Internal problem ID [1527]
Internal file name [OUTPUT/1528_Sunday_June_05_2022_02_20_45_AM_76185064/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 30.
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]]
\[ \boxed {y^{\prime \prime \prime }+y^{\prime \prime }-4 y^{\prime }-4 y={\mathrm e}^{-x} \left (\left (1-22 x \right ) \cos \left (2 x \right )-\left (1+6 x \right ) \sin \left (2 x \right )\right )} \] 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 \[ y^{\prime \prime \prime }+y^{\prime \prime }-4 y^{\prime }-4 y = 0 \] The characteristic equation is \[ \lambda ^{3}+\lambda ^{2}-4 \lambda -4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{-2 x}+{\mathrm e}^{2 x} c_{3} \] 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}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+y^{\prime \prime }-4 y^{\prime }-4 y = {\mathrm e}^{-x} \left (\left (1-22 x \right ) \cos \left (2 x \right )-\left (1+6 x \right ) \sin \left (2 x \right )\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{-x} \left (\left (1-22 x \right ) \cos \left (2 x \right )-\left (1+6 x \right ) \sin \left (2 x \right )\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-x} \cos \left (2 x \right ), {\mathrm e}^{-x} \sin \left (2 x \right ), {\mathrm e}^{-x} \cos \left (2 x \right ) x, {\mathrm e}^{-x} \sin \left (2 x \right ) x\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{-2 x}, {\mathrm e}^{-x}, {\mathrm e}^{2 x}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{-x} \cos \left (2 x \right )+A_{2} {\mathrm e}^{-x} \sin \left (2 x \right )+A_{3} {\mathrm e}^{-x} \cos \left (2 x \right ) x +A_{4} {\mathrm e}^{-x} \sin \left (2 x \right ) x \] 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 \[ -8 A_{4} {\mathrm e}^{-x} \cos \left (2 x \right )+8 A_{3} {\mathrm e}^{-x} \sin \left (2 x \right )+8 A_{4} {\mathrm e}^{-x} \sin \left (2 x \right ) x +8 A_{3} {\mathrm e}^{-x} \cos \left (2 x \right ) x +14 A_{3} {\mathrm e}^{-x} \sin \left (2 x \right ) x -15 A_{4} {\mathrm e}^{-x} \sin \left (2 x \right )-14 A_{4} {\mathrm e}^{-x} \cos \left (2 x \right ) x -15 A_{3} {\mathrm e}^{-x} \cos \left (2 x \right )+8 A_{1} {\mathrm e}^{-x} \cos \left (2 x \right )+8 A_{2} {\mathrm e}^{-x} \sin \left (2 x \right )-14 A_{2} {\mathrm e}^{-x} \cos \left (2 x \right )+14 A_{1} {\mathrm e}^{-x} \sin \left (2 x \right ) = {\mathrm e}^{-x} \left (\left (1-22 x \right ) \cos \left (2 x \right )-\left (1+6 x \right ) \sin \left (2 x \right )\right ) \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 1, A_{2} = 1, A_{3} = -1, A_{4} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = {\mathrm e}^{-x} \cos \left (2 x \right )+{\mathrm e}^{-x} \sin \left (2 x \right )-{\mathrm e}^{-x} \cos \left (2 x \right ) x +{\mathrm e}^{-x} \sin \left (2 x \right ) x \] 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}^{2 x} c_{3}\right ) + \left ({\mathrm e}^{-x} \cos \left (2 x \right )+{\mathrm e}^{-x} \sin \left (2 x \right )-{\mathrm e}^{-x} \cos \left (2 x \right ) x +{\mathrm e}^{-x} \sin \left (2 x \right ) x\right ) \\ \end{align*} Which simplifies to \[ y = \left ({\mathrm e}^{4 x} c_{3} +c_{1} {\mathrm e}^{x}+c_{2} \right ) {\mathrm e}^{-2 x}+{\mathrm e}^{-x} \cos \left (2 x \right )+{\mathrm e}^{-x} \sin \left (2 x \right )-{\mathrm e}^{-x} \cos \left (2 x \right ) x +{\mathrm e}^{-x} \sin \left (2 x \right ) x \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left ({\mathrm e}^{4 x} c_{3} +c_{1} {\mathrm e}^{x}+c_{2} \right ) {\mathrm e}^{-2 x}+{\mathrm e}^{-x} \cos \left (2 x \right )+{\mathrm e}^{-x} \sin \left (2 x \right )-{\mathrm e}^{-x} \cos \left (2 x \right ) x +{\mathrm e}^{-x} \sin \left (2 x \right ) x \\ \end{align*}
Verification of solutions
\[ y = \left ({\mathrm e}^{4 x} c_{3} +c_{1} {\mathrm e}^{x}+c_{2} \right ) {\mathrm e}^{-2 x}+{\mathrm e}^{-x} \cos \left (2 x \right )+{\mathrm e}^{-x} \sin \left (2 x \right )-{\mathrm e}^{-x} \cos \left (2 x \right ) x +{\mathrm e}^{-x} \sin \left (2 x \right ) x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+y^{\prime \prime }-4 y^{\prime }-4 y={\mathrm e}^{-x} \left (\left (1-22 x \right ) \cos \left (2 x \right )-\left (1+6 x \right ) \sin \left (2 x \right )\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=4 y-6 \,{\mathrm e}^{-x} \sin \left (2 x \right ) x -22 \,{\mathrm e}^{-x} \cos \left (2 x \right ) x -{\mathrm e}^{-x} \sin \left (2 x \right )+{\mathrm e}^{-x} \cos \left (2 x \right )-y^{\prime \prime }+4 y^{\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 }+y^{\prime \prime }-4 y^{\prime }-4 y=-{\mathrm e}^{-x} \left (22 x \cos \left (2 x \right )+6 x \sin \left (2 x \right )-\cos \left (2 x \right )+\sin \left (2 x \right )\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 {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=-6 \,{\mathrm e}^{-x} \sin \left (2 x \right ) x -22 \,{\mathrm e}^{-x} \cos \left (2 x \right ) x -{\mathrm e}^{-x} \sin \left (2 x \right )+{\mathrm e}^{-x} \cos \left (2 x \right )-y_{3}\left (x \right )+4 y_{2}\left (x \right )+4 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 )=-6 \,{\mathrm e}^{-x} \sin \left (2 x \right ) x -22 \,{\mathrm e}^{-x} \cos \left (2 x \right ) x -{\mathrm e}^{-x} \sin \left (2 x \right )+{\mathrm e}^{-x} \cos \left (2 x \right )-y_{3}\left (x \right )+4 y_{2}\left (x \right )+4 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 \\ 4 & 4 & -1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ -22 \,{\mathrm e}^{-x} \cos \left (2 x \right ) x -6 \,{\mathrm e}^{-x} \sin \left (2 x \right ) x +{\mathrm e}^{-x} \cos \left (2 x \right )-{\mathrm e}^{-x} \sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ -22 \,{\mathrm e}^{-x} \cos \left (2 x \right ) x -6 \,{\mathrm e}^{-x} \sin \left (2 x \right ) x +{\mathrm e}^{-x} \cos \left (2 x \right )-{\mathrm e}^{-x} \sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & 4 & -1 \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}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \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 ]\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}}_{1}={\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}}_{2}={\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}}_{3}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \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}+{\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}^{-2 x}}{4} & {\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{4} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -{\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-2 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}{ccc} \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{4} \\ -\frac {{\mathrm e}^{-2 x}}{2} & -{\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{-x} & {\mathrm e}^{2 x} \end {array}\right ]\cdot \left (\left [\begin {array}{ccc} \frac {1}{4} & 1 & \frac {1}{4} \\ -\frac {1}{2} & -1 & \frac {1}{2} \\ 1 & 1 & 1 \end {array}\right ]\right )^{-\mathrm {1}} \\ {} & \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}+8 \,{\mathrm e}^{x}-3\right ) {\mathrm e}^{-2 x}}{6} & -\frac {{\mathrm e}^{-2 x}}{4}+\frac {{\mathrm e}^{2 x}}{4} & \frac {\left ({\mathrm e}^{4 x}-4 \,{\mathrm e}^{x}+3\right ) {\mathrm e}^{-2 x}}{12} \\ \frac {\left ({\mathrm e}^{4 x}-4 \,{\mathrm e}^{x}+3\right ) {\mathrm e}^{-2 x}}{3} & \frac {{\mathrm e}^{-2 x}}{2}+\frac {{\mathrm e}^{2 x}}{2} & \frac {\left ({\mathrm e}^{4 x}+2 \,{\mathrm e}^{x}-3\right ) {\mathrm e}^{-2 x}}{6} \\ \frac {2 \left ({\mathrm e}^{4 x}+2 \,{\mathrm e}^{x}-3\right ) {\mathrm e}^{-2 x}}{3} & -{\mathrm e}^{-2 x}+{\mathrm e}^{2 x} & \frac {\left ({\mathrm e}^{4 x}-{\mathrm e}^{x}+3\right ) {\mathrm e}^{-2 x}}{3} \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}^{-2 x} \left (\left (x -1\right ) {\mathrm e}^{x} \cos \left (2 x \right )-\left (x +1\right ) {\mathrm e}^{x} \sin \left (2 x \right )+\frac {5 \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{4 x}}{12}-\frac {3}{4}\right ) \\ {\mathrm e}^{-2 x} \left ({\mathrm e}^{x} \left (-2+x \right ) \sin \left (2 x \right )+3 \,{\mathrm e}^{x} \cos \left (2 x \right ) x +\frac {5 \,{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{4 x}}{6}-\frac {3}{2}\right ) \\ -{\mathrm e}^{-2 x} \left (\left (x +1\right ) {\mathrm e}^{x} \cos \left (2 x \right )+{\mathrm e}^{x} \left (7 x -3\right ) \sin \left (2 x \right )+\frac {5 \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{4 x}}{3}-3\right ) \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}^{-2 x} \left (\left (x -1\right ) {\mathrm e}^{x} \cos \left (2 x \right )-\left (x +1\right ) {\mathrm e}^{x} \sin \left (2 x \right )+\frac {5 \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{4 x}}{12}-\frac {3}{4}\right ) \\ {\mathrm e}^{-2 x} \left ({\mathrm e}^{x} \left (-2+x \right ) \sin \left (2 x \right )+3 \,{\mathrm e}^{x} \cos \left (2 x \right ) x +\frac {5 \,{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{4 x}}{6}-\frac {3}{2}\right ) \\ -{\mathrm e}^{-2 x} \left (\left (x +1\right ) {\mathrm e}^{x} \cos \left (2 x \right )+{\mathrm e}^{x} \left (7 x -3\right ) \sin \left (2 x \right )+\frac {5 \,{\mathrm e}^{x}}{3}+\frac {{\mathrm e}^{4 x}}{3}-3\right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{-2 x} \left (\left (x -1\right ) {\mathrm e}^{x} \cos \left (2 x \right )+\left (-\frac {c_{3}}{4}+\frac {1}{12}\right ) {\mathrm e}^{4 x}-\left (x +1\right ) {\mathrm e}^{x} \sin \left (2 x \right )+\left (-c_{2} +\frac {5}{3}\right ) {\mathrm e}^{x}-\frac {c_{1}}{4}-\frac {3}{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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 50
dsolve(1*diff(y(x),x$3)+1*diff(y(x),x$2)-4*diff(y(x),x)-4*y(x)=exp(-x)*((1-22*x)*cos(2*x)-(1+6*x)*sin(2*x)),y(x), singsol=all)
\[ y \left (x \right ) = -\left ({\mathrm e}^{x} \left (x -1\right ) \cos \left (2 x \right )-\left (x +1\right ) {\mathrm e}^{x} \sin \left (2 x \right )-c_{3} {\mathrm e}^{4 x}+\left (-c_{2} +\frac {5}{3}\right ) {\mathrm e}^{x}-c_{1} \right ) {\mathrm e}^{-2 x} \]
✓ Solution by Mathematica
Time used: 0.02 (sec). Leaf size: 51
DSolve[1*y'''[x]+1*y''[x]-4*y'[x]-4*y[x]==Exp[-x]*((1-22*x)*Cos[2*x]-(1+6*x)*Sin[2*x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-2 x} \left (e^x (x+1) \sin (2 x)-e^x (x-1) \cos (2 x)+c_2 e^x+c_3 e^{4 x}+c_1\right ) \]