {\left (35 \,{\mathrm e}^{4 x}-72 \,{\mathrm e}^{3 x}+64 \,{\mathrm e}^{x}-27\right ) {\mathrm e}^{-3 x}}{12} \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} {\mathrm e}^{-3 x} \left (\left (3 x^{2}+x +4\right ) {\mathrm e}^{3 x}-\frac {4 \,{\mathrm e}^{x}}{3}-\frac {35 \,{\mathrm e}^{4 x}}{12}+\frac {1}{4}\right ) \\ -\frac {\left (35 \,{\mathrm e}^{4 x}-72 x \,{\mathrm e}^{3 x}-12 \,{\mathrm e}^{3 x}-32 \,{\mathrm e}^{x}+9\right ) {\mathrm e}^{-3 x}}{12} \\ -\frac {\left (35 \,{\mathrm e}^{4 x}-72 \,{\mathrm e}^{3 x}+64 \,{\mathrm e}^{x}-27\right ) {\mathrm e}^{-3 x}}{12} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-3 x} \left (\left (3 x^{2}+x +4\right ) {\mathrm e}^{3 x}+\frac {c_{2} {\mathrm e}^{x}}{4}+{\mathrm e}^{4 x} c_{3} +\frac {c_{1}}{9}-\frac {4 \,{\mathrm e}^{x}}{3}-\frac {35 \,{\mathrm e}^{4 x}}{12}+\frac {1}{4}\right ) \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`

\[ y \left (x \right ) = {\mathrm e}^{-3 x} \left (\left (3 x^{2}+x +4\right ) {\mathrm e}^{3 x}+c_{1} {\mathrm e}^{4 x}+c_{3} {\mathrm e}^{x}+c_{2} \right ) \]

11.9 problem 9

11.9.1 Maple step by step solution