Internal problem ID [1506]
Internal file name [OUTPUT/1507_Sunday_June_05_2022_02_20_00_AM_31149122/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 9.
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]]
\[ \boxed {2 y^{\prime \prime \prime }-7 y^{\prime \prime }+4 y^{\prime }+4 y={\mathrm e}^{2 x} \left (17+30 x \right )} \] 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 \[ 2 y^{\prime \prime \prime }-7 y^{\prime \prime }+4 y^{\prime }+4 y = 0 \] The characteristic equation is \[ 2 \lambda ^{3}-7 \lambda ^{2}+4 \lambda +4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -{\frac {1}{2}}\\ \lambda _2 &= 2\\ \lambda _3 &= 2 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{2 x}+c_{2} x \,{\mathrm e}^{2 x}+{\mathrm e}^{-\frac {x}{2}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{2 x} \\ y_2 &= x \,{\mathrm e}^{2 x} \\ y_3 &= {\mathrm e}^{-\frac {x}{2}} \\ \end{align*} Now the particular solution to the given ODE is found \[ 2 y^{\prime \prime \prime }-7 y^{\prime \prime }+4 y^{\prime }+4 y = {\mathrm e}^{2 x} \left (17+30 x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{2 x} \left (17+30 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{2 x}, {\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{x \,{\mathrm e}^{2 x}, {\mathrm e}^{2 x}, {\mathrm e}^{-\frac {x}{2}}\right \} \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{2 x}, x^{2} {\mathrm e}^{2 x}\}] \] Since \(x \,{\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x^{3}\}] \] 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. \[ y_p = A_{1} x^{2} {\mathrm e}^{2 x}+A_{2} {\mathrm e}^{2 x} x^{3} \] 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 \[ 10 A_{1} {\mathrm e}^{2 x}+30 A_{2} {\mathrm e}^{2 x} x +12 A_{2} {\mathrm e}^{2 x} = {\mathrm e}^{2 x} \left (17+30 x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{2}}, A_{2} = 1\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{2} {\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x} x^{3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{2 x}+c_{2} x \,{\mathrm e}^{2 x}+{\mathrm e}^{-\frac {x}{2}} c_{3}\right ) + \left (\frac {x^{2} {\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x} x^{3}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{-\frac {x}{2}} c_{3} +{\mathrm e}^{2 x} \left (c_{2} x +c_{1} \right )+\frac {x^{2} {\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x} x^{3} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\frac {x}{2}} c_{3} +{\mathrm e}^{2 x} \left (c_{2} x +c_{1} \right )+\frac {x^{2} {\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x} x^{3} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-\frac {x}{2}} c_{3} +{\mathrm e}^{2 x} \left (c_{2} x +c_{1} \right )+\frac {x^{2} {\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x} x^{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime \prime }-7 y^{\prime \prime }+4 y^{\prime }+4 y={\mathrm e}^{2 x} \left (17+30 x \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 }=-2 y+\frac {7 y^{\prime \prime }}{2}-2 y^{\prime }+15 x \,{\mathrm e}^{2 x}+\frac {17 \,{\mathrm e}^{2 x}}{2} \\ \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 }-\frac {7 y^{\prime \prime }}{2}+2 y^{\prime }+2 y=\frac {{\mathrm e}^{2 x} \left (17+30 x \right )}{2} \\ \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 )=15 x \,{\mathrm e}^{2 x}+\frac {17 \,{\mathrm e}^{2 x}}{2}+\frac {7 y_{3}\left (x \right )}{2}-2 y_{2}\left (x \right )-2 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 )=15 x \,{\mathrm e}^{2 x}+\frac {17 \,{\mathrm e}^{2 x}}{2}+\frac {7 y_{3}\left (x \right )}{2}-2 y_{2}\left (x \right )-2 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 \\ -2 & -2 & \frac {7}{2} \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 15 x \,{\mathrm e}^{2 x}+\frac {17 \,{\mathrm e}^{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 \\ 15 x \,{\mathrm e}^{2 x}+\frac {17 \,{\mathrm e}^{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 \\ -2 & -2 & \frac {7}{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 [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} 2 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{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 =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 =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} 2 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -2 & \frac {7}{2} \end {array}\right ]-2\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -\frac {1}{8} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} 2 \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{2 x}\cdot \left (x \cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {1}{8} \\ 0 \\ 0 \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} 4 \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{2 x}}{4} & {\mathrm e}^{2 x} \left (\frac {x}{4}-\frac {1}{8}\right ) \\ -2 \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{2 x}}{2} & \frac {x \,{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{2 x} & 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} 4 \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{2 x}}{4} & {\mathrm e}^{2 x} \left (\frac {x}{4}-\frac {1}{8}\right ) \\ -2 \,{\mathrm e}^{-\frac {x}{2}} & \frac {{\mathrm e}^{2 x}}{2} & \frac {x \,{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{2 x} & x \,{\mathrm e}^{2 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 4 & \frac {1}{4} & -\frac {1}{8} \\ -2 & \frac {1}{2} & 0 \\ 1 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \left (1-2 x \right ) {\mathrm e}^{2 x} & -\frac {8 \,{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {8 \,{\mathrm e}^{2 x}}{5}-3 x \,{\mathrm e}^{2 x} & \frac {4 \,{\mathrm e}^{-\frac {x}{2}}}{5}-\frac {4 \,{\mathrm e}^{2 x}}{5}+2 x \,{\mathrm e}^{2 x} \\ -4 x \,{\mathrm e}^{2 x} & \frac {4 \,{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {{\mathrm e}^{2 x}}{5}-6 x \,{\mathrm e}^{2 x} & -\frac {2 \,{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {2 \,{\mathrm e}^{2 x}}{5}+4 x \,{\mathrm e}^{2 x} \\ -8 x \,{\mathrm e}^{2 x} & -\frac {2 \,{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {2 \,{\mathrm e}^{2 x}}{5}-12 x \,{\mathrm e}^{2 x} & \frac {{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {4 \,{\mathrm e}^{2 x}}{5}+8 x \,{\mathrm e}^{2 x} \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 {4 \,{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {\left (50 x^{3}+25 x^{2}-20 x +8\right ) {\mathrm e}^{2 x}}{10} \\ \frac {2 \,{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {\left (50 x^{3}+100 x^{2}+5 x -2\right ) {\mathrm e}^{2 x}}{5} \\ -\frac {{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {\left (100 x^{3}+200 x^{2}+40 x +1\right ) {\mathrm e}^{2 x}}{5} \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 {4 \,{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {\left (50 x^{3}+25 x^{2}-20 x +8\right ) {\mathrm e}^{2 x}}{10} \\ \frac {2 \,{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {\left (50 x^{3}+100 x^{2}+5 x -2\right ) {\mathrm e}^{2 x}}{5} \\ -\frac {{\mathrm e}^{-\frac {x}{2}}}{5}+\frac {\left (100 x^{3}+200 x^{2}+40 x +1\right ) {\mathrm e}^{2 x}}{5} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (160 c_{1} -32\right ) {\mathrm e}^{-\frac {x}{2}}}{40}+5 \left (x^{3}+\frac {x^{2}}{2}+\left (\frac {c_{3}}{20}-\frac {2}{5}\right ) x +\frac {c_{2}}{20}-\frac {c_{3}}{40}+\frac {4}{25}\right ) {\mathrm e}^{2 x} \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: 29
dsolve(2*diff(y(x),x$3)-7*diff(y(x),x$2)+4*diff(y(x),x)+4*y(x)=exp(2*x)*(17+30*x),y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {x}{2}} c_{2} +{\mathrm e}^{2 x} \left (x^{3}+\frac {1}{2} x^{2}+c_{1} +c_{3} x \right ) \]
✓ Solution by Mathematica
Time used: 0.03 (sec). Leaf size: 46
DSolve[2*y'''[x]-7*y''[x]+4*y'[x]+4*y[x]==Exp[2*x]*(17+30*x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{2 x} \left (x^3+\frac {x^2}{2}+\left (-\frac {2}{5}+c_3\right ) x+\frac {4}{25}+c_2\right )+c_1 e^{-x/2} \]