2.19 problem 19

Internal problem ID [3260]
Internal file name [OUTPUT/2752_Sunday_June_05_2022_08_40_02_AM_40663654/index.tex]

Book: Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 4. Linear Differential Equations. Page 183
Problem number: 19.
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]]

\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime \prime }+4 y^{\prime }-8 y={\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2}} \] 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 \[ y^{\prime \prime \prime }-2 y^{\prime \prime }+4 y^{\prime }-8 y = 0 \] The characteristic equation is \[ \lambda ^{3}-2 \lambda ^{2}+4 \lambda -8 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 2 i\\ \lambda _3 &= -2 i \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-2 i x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{2 i x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 i x} \\ y_2 &= {\mathrm e}^{2 x} \\ y_3 &= {\mathrm e}^{2 i x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-2 y^{\prime \prime }+4 y^{\prime }-8 y = {\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{2 x} \cos \left (2 x \right ), {\mathrm e}^{2 x} \sin \left (2 x \right )\}, \{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{2 x}, {\mathrm e}^{-2 i x}, {\mathrm e}^{2 i x}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{2 x} \cos \left (2 x \right )+A_{2} {\mathrm e}^{2 x} \sin \left (2 x \right )+A_{3}+A_{4} x +A_{5} x^{2} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\}\) 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 \[ -8 A_{1} {\mathrm e}^{2 x} \sin \left (2 x \right )-16 A_{1} {\mathrm e}^{2 x} \cos \left (2 x \right )+8 A_{2} {\mathrm e}^{2 x} \cos \left (2 x \right )-16 A_{2} {\mathrm e}^{2 x} \sin \left (2 x \right )-4 A_{5}+4 A_{4}+8 A_{5} x -8 A_{3}-8 A_{4} x -8 A_{5} x^{2} = {\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{40}}, A_{2} = -{\frac {1}{20}}, A_{3} = 0, A_{4} = -{\frac {1}{4}}, A_{5} = -{\frac {1}{4}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {{\mathrm e}^{2 x} \cos \left (2 x \right )}{40}-\frac {{\mathrm e}^{2 x} \sin \left (2 x \right )}{20}-\frac {x}{4}-\frac {x^{2}}{4} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-2 i x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{2 i x} c_{3}\right ) + \left (-\frac {{\mathrm e}^{2 x} \cos \left (2 x \right )}{40}-\frac {{\mathrm e}^{2 x} \sin \left (2 x \right )}{20}-\frac {x}{4}-\frac {x^{2}}{4}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-2 i x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{2 i x} c_{3} -\frac {{\mathrm e}^{2 x} \cos \left (2 x \right )}{40}-\frac {{\mathrm e}^{2 x} \sin \left (2 x \right )}{20}-\frac {x}{4}-\frac {x^{2}}{4} \\ \end{align*}

Verification of solutions

\[ y = {\mathrm e}^{-2 i x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{2 i x} c_{3} -\frac {{\mathrm e}^{2 x} \cos \left (2 x \right )}{40}-\frac {{\mathrm e}^{2 x} \sin \left (2 x \right )}{20}-\frac {x}{4}-\frac {x^{2}}{4} \] Verified OK.

2.19.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime \prime }-2 \frac {d}{d x}y^{\prime }+4 y^{\prime }-8 y={\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\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 )=\frac {d}{d x}y^{\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 )={\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2}+2 y_{3}\left (x \right )-4 y_{2}\left (x \right )+8 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 )={\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2}+2 y_{3}\left (x \right )-4 y_{2}\left (x \right )+8 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 \\ 8 & -4 & 2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{2 x} \sin \left (2 x \right )+2 x^{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 8 & -4 & 2 \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 [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ -\frac {\mathrm {I}}{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 complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} x}\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right )\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{4}+\frac {\mathrm {I} \sin \left (2 x \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (x \right )=\left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{4} \\ \frac {\sin \left (2 x \right )}{2} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} \frac {\sin \left (2 x \right )}{4} \\ \frac {\cos \left (2 x \right )}{2} \\ -\sin \left (2 x \right ) \end {array}\right ]\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 )+{\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} & -\frac {\cos \left (2 x \right )}{4} & \frac {\sin \left (2 x \right )}{4} \\ \frac {{\mathrm e}^{2 x}}{2} & \frac {\sin \left (2 x \right )}{2} & \frac {\cos \left (2 x \right )}{2} \\ {\mathrm e}^{2 x} & \cos \left (2 x \right ) & -\sin \left (2 x \right ) \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 \phi \left (0\right )^{-1} \\ {} & \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} & -\frac {\cos \left (2 x \right )}{4} & \frac {\sin \left (2 x \right )}{4} \\ \frac {{\mathrm e}^{2 x}}{2} & \frac {\sin \left (2 x \right )}{2} & \frac {\cos \left (2 x \right )}{2} \\ {\mathrm e}^{2 x} & \cos \left (2 x \right ) & -\sin \left (2 x \right ) \end {array}\right ]\cdot \left [\begin {array}{ccc} \frac {1}{4} & -\frac {1}{4} & 0 \\ \frac {1}{2} & 0 & \frac {1}{2} \\ 1 & 1 & 0 \end {array}\right ]^{-1} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{2 x}}{2}+\frac {\cos \left (2 x \right )}{2}-\frac {\sin \left (2 x \right )}{2} & \frac {\sin \left (2 x \right )}{2} & \frac {{\mathrm e}^{2 x}}{8}-\frac {\cos \left (2 x \right )}{8}-\frac {\sin \left (2 x \right )}{8} \\ {\mathrm e}^{2 x}-\sin \left (2 x \right )-\cos \left (2 x \right ) & \cos \left (2 x \right ) & \frac {{\mathrm e}^{2 x}}{4}+\frac {\sin \left (2 x \right )}{4}-\frac {\cos \left (2 x \right )}{4} \\ 2 \,{\mathrm e}^{2 x}-2 \cos \left (2 x \right )+2 \sin \left (2 x \right ) & -2 \sin \left (2 x \right ) & \frac {{\mathrm e}^{2 x}}{2}+\frac {\cos \left (2 x \right )}{2}+\frac {\sin \left (2 x \right )}{2} \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 )=\Phi \left (x \right )^{-1}\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}\Phi \left (s \right )^{-1}\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}\Phi \left (s \right )^{-1}\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 (-\cos \left (2 x \right )+5-2 \sin \left (2 x \right )\right ) {\mathrm e}^{2 x}}{40}-\frac {x^{2}}{4}-\frac {x}{4}-\frac {\cos \left (2 x \right )}{10}+\frac {3 \sin \left (2 x \right )}{40} \\ \frac {\left (-3 \cos \left (2 x \right )+5-\sin \left (2 x \right )\right ) {\mathrm e}^{2 x}}{20}-\frac {x}{2}+\frac {3 \cos \left (2 x \right )}{20}+\frac {\sin \left (2 x \right )}{5}-\frac {1}{4} \\ \frac {\left (-4 \cos \left (2 x \right )+5+2 \sin \left (2 x \right )\right ) {\mathrm e}^{2 x}}{10}+\frac {2 \cos \left (2 x \right )}{5}-\frac {3 \sin \left (2 x \right )}{10}-\frac {1}{2} \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 )+\left [\begin {array}{c} \frac {\left (-\cos \left (2 x \right )+5-2 \sin \left (2 x \right )\right ) {\mathrm e}^{2 x}}{40}-\frac {x^{2}}{4}-\frac {x}{4}-\frac {\cos \left (2 x \right )}{10}+\frac {3 \sin \left (2 x \right )}{40} \\ \frac {\left (-3 \cos \left (2 x \right )+5-\sin \left (2 x \right )\right ) {\mathrm e}^{2 x}}{20}-\frac {x}{2}+\frac {3 \cos \left (2 x \right )}{20}+\frac {\sin \left (2 x \right )}{5}-\frac {1}{4} \\ \frac {\left (-4 \cos \left (2 x \right )+5+2 \sin \left (2 x \right )\right ) {\mathrm e}^{2 x}}{10}+\frac {2 \cos \left (2 x \right )}{5}-\frac {3 \sin \left (2 x \right )}{10}-\frac {1}{2} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (10 c_{1} -\cos \left (2 x \right )-2 \sin \left (2 x \right )+5\right ) {\mathrm e}^{2 x}}{40}+\frac {\left (-2-5 c_{2} \right ) \cos \left (2 x \right )}{20}+\frac {\left (10 c_{3} +3\right ) \sin \left (2 x \right )}{40}-\frac {x^{2}}{4}-\frac {x}{4} \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: 58

dsolve(diff(y(x),x$3)-2*diff(y(x),x$2)+4*diff(y(x),x)-8*y(x)=exp(2*x)*sin(2*x)+2*x^2,y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\left (80 c_{2} -2 \cos \left (2 x \right )-4 \sin \left (2 x \right )-5\right ) {\mathrm e}^{2 x}}{80}+\frac {\left (80 c_{1} -5\right ) \cos \left (2 x \right )}{80}+\frac {\left (80 c_{3} +5\right ) \sin \left (2 x \right )}{80}-\frac {x^{2}}{4}-\frac {x}{4} \]

✓ Solution by Mathematica

Time used: 0.241 (sec). Leaf size: 61

DSolve[y'''[x]-2*y''[x]+4*y'[x]-8*y[x]==Exp[2*x]*Sin[2*x]+2*x^2,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{80} \left (-20 x (x+1)+5 (-1+16 c_3) e^{2 x}-2 \left (e^{2 x}-40 c_1\right ) \cos (2 x)-4 \left (e^{2 x}-20 c_2\right ) \sin (2 x)\right ) \]