Internal problem ID [1513]
Internal file name [OUTPUT/1514_Sunday_June_05_2022_02_20_14_AM_64334905/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 16.
ODE order: 4.
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
[[_high_order, _linear, _nonhomogeneous]]
\[ \boxed {4 y^{\prime \prime \prime \prime }-11 y^{\prime \prime }-9 y^{\prime }-2 y=-{\mathrm e}^{x} \left (1-6 x \right )} \] 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 \[ 4 y^{\prime \prime \prime \prime }-11 y^{\prime \prime }-9 y^{\prime }-2 y = 0 \] The characteristic equation is \[ 4 \lambda ^{4}-11 \lambda ^{2}-9 \lambda -2 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= 2\\ \lambda _3 &= -{\frac {1}{2}}\\ \lambda _4 &= -{\frac {1}{2}} \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{-\frac {x}{2}} c_{3} +x \,{\mathrm e}^{-\frac {x}{2}} c_{4} \] 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}^{-\frac {x}{2}} \\ y_4 &= x \,{\mathrm e}^{-\frac {x}{2}} \\ \end{align*} Now the particular solution to the given ODE is found \[ 4 y^{\prime \prime \prime \prime }-11 y^{\prime \prime }-9 y^{\prime }-2 y = -{\mathrm e}^{x} \left (1-6 x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ -{\mathrm e}^{x} \left (1-6 x \right ) \] 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 \[ \left \{x \,{\mathrm e}^{-\frac {x}{2}}, {\mathrm e}^{-x}, {\mathrm e}^{2 x}, {\mathrm e}^{-\frac {x}{2}}\right \} \] 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 \[ -15 A_{1} {\mathrm e}^{x}-18 A_{1} x \,{\mathrm e}^{x}-18 A_{2} {\mathrm e}^{x} = -{\mathrm e}^{x} \left (1-6 x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{3}}, A_{2} = {\frac {1}{3}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {x \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{x}}{3} \] 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}^{-\frac {x}{2}} c_{3} +x \,{\mathrm e}^{-\frac {x}{2}} c_{4}\right ) + \left (-\frac {x \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{x}}{3}\right ) \\ \end{align*} Which simplifies to \[ y = \left (\left (c_{4} x +c_{3} \right ) {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{3 x}+c_{1} \right ) {\mathrm e}^{-x}-\frac {x \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{x}}{3} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\left (c_{4} x +c_{3} \right ) {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{3 x}+c_{1} \right ) {\mathrm e}^{-x}-\frac {x \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{x}}{3} \\ \end{align*}
Verification of solutions
\[ y = \left (\left (c_{4} x +c_{3} \right ) {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{3 x}+c_{1} \right ) {\mathrm e}^{-x}-\frac {x \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{x}}{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime \prime \prime }-11 y^{\prime \prime }-9 y^{\prime }-2 y=-{\mathrm e}^{x} \left (1-6 x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=\frac {y}{2}+\frac {11 y^{\prime \prime }}{4}+\frac {9 y^{\prime }}{4}+\frac {3 x \,{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x}}{4} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }-\frac {11 y^{\prime \prime }}{4}-\frac {9 y^{\prime }}{4}-\frac {y}{2}=\frac {{\mathrm e}^{x} \left (-1+6 x \right )}{4} \\ \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 {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=\frac {3 x \,{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x}}{4}+\frac {11 y_{3}\left (x \right )}{4}+\frac {9 y_{2}\left (x \right )}{4}+\frac {y_{1}\left (x \right )}{2} \\ & {} & \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_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{4}^{\prime }\left (x \right )=\frac {3 x \,{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x}}{4}+\frac {11 y_{3}\left (x \right )}{4}+\frac {9 y_{2}\left (x \right )}{4}+\frac {y_{1}\left (x \right )}{2}\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 ) \\ y_{4}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \frac {1}{2} & \frac {9}{4} & \frac {11}{4} & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ \frac {3 x \,{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x}}{4} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ \frac {3 x \,{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x}}{4} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \frac {1}{2} & \frac {9}{4} & \frac {11}{4} & 0 \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 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} -1 \\ 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 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -\frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =-\frac {1}{2}\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =-\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} -\frac {1}{2} \\ {} & {} & \left (\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \frac {1}{2} & \frac {9}{4} & \frac {11}{4} & 0 \end {array}\right ]--\frac {1}{2}\cdot \left [\begin {array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -16 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -\frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left (x \cdot \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -16 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 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}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}+{\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}{cccc} -{\mathrm e}^{-x} & -8 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} \left (-8 x -16\right ) & \frac {{\mathrm e}^{2 x}}{8} \\ {\mathrm e}^{-x} & 4 \,{\mathrm e}^{-\frac {x}{2}} & 4 x \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{2 x}}{4} \\ -{\mathrm e}^{-x} & -2 \,{\mathrm e}^{-\frac {x}{2}} & -2 x \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-x} & {\mathrm e}^{-\frac {x}{2}} & x \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{2 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}{cccc} -{\mathrm e}^{-x} & -8 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} \left (-8 x -16\right ) & \frac {{\mathrm e}^{2 x}}{8} \\ {\mathrm e}^{-x} & 4 \,{\mathrm e}^{-\frac {x}{2}} & 4 x \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{2 x}}{4} \\ -{\mathrm e}^{-x} & -2 \,{\mathrm e}^{-\frac {x}{2}} & -2 x \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-x} & {\mathrm e}^{-\frac {x}{2}} & x \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{2 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -1 & -8 & -16 & \frac {1}{8} \\ 1 & 4 & 0 & \frac {1}{4} \\ -1 & -2 & 0 & \frac {1}{2} \\ 1 & 1 & 0 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{cccc} \frac {{\mathrm e}^{-\frac {x}{2}} \left (2+x \right )}{2} & \frac {\left (2 \,{\mathrm e}^{3 x}+75 \,{\mathrm e}^{\frac {x}{2}} x -42 \,{\mathrm e}^{\frac {x}{2}}+40\right ) {\mathrm e}^{-x}}{60} & \frac {\left (2 \,{\mathrm e}^{3 x}+5 \,{\mathrm e}^{\frac {x}{2}} x -22 \,{\mathrm e}^{\frac {x}{2}}+20\right ) {\mathrm e}^{-x}}{20} & \frac {\left (2 \,{\mathrm e}^{3 x}-15 \,{\mathrm e}^{\frac {x}{2}} x +18 \,{\mathrm e}^{\frac {x}{2}}-20\right ) {\mathrm e}^{-x}}{30} \\ -\frac {x \,{\mathrm e}^{-\frac {x}{2}}}{4} & \frac {\left (8 \,{\mathrm e}^{3 x}-75 \,{\mathrm e}^{\frac {x}{2}} x +192 \,{\mathrm e}^{\frac {x}{2}}-80\right ) {\mathrm e}^{-x}}{120} & \frac {\left (8 \,{\mathrm e}^{3 x}-5 \,{\mathrm e}^{\frac {x}{2}} x +32 \,{\mathrm e}^{\frac {x}{2}}-40\right ) {\mathrm e}^{-x}}{40} & \frac {\left (8 \,{\mathrm e}^{3 x}+15 \,{\mathrm e}^{\frac {x}{2}} x -48 \,{\mathrm e}^{\frac {x}{2}}+40\right ) {\mathrm e}^{-x}}{60} \\ \frac {x \,{\mathrm e}^{-\frac {x}{2}}}{8} & \frac {\left (32 \,{\mathrm e}^{3 x}+75 \,{\mathrm e}^{\frac {x}{2}} x -192 \,{\mathrm e}^{\frac {x}{2}}+160\right ) {\mathrm e}^{-x}}{240} & \frac {\left (32 \,{\mathrm e}^{3 x}+5 \,{\mathrm e}^{\frac {x}{2}} x -32 \,{\mathrm e}^{\frac {x}{2}}+80\right ) {\mathrm e}^{-x}}{80} & \frac {\left (32 \,{\mathrm e}^{3 x}-15 \,{\mathrm e}^{\frac {x}{2}} x +48 \,{\mathrm e}^{\frac {x}{2}}-80\right ) {\mathrm e}^{-x}}{120} \\ -\frac {x \,{\mathrm e}^{-\frac {x}{2}}}{16} & \frac {\left (128 \,{\mathrm e}^{3 x}-75 \,{\mathrm e}^{\frac {x}{2}} x +192 \,{\mathrm e}^{\frac {x}{2}}-320\right ) {\mathrm e}^{-x}}{480} & \frac {\left (128 \,{\mathrm e}^{3 x}-5 \,{\mathrm e}^{\frac {x}{2}} x +32 \,{\mathrm e}^{\frac {x}{2}}-160\right ) {\mathrm e}^{-x}}{160} & \frac {\left (128 \,{\mathrm e}^{3 x}+15 \,{\mathrm e}^{\frac {x}{2}} x -48 \,{\mathrm e}^{\frac {x}{2}}+160\right ) {\mathrm e}^{-x}}{240} \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 ({\mathrm e}^{3 x}-4 x \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{2 x}-5 \,{\mathrm e}^{\frac {x}{2}} x -4\right ) {\mathrm e}^{-x}}{12} \\ \frac {\left (4 \,{\mathrm e}^{3 x}-8 x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{2 x}+5 \,{\mathrm e}^{\frac {x}{2}} x -10 \,{\mathrm e}^{\frac {x}{2}}+8\right ) {\mathrm e}^{-x}}{24} \\ \frac {{\mathrm e}^{-x} \left (-\frac {5 \left (-2+x \right ) {\mathrm e}^{\frac {x}{2}}}{2}-14 x \,{\mathrm e}^{2 x}-5 \,{\mathrm e}^{2 x}+8 \,{\mathrm e}^{3 x}-8\right )}{24} \\ \frac {{\mathrm e}^{-x} \left (\frac {5 \left (-2+x \right ) {\mathrm e}^{\frac {x}{2}}}{2}-22 x \,{\mathrm e}^{2 x}-43 \,{\mathrm e}^{2 x}+32 \,{\mathrm e}^{3 x}+16\right )}{48} \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}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}+\left [\begin {array}{c} \frac {\left ({\mathrm e}^{3 x}-4 x \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{2 x}-5 \,{\mathrm e}^{\frac {x}{2}} x -4\right ) {\mathrm e}^{-x}}{12} \\ \frac {\left (4 \,{\mathrm e}^{3 x}-8 x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{2 x}+5 \,{\mathrm e}^{\frac {x}{2}} x -10 \,{\mathrm e}^{\frac {x}{2}}+8\right ) {\mathrm e}^{-x}}{24} \\ \frac {{\mathrm e}^{-x} \left (-\frac {5 \left (-2+x \right ) {\mathrm e}^{\frac {x}{2}}}{2}-14 x \,{\mathrm e}^{2 x}-5 \,{\mathrm e}^{2 x}+8 \,{\mathrm e}^{3 x}-8\right )}{24} \\ \frac {{\mathrm e}^{-x} \left (\frac {5 \left (-2+x \right ) {\mathrm e}^{\frac {x}{2}}}{2}-22 x \,{\mathrm e}^{2 x}-43 \,{\mathrm e}^{2 x}+32 \,{\mathrm e}^{3 x}+16\right )}{48} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-8 \left (\left (\left (c_{3} +\frac {5}{96}\right ) x +c_{2} +2 c_{3} \right ) {\mathrm e}^{\frac {x}{2}}+\left (\frac {x}{24}-\frac {1}{32}\right ) {\mathrm e}^{2 x}+\left (-\frac {c_{4}}{64}-\frac {1}{96}\right ) {\mathrm e}^{3 x}+\frac {c_{1}}{8}+\frac {1}{24}\right ) {\mathrm e}^{-x} \end {array} \]
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 4; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 40
dsolve(4*diff(y(x),x$4)-11*diff(y(x),x$2)-9*diff(y(x),x)-2*y(x)=-exp(x)*(1-6*x),y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-x} \left (\left (c_{4} x +c_{3} \right ) {\mathrm e}^{\frac {x}{2}}-\frac {{\mathrm e}^{2 x} x}{3}+c_{2} {\mathrm e}^{3 x}+c_{1} +\frac {{\mathrm e}^{2 x}}{3}\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 47
DSolve[4*y''''[x]-11*y''[x]-9*y'[x]-2*y[x]==-Exp[x]*(1-6*x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {1}{3} e^x (x-1)+e^{-x/2} (c_2 x+c_1)+c_3 e^{-x}+c_4 e^{2 x} \]