Internal problem ID [1502]
Internal file name [OUTPUT/1503_Sunday_June_05_2022_02_19_52_AM_4034753/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.3. Undetermined
Coefficients for Higher Order Equations. Page 495
Problem number: section 9.3, problem 5.
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]]
\[ \boxed {y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime }-3 y={\mathrm e}^{x} \left (16 x^{3}+24 x^{2}+2 x -1\right )} \] 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 \[ y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime }-3 y = 0 \] The characteristic equation is \[ \lambda ^{3}+3 \lambda ^{2}-\lambda -3 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -3\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{-3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime }-3 y = {\mathrm e}^{x} \left (16 x^{3}+24 x^{2}+2 x -1\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x} \left (16 x^{3}+24 x^{2}+2 x -1\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x} x^{3}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{-3 x}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x} x^{3}, {\mathrm e}^{x} x^{4}\}] \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{x}+A_{2} x^{2} {\mathrm e}^{x}+A_{3} {\mathrm e}^{x} x^{3}+A_{4} {\mathrm e}^{x} x^{4} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}\}\) 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 \[ 72 A_{4} {\mathrm e}^{x} x^{2}+24 A_{4} {\mathrm e}^{x} x +16 A_{2} x \,{\mathrm e}^{x}+24 A_{3} {\mathrm e}^{x} x^{2}+32 A_{4} {\mathrm e}^{x} x^{3}+36 A_{3} {\mathrm e}^{x} x +8 A_{1} {\mathrm e}^{x}+12 A_{2} {\mathrm e}^{x}+6 A_{3} {\mathrm e}^{x} = {\mathrm e}^{x} \left (16 x^{3}+24 x^{2}+2 x -1\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{2}}, A_{2} = {\frac {1}{2}}, A_{3} = -{\frac {1}{2}}, A_{4} = {\frac {1}{2}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {x \,{\mathrm e}^{x}}{2}+\frac {x^{2} {\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x} x^{3}}{2}+\frac {{\mathrm e}^{x} x^{4}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{3}\right ) + \left (-\frac {x \,{\mathrm e}^{x}}{2}+\frac {x^{2} {\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x} x^{3}}{2}+\frac {{\mathrm e}^{x} x^{4}}{2}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{3} -\frac {x \,{\mathrm e}^{x}}{2}+\frac {x^{2} {\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x} x^{3}}{2}+\frac {{\mathrm e}^{x} x^{4}}{2} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{3} -\frac {x \,{\mathrm e}^{x}}{2}+\frac {x^{2} {\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x} x^{3}}{2}+\frac {{\mathrm e}^{x} x^{4}}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime }-3 y={\mathrm e}^{x} \left (16 x^{3}+24 x^{2}+2 x -1\right ) \\ \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 )=16 \,{\mathrm e}^{x} x^{3}+24 x^{2} {\mathrm e}^{x}+2 x \,{\mathrm e}^{x}-3 y_{3}\left (x \right )+y_{2}\left (x \right )+3 y_{1}\left (x \right )-{\mathrm e}^{x} \\ & {} & \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 )=16 \,{\mathrm e}^{x} x^{3}+24 x^{2} {\mathrm e}^{x}+2 x \,{\mathrm e}^{x}-3 y_{3}\left (x \right )+y_{2}\left (x \right )+3 y_{1}\left (x \right )-{\mathrm e}^{x}\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 \\ 3 & 1 & -3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 16 \,{\mathrm e}^{x} x^{3}+24 x^{2} {\mathrm e}^{x}+2 x \,{\mathrm e}^{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 \\ 16 \,{\mathrm e}^{x} x^{3}+24 x^{2} {\mathrm e}^{x}+2 x \,{\mathrm e}^{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 \\ 3 & 1 & -3 \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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 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 [-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}}_{2}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 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} & {\mathrm e}^{-x} & {\mathrm e}^{x} \\ -\frac {{\mathrm e}^{-3 x}}{3} & -{\mathrm e}^{-x} & {\mathrm e}^{x} \\ {\mathrm e}^{-3 x} & {\mathrm e}^{-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} & {\mathrm e}^{-x} & {\mathrm e}^{x} \\ -\frac {{\mathrm e}^{-3 x}}{3} & -{\mathrm e}^{-x} & {\mathrm e}^{x} \\ {\mathrm e}^{-3 x} & {\mathrm e}^{-x} & {\mathrm e}^{x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{9} & 1 & 1 \\ -\frac {1}{3} & -1 & 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 (3 \,{\mathrm e}^{4 x}+6 \,{\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-3 x}}{8} & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & \frac {\left ({\mathrm e}^{4 x}-2 \,{\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-3 x}}{8} \\ \frac {3 \left ({\mathrm e}^{4 x}-2 \,{\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-3 x}}{8} & \frac {{\mathrm e}^{x}}{2}+\frac {{\mathrm e}^{-x}}{2} & \frac {\left ({\mathrm e}^{4 x}+2 \,{\mathrm e}^{2 x}-3\right ) {\mathrm e}^{-3 x}}{8} \\ \frac {3 \left ({\mathrm e}^{4 x}+2 \,{\mathrm e}^{2 x}-3\right ) {\mathrm e}^{-3 x}}{8} & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & \frac {\left ({\mathrm e}^{4 x}-2 \,{\mathrm e}^{2 x}+9\right ) {\mathrm e}^{-3 x}}{8} \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 {{\mathrm e}^{-x}}{4}+\frac {\left (2 x^{4}-2 x^{3}+2 x^{2}-2 x +1\right ) {\mathrm e}^{x}}{4} \\ \frac {{\mathrm e}^{-x}}{4}+\frac {\left (2 x^{4}+6 x^{3}-4 x^{2}+2 x -1\right ) {\mathrm e}^{x}}{4} \\ -\frac {{\mathrm e}^{-x}}{4}+\frac {\left (2 x^{4}+14 x^{3}+14 x^{2}-6 x +1\right ) {\mathrm e}^{x}}{4} \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 {{\mathrm e}^{-x}}{4}+\frac {\left (2 x^{4}-2 x^{3}+2 x^{2}-2 x +1\right ) {\mathrm e}^{x}}{4} \\ \frac {{\mathrm e}^{-x}}{4}+\frac {\left (2 x^{4}+6 x^{3}-4 x^{2}+2 x -1\right ) {\mathrm e}^{x}}{4} \\ -\frac {{\mathrm e}^{-x}}{4}+\frac {\left (2 x^{4}+14 x^{3}+14 x^{2}-6 x +1\right ) {\mathrm e}^{x}}{4} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (x^{4}-x^{3}+x^{2}-x +2 c_{3} +\frac {1}{2}\right ) {\mathrm e}^{4 x}+\left (2 c_{2} -\frac {1}{2}\right ) {\mathrm e}^{2 x}+\frac {2 c_{1}}{9}\right ) {\mathrm e}^{-3 x}}{2} \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: 44
dsolve(diff(y(x),x$3)+3*diff(y(x),x$2)-diff(y(x),x)-3*y(x)=exp(x)*(-1+2*x+24*x^2+16*x^3),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (x^{4}-x^{3}+x^{2}+2 c_{1} -x \right ) {\mathrm e}^{4 x}+2 c_{3} {\mathrm e}^{2 x}+2 c_{2} \right ) {\mathrm e}^{-3 x}}{2} \]
✓ Solution by Mathematica
Time used: 0.071 (sec). Leaf size: 53
DSolve[y'''[x]+3*y''[x]-y'[x]-3*y[x]==Exp[x]*(-1+2*x+24*x^2+16*x^3),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} e^x \left (2 x^4-2 x^3+2 x^2-2 x+1+4 c_3\right )+c_1 e^{-3 x}+c_2 e^{-x} \]