Internal problem ID [11812]
Internal file name [OUTPUT/11822_Thursday_April_11_2024_08_50_02_PM_7433673/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 4, Section 4.3. The method of undetermined coefficients. Exercises page
151
Problem number: 39.
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]]
\[ \boxed {y^{\prime \prime \prime }-4 y^{\prime \prime }+y^{\prime }+6 y=3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (x \right )} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {33}{40}}, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 0\right ] \end {align*}
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 \[ y^{\prime \prime \prime }-4 y^{\prime \prime }+y^{\prime }+6 y = 0 \] The characteristic equation is \[ \lambda ^{3}-4 \lambda ^{2}+\lambda +6 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 3\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{3 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= {\mathrm e}^{2 x} \\ y_3 &= {\mathrm e}^{3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-4 y^{\prime \prime }+y^{\prime }+6 y = 3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}, \{\cos \left (x \right ), \sin \left (x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{-x}, {\mathrm e}^{2 x}, {\mathrm e}^{3 x}\} \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{x}+A_{2} {\mathrm e}^{x}+A_{3} \cos \left (x \right )+A_{4} \sin \left (x \right ) \] 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 \[ -4 A_{1} {\mathrm e}^{x}+4 A_{1} x \,{\mathrm e}^{x}+4 A_{2} {\mathrm e}^{x}+10 A_{3} \cos \left (x \right )+10 A_{4} \sin \left (x \right ) = 3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {3}{4}}, A_{2} = {\frac {5}{4}}, A_{3} = 0, A_{4} = -{\frac {1}{10}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {3 x \,{\mathrm e}^{x}}{4}+\frac {5 \,{\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \] 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}^{3 x} c_{3}\right ) + \left (\frac {3 x \,{\mathrm e}^{x}}{4}+\frac {5 \,{\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10}\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{3 x} c_{3} +\frac {3 x \,{\mathrm e}^{x}}{4}+\frac {5 \,{\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = {\frac {33}{40}}\) and \(x = 0\) in the above gives \begin {align*} {\frac {33}{40}} = c_{1} +c_{2} +c_{3} +\frac {5}{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} {\mathrm e}^{-x}+2 c_{2} {\mathrm e}^{2 x}+3 \,{\mathrm e}^{3 x} c_{3} +\frac {3 x \,{\mathrm e}^{x}}{4}+2 \,{\mathrm e}^{x}-\frac {\cos \left (x \right )}{10} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = -c_{1} +2 c_{2} +3 c_{3} +\frac {19}{10}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{-x}+4 c_{2} {\mathrm e}^{2 x}+9 \,{\mathrm e}^{3 x} c_{3} +\frac {3 x \,{\mathrm e}^{x}}{4}+\frac {11 \,{\mathrm e}^{x}}{4}+\frac {\sin \left (x \right )}{10} \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +4 c_{2} +9 c_{3} +\frac {11}{4}\tag {3A} \end {align*}
Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&={\frac {7}{20}}\\ c_{2}&=-{\frac {31}{40}}\\ c_{3}&=0 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {7 \,{\mathrm e}^{-x}}{20}-\frac {31 \,{\mathrm e}^{2 x}}{40}+\frac {3 x \,{\mathrm e}^{x}}{4}+\frac {5 \,{\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {7 \,{\mathrm e}^{-x}}{20}-\frac {31 \,{\mathrm e}^{2 x}}{40}+\frac {3 x \,{\mathrm e}^{x}}{4}+\frac {5 \,{\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \\ \end{align*}
Verification of solutions
\[ y = \frac {7 \,{\mathrm e}^{-x}}{20}-\frac {31 \,{\mathrm e}^{2 x}}{40}+\frac {3 x \,{\mathrm e}^{x}}{4}+\frac {5 \,{\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-4 y^{\prime \prime }+y^{\prime }+6 y=3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (x \right ), y \left (0\right )=\frac {33}{40}, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\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 )=3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (x \right )+4 y_{3}\left (x \right )-y_{2}\left (x \right )-6 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 )=3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (x \right )+4 y_{3}\left (x \right )-y_{2}\left (x \right )-6 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 \\ -6 & -1 & 4 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (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 \\ 3 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}-\sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -1 & 4 \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 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ]\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}}_{1}={\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}}_{2}={\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 [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}}_{3}={\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 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} {\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{4} & \frac {{\mathrm e}^{3 x}}{9} \\ -{\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{2} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-x} & {\mathrm e}^{2 x} & {\mathrm e}^{3 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} {\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{4} & \frac {{\mathrm e}^{3 x}}{9} \\ -{\mathrm e}^{-x} & \frac {{\mathrm e}^{2 x}}{2} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-x} & {\mathrm e}^{2 x} & {\mathrm e}^{3 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & \frac {1}{4} & \frac {1}{9} \\ -1 & \frac {1}{2} & \frac {1}{3} \\ 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 {{\mathrm e}^{-x}}{2}+{\mathrm e}^{2 x}-\frac {{\mathrm e}^{3 x}}{2} & -\frac {5 \,{\mathrm e}^{-x}}{12}+\frac {2 \,{\mathrm e}^{2 x}}{3}-\frac {{\mathrm e}^{3 x}}{4} & \frac {{\mathrm e}^{-x}}{12}-\frac {{\mathrm e}^{2 x}}{3}+\frac {{\mathrm e}^{3 x}}{4} \\ -\frac {{\mathrm e}^{-x}}{2}+2 \,{\mathrm e}^{2 x}-\frac {3 \,{\mathrm e}^{3 x}}{2} & \frac {5 \,{\mathrm e}^{-x}}{12}+\frac {4 \,{\mathrm e}^{2 x}}{3}-\frac {3 \,{\mathrm e}^{3 x}}{4} & -\frac {{\mathrm e}^{-x}}{12}-\frac {2 \,{\mathrm e}^{2 x}}{3}+\frac {3 \,{\mathrm e}^{3 x}}{4} \\ \frac {{\mathrm e}^{-x}}{2}+4 \,{\mathrm e}^{2 x}-\frac {9 \,{\mathrm e}^{3 x}}{2} & -\frac {5 \,{\mathrm e}^{-x}}{12}+\frac {8 \,{\mathrm e}^{2 x}}{3}-\frac {9 \,{\mathrm e}^{3 x}}{4} & \frac {{\mathrm e}^{-x}}{12}-\frac {4 \,{\mathrm e}^{2 x}}{3}+\frac {9 \,{\mathrm e}^{3 x}}{4} \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}}{16}-\frac {8 \,{\mathrm e}^{2 x}}{5}+\frac {33 \,{\mathrm e}^{3 x}}{80}+\frac {\left (5+3 x \right ) {\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \\ \frac {3 x \,{\mathrm e}^{x}}{4}-\frac {\cos \left (x \right )}{10}+\frac {99 \,{\mathrm e}^{3 x}}{80}-\frac {16 \,{\mathrm e}^{2 x}}{5}+2 \,{\mathrm e}^{x}+\frac {{\mathrm e}^{-x}}{16} \\ -\frac {{\mathrm e}^{-x}}{16}-\frac {32 \,{\mathrm e}^{2 x}}{5}+\frac {297 \,{\mathrm e}^{3 x}}{80}+\frac {\left (3 x +11\right ) {\mathrm e}^{x}}{4}+\frac {\sin \left (x \right )}{10} \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}}{16}-\frac {8 \,{\mathrm e}^{2 x}}{5}+\frac {33 \,{\mathrm e}^{3 x}}{80}+\frac {\left (5+3 x \right ) {\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \\ \frac {3 x \,{\mathrm e}^{x}}{4}-\frac {\cos \left (x \right )}{10}+\frac {99 \,{\mathrm e}^{3 x}}{80}-\frac {16 \,{\mathrm e}^{2 x}}{5}+2 \,{\mathrm e}^{x}+\frac {{\mathrm e}^{-x}}{16} \\ -\frac {{\mathrm e}^{-x}}{16}-\frac {32 \,{\mathrm e}^{2 x}}{5}+\frac {297 \,{\mathrm e}^{3 x}}{80}+\frac {\left (3 x +11\right ) {\mathrm e}^{x}}{4}+\frac {\sin \left (x \right )}{10} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (720 c_{1} -45\right ) {\mathrm e}^{-x}}{720}+\frac {\left (180 c_{2} -1152\right ) {\mathrm e}^{2 x}}{720}+\frac {\left (80 c_{3} +297\right ) {\mathrm e}^{3 x}}{720}+\frac {\left (540 x +900\right ) {\mathrm e}^{x}}{720}-\frac {\sin \left (x \right )}{10} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=\frac {33}{40} \\ {} & {} & \frac {33}{40}=c_{1} +\frac {c_{2}}{4}+\frac {c_{3}}{9} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\left (720 c_{1} -45\right ) {\mathrm e}^{-x}}{720}+\frac {\left (180 c_{2} -1152\right ) {\mathrm e}^{2 x}}{360}+\frac {\left (80 c_{3} +297\right ) {\mathrm e}^{3 x}}{240}+\frac {3 \,{\mathrm e}^{x}}{4}+\frac {\left (540 x +900\right ) {\mathrm e}^{x}}{720}-\frac {\cos \left (x \right )}{10} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-c_{1} +\frac {c_{2}}{2}+\frac {c_{3}}{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (720 c_{1} -45\right ) {\mathrm e}^{-x}}{720}+\frac {\left (180 c_{2} -1152\right ) {\mathrm e}^{2 x}}{180}+\frac {\left (80 c_{3} +297\right ) {\mathrm e}^{3 x}}{80}+\frac {3 \,{\mathrm e}^{x}}{2}+\frac {\left (540 x +900\right ) {\mathrm e}^{x}}{720}+\frac {\sin \left (x \right )}{10} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {33}{80}, c_{2} =\frac {33}{10}, c_{3} =-\frac {297}{80}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {7 \,{\mathrm e}^{-x}}{20}-\frac {31 \,{\mathrm e}^{2 x}}{40}+\frac {\left (5+3 x \right ) {\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \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: 0.031 (sec). Leaf size: 30
dsolve([diff(y(x),x$3)-4*diff(y(x),x$2)+diff(y(x),x)+6*y(x)=3*x*exp(x)+2*exp(x)-sin(x),y(0) = 33/40, D(y)(0) = 0, (D@@2)(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = \frac {7 \,{\mathrm e}^{-x}}{20}-\frac {31 \,{\mathrm e}^{2 x}}{40}+\frac {\left (3 x +5\right ) {\mathrm e}^{x}}{4}-\frac {\sin \left (x \right )}{10} \]
✓ Solution by Mathematica
Time used: 0.285 (sec). Leaf size: 38
DSolve[{y'''[x]-4*y''[x]+y'[x]+6*y[x]==3*x*Exp[x]+2*Exp[x]-Sin[x],{y[0]==33/40,y'[0]==0,y''[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{40} \left (10 e^x (3 x+5)+14 e^{-x}-31 e^{2 x}-4 \sin (x)\right ) \]