Internal problem ID [1519]
Internal file name [OUTPUT/1520_Sunday_June_05_2022_02_20_28_AM_83982495/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 22.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }-5 y^{\prime \prime }+4 y={\mathrm e}^{x} \left (-3 x^{2}+x +3\right )} \] 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 \[ y^{\prime \prime \prime \prime }-5 y^{\prime \prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{4}-5 \lambda ^{2}+4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2\\ \lambda _3 &= 1\\ \lambda _4 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{-2 x}+{\mathrm e}^{x} c_{3} +{\mathrm e}^{2 x} c_{4} \] 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}^{x} \\ y_4 &= {\mathrm e}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-5 y^{\prime \prime }+4 y = {\mathrm e}^{x} \left (-3 x^{2}+x +3\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x} \left (-3 x^{2}+x +3\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{-2 x}, {\mathrm e}^{-x}, {\mathrm e}^{2 x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x} x^{3}\}] \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{x}+A_{2} x^{2} {\mathrm e}^{x}+A_{3} {\mathrm e}^{x} x^{3} \] 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 \[ -6 A_{1} {\mathrm e}^{x}+2 A_{2} {\mathrm e}^{x}-12 A_{2} x \,{\mathrm e}^{x}-18 A_{3} {\mathrm e}^{x} x^{2}+6 A_{3} {\mathrm e}^{x} x +24 A_{3} {\mathrm e}^{x} = {\mathrm e}^{x} \left (-3 x^{2}+x +3\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{6}}, A_{2} = 0, A_{3} = {\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x \,{\mathrm e}^{x}}{6}+\frac {{\mathrm e}^{x} x^{3}}{6} \] 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}^{x} c_{3} +{\mathrm e}^{2 x} c_{4}\right ) + \left (\frac {x \,{\mathrm e}^{x}}{6}+\frac {{\mathrm e}^{x} x^{3}}{6}\right ) \\ \end{align*}
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}^{x} c_{3} +{\mathrm e}^{2 x} c_{4} +\frac {x \,{\mathrm e}^{x}}{6}+\frac {{\mathrm e}^{x} x^{3}}{6} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{-2 x}+{\mathrm e}^{x} c_{3} +{\mathrm e}^{2 x} c_{4} +\frac {x \,{\mathrm e}^{x}}{6}+\frac {{\mathrm e}^{x} x^{3}}{6} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }-5 y^{\prime \prime }+4 y={\mathrm e}^{x} \left (-3 x^{2}+x +3\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 }=-4 y-3 x^{2} {\mathrm e}^{x}+x \,{\mathrm e}^{x}+3 \,{\mathrm e}^{x}+5 y^{\prime \prime } \\ \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 }-5 y^{\prime \prime }+4 y=-{\mathrm e}^{x} \left (3 x^{2}-x -3\right ) \\ \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 )=-3 x^{2} {\mathrm e}^{x}+x \,{\mathrm e}^{x}+5 y_{3}\left (x \right )-4 y_{1}\left (x \right )+3 \,{\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_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{4}^{\prime }\left (x \right )=-3 x^{2} {\mathrm e}^{x}+x \,{\mathrm e}^{x}+5 y_{3}\left (x \right )-4 y_{1}\left (x \right )+3 \,{\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 ) \\ 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 \\ -4 & 0 & 5 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ -3 x^{2} {\mathrm e}^{x}+x \,{\mathrm e}^{x}+3 \,{\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 \\ 0 \\ -3 x^{2} {\mathrm e}^{x}+x \,{\mathrm e}^{x}+3 \,{\mathrm e}^{x} \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 \\ -4 & 0 & 5 & 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 [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \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 [-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}}_{1}={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \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 \\ 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 \\ 1 \end {array}\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}}_{3}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\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}+c_{3} {\moverset {\rightarrow }{y}}_{3}+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} -\frac {{\mathrm e}^{-2 x}}{8} & -{\mathrm e}^{-x} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{8} \\ \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{-x} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{4} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -{\mathrm e}^{-x} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-x} & {\mathrm e}^{x} & {\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} -\frac {{\mathrm e}^{-2 x}}{8} & -{\mathrm e}^{-x} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{8} \\ \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{-x} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{4} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -{\mathrm e}^{-x} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-x} & {\mathrm e}^{x} & {\mathrm e}^{2 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -\frac {1}{8} & -1 & 1 & \frac {1}{8} \\ \frac {1}{4} & 1 & 1 & \frac {1}{4} \\ -\frac {1}{2} & -1 & 1 & \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}{cccc} -\frac {\left ({\mathrm e}^{4 x}-4 \,{\mathrm e}^{3 x}-4 \,{\mathrm e}^{x}+1\right ) {\mathrm e}^{-2 x}}{6} & -\frac {\left ({\mathrm e}^{4 x}-8 \,{\mathrm e}^{3 x}+8 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{12} & -\frac {\left (-{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{6} & \frac {\left ({\mathrm e}^{4 x}-2 \,{\mathrm e}^{3 x}+2 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{12} \\ -\frac {\left ({\mathrm e}^{4 x}-2 \,{\mathrm e}^{3 x}+2 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{3} & -\frac {\left ({\mathrm e}^{4 x}-4 \,{\mathrm e}^{3 x}-4 \,{\mathrm e}^{x}+1\right ) {\mathrm e}^{-2 x}}{6} & \frac {\left (2 \,{\mathrm e}^{4 x}-{\mathrm e}^{3 x}+{\mathrm e}^{x}-2\right ) {\mathrm e}^{-2 x}}{6} & -\frac {\left (-{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{6} \\ \frac {2 \left (-{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{3} & -\frac {\left ({\mathrm e}^{4 x}-2 \,{\mathrm e}^{3 x}+2 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{3} & -\frac {\left (-4 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+{\mathrm e}^{x}-4\right ) {\mathrm e}^{-2 x}}{6} & \frac {\left (2 \,{\mathrm e}^{4 x}-{\mathrm e}^{3 x}+{\mathrm e}^{x}-2\right ) {\mathrm e}^{-2 x}}{6} \\ -\frac {2 \left (2 \,{\mathrm e}^{4 x}-{\mathrm e}^{3 x}+{\mathrm e}^{x}-2\right ) {\mathrm e}^{-2 x}}{3} & \frac {2 \left (-{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2 x}}{3} & \frac {\left (8 \,{\mathrm e}^{4 x}-{\mathrm e}^{3 x}+{\mathrm e}^{x}-8\right ) {\mathrm e}^{-2 x}}{6} & -\frac {\left (-4 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}+{\mathrm e}^{x}-4\right ) {\mathrm e}^{-2 x}}{6} \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}^{-2 x} \left (\left (6 x^{3}+6 x +7\right ) {\mathrm e}^{3 x}-3 \,{\mathrm e}^{x}-6 \,{\mathrm e}^{4 x}+2\right )}{36} \\ \frac {{\mathrm e}^{-2 x} \left (\left (x^{3}+3 x^{2}+x +\frac {13}{6}\right ) {\mathrm e}^{3 x}+\frac {{\mathrm e}^{x}}{2}-2 \,{\mathrm e}^{4 x}-\frac {2}{3}\right )}{6} \\ \frac {{\mathrm e}^{-2 x} \left (\left (x^{3}+6 x^{2}+7 x +\frac {19}{6}\right ) {\mathrm e}^{3 x}-\frac {{\mathrm e}^{x}}{2}-4 \,{\mathrm e}^{4 x}+\frac {4}{3}\right )}{6} \\ \frac {{\mathrm e}^{-2 x} \left (\left (x^{3}+9 x^{2}+19 x +\frac {61}{6}\right ) {\mathrm e}^{3 x}+\frac {{\mathrm e}^{x}}{2}-8 \,{\mathrm e}^{4 x}-\frac {8}{3}\right )}{6} \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}+c_{4} {\moverset {\rightarrow }{y}}_{4}+\left [\begin {array}{c} \frac {{\mathrm e}^{-2 x} \left (\left (6 x^{3}+6 x +7\right ) {\mathrm e}^{3 x}-3 \,{\mathrm e}^{x}-6 \,{\mathrm e}^{4 x}+2\right )}{36} \\ \frac {{\mathrm e}^{-2 x} \left (\left (x^{3}+3 x^{2}+x +\frac {13}{6}\right ) {\mathrm e}^{3 x}+\frac {{\mathrm e}^{x}}{2}-2 \,{\mathrm e}^{4 x}-\frac {2}{3}\right )}{6} \\ \frac {{\mathrm e}^{-2 x} \left (\left (x^{3}+6 x^{2}+7 x +\frac {19}{6}\right ) {\mathrm e}^{3 x}-\frac {{\mathrm e}^{x}}{2}-4 \,{\mathrm e}^{4 x}+\frac {4}{3}\right )}{6} \\ \frac {{\mathrm e}^{-2 x} \left (\left (x^{3}+9 x^{2}+19 x +\frac {61}{6}\right ) {\mathrm e}^{3 x}+\frac {{\mathrm e}^{x}}{2}-8 \,{\mathrm e}^{4 x}-\frac {8}{3}\right )}{6} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 x} \left (\left (x^{3}+x +6 c_{3} +\frac {7}{6}\right ) {\mathrm e}^{3 x}+\left (\frac {3 c_{4}}{4}-1\right ) {\mathrm e}^{4 x}+\left (-6 c_{2} -\frac {1}{2}\right ) {\mathrm e}^{x}-\frac {3 c_{1}}{4}+\frac {1}{3}\right )}{6} \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: 0.016 (sec). Leaf size: 39
dsolve(1*diff(y(x),x$4)+0*diff(y(x),x$3)-5*diff(y(x),x$2)-0*diff(y(x),x)+4*y(x)=exp(x)*(3+x-3*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (x^{3}+6 c_{1} +x \right ) {\mathrm e}^{3 x}+6 c_{3} {\mathrm e}^{x}+6 c_{4} {\mathrm e}^{4 x}+6 c_{2} \right ) {\mathrm e}^{-2 x}}{6} \]
✓ Solution by Mathematica
Time used: 0.111 (sec). Leaf size: 51
DSolve[1*y''''[x]+0*y'''[x]-5*y''[x]-0*y'[x]+4*y[x]==Exp[x]*(3+x-3*x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{36} e^x \left (6 x^3+6 x+7+36 c_3\right )+c_1 e^{-2 x}+c_2 e^{-x}+c_4 e^{2 x} \]