Internal problem ID [11813]
Internal file name [OUTPUT/11823_Thursday_April_11_2024_08_50_03_PM_30210734/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: 40.
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]]
\[ \boxed {y^{\prime \prime \prime }-6 y^{\prime \prime }+9 y^{\prime }-4 y=8 x^{2}+3-6 \,{\mathrm e}^{2 x}} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 7, y^{\prime \prime }\left (0\right ) = 0] \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 \[ y^{\prime \prime \prime }-6 y^{\prime \prime }+9 y^{\prime }-4 y = 0 \] The characteristic equation is \[ \lambda ^{3}-6 \lambda ^{2}+9 \lambda -4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 4\\ \lambda _2 &= 1\\ \lambda _3 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+{\mathrm e}^{4 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= x \,{\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{4 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-6 y^{\prime \prime }+9 y^{\prime }-4 y = 8 x^{2}+3-6 \,{\mathrm e}^{2 x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 8 x^{2}+3-6 \,{\mathrm e}^{2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{2 x}\}, \{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{x}, {\mathrm e}^{x}, {\mathrm e}^{4 x}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{2 x}+A_{2}+A_{3} x +A_{4} x^{2} \] 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 \[ -2 A_{1} {\mathrm e}^{2 x}-12 A_{4}+9 A_{3}+18 A_{4} x -4 A_{2}-4 A_{3} x -4 A_{4} x^{2} = 8 x^{2}+3-6 \,{\mathrm e}^{2 x} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 3, A_{2} = -15, A_{3} = -9, A_{4} = -2] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = 3 \,{\mathrm e}^{2 x}-15-9 x -2 x^{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+{\mathrm e}^{4 x} c_{3}\right ) + \left (3 \,{\mathrm e}^{2 x}-15-9 x -2 x^{2}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{4 x} c_{3} +{\mathrm e}^{x} \left (c_{2} x +c_{1} \right )+3 \,{\mathrm e}^{2 x}-15-9 x -2 x^{2} \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = {\mathrm e}^{4 x} c_{3} +{\mathrm e}^{x} \left (c_{2} x +c_{1} \right )+3 \,{\mathrm e}^{2 x}-15-9 x -2 x^{2} \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 \(y = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = -12+c_{3} +c_{1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 4 \,{\mathrm e}^{4 x} c_{3} +{\mathrm e}^{x} \left (c_{2} x +c_{1} \right )+c_{2} {\mathrm e}^{x}+6 \,{\mathrm e}^{2 x}-9-4 x \end {align*}
substituting \(y^{\prime } = 7\) and \(x = 0\) in the above gives \begin {align*} 7 = -3+4 c_{3} +c_{1} +c_{2}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 16 \,{\mathrm e}^{4 x} c_{3} +{\mathrm e}^{x} \left (c_{2} x +c_{1} \right )+2 c_{2} {\mathrm e}^{x}+12 \,{\mathrm e}^{2 x}-4 \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = 8+16 c_{3} +c_{1} +2 c_{2}\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 {44}{3}}\\ c_{2}&=2\\ c_{3}&=-{\frac {5}{3}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -15-\frac {5 \,{\mathrm e}^{4 x}}{3}+2 x \,{\mathrm e}^{x}+\frac {44 \,{\mathrm e}^{x}}{3}+3 \,{\mathrm e}^{2 x}-9 x -2 x^{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -15-\frac {5 \,{\mathrm e}^{4 x}}{3}+2 x \,{\mathrm e}^{x}+\frac {44 \,{\mathrm e}^{x}}{3}+3 \,{\mathrm e}^{2 x}-9 x -2 x^{2} \\ \end{align*}
Verification of solutions
\[ y = -15-\frac {5 \,{\mathrm e}^{4 x}}{3}+2 x \,{\mathrm e}^{x}+\frac {44 \,{\mathrm e}^{x}}{3}+3 \,{\mathrm e}^{2 x}-9 x -2 x^{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-6 y^{\prime \prime }+9 y^{\prime }-4 y=8 x^{2}+3-6 \,{\mathrm e}^{2 x}, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=7, 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 )=8 x^{2}+6 y_{3}\left (x \right )-9 y_{2}\left (x \right )+4 y_{1}\left (x \right )-6 \,{\mathrm e}^{2 x}+3 \\ & {} & \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 )=8 x^{2}+6 y_{3}\left (x \right )-9 y_{2}\left (x \right )+4 y_{1}\left (x \right )-6 \,{\mathrm e}^{2 x}+3\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 & -9 & 6 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 8 x^{2}+3-6 \,{\mathrm e}^{2 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 8 x^{2}+3-6 \,{\mathrm e}^{2 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & -9 & 6 \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 [1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [4, \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} 1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 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 =1\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\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 =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\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}}_{2}\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} 1 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & -9 & 6 \end {array}\right ]-1\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} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} 1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{x}\cdot \left (x \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [4, \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{4 x}\cdot \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+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} & \left (x -1\right ) {\mathrm e}^{x} & \frac {{\mathrm e}^{4 x}}{16} \\ {\mathrm e}^{x} & x \,{\mathrm e}^{x} & \frac {{\mathrm e}^{4 x}}{4} \\ {\mathrm e}^{x} & x \,{\mathrm e}^{x} & {\mathrm e}^{4 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} & \left (x -1\right ) {\mathrm e}^{x} & \frac {{\mathrm e}^{4 x}}{16} \\ {\mathrm e}^{x} & x \,{\mathrm e}^{x} & \frac {{\mathrm e}^{4 x}}{4} \\ {\mathrm e}^{x} & x \,{\mathrm e}^{x} & {\mathrm e}^{4 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & -1 & \frac {1}{16} \\ 1 & 0 & \frac {1}{4} \\ 1 & 0 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} -\left (x -1\right ) {\mathrm e}^{x} & -\frac {{\mathrm e}^{4 x}}{12}+\frac {\left (15 x +1\right ) {\mathrm e}^{x}}{12} & \frac {{\mathrm e}^{4 x}}{12}+\frac {\left (-3 x -1\right ) {\mathrm e}^{x}}{12} \\ -x \,{\mathrm e}^{x} & \frac {4 \,{\mathrm e}^{x}}{3}+\frac {5 x \,{\mathrm e}^{x}}{4}-\frac {{\mathrm e}^{4 x}}{3} & \frac {{\mathrm e}^{4 x}}{3}+\frac {\left (-3 x -4\right ) {\mathrm e}^{x}}{12} \\ -x \,{\mathrm e}^{x} & \frac {4 \,{\mathrm e}^{x}}{3}+\frac {5 x \,{\mathrm e}^{x}}{4}-\frac {4 \,{\mathrm e}^{4 x}}{3} & \frac {4 \,{\mathrm e}^{4 x}}{3}+\frac {\left (-3 x -4\right ) {\mathrm e}^{x}}{12} \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 {45}{4}+\frac {9 \,{\mathrm e}^{2 x}}{4}-\frac {{\mathrm e}^{4 x}}{6}+\frac {5 \left (22-15 x \right ) {\mathrm e}^{x}}{12}-\frac {3 x^{2}}{2}-\frac {27 x}{4} \\ -\frac {27}{4}+\frac {9 \,{\mathrm e}^{2 x}}{2}-\frac {2 \,{\mathrm e}^{4 x}}{3}+\frac {5 \left (7-15 x \right ) {\mathrm e}^{x}}{12}-3 x \\ -\frac {31}{4}+\frac {15 \,{\mathrm e}^{2 x}}{2}-\frac {8 \,{\mathrm e}^{4 x}}{3}+\frac {5 \left (7-15 x \right ) {\mathrm e}^{x}}{12}-2 x^{2}-4 x \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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} -\frac {45}{4}+\frac {9 \,{\mathrm e}^{2 x}}{4}-\frac {{\mathrm e}^{4 x}}{6}+\frac {5 \left (22-15 x \right ) {\mathrm e}^{x}}{12}-\frac {3 x^{2}}{2}-\frac {27 x}{4} \\ -\frac {27}{4}+\frac {9 \,{\mathrm e}^{2 x}}{2}-\frac {2 \,{\mathrm e}^{4 x}}{3}+\frac {5 \left (7-15 x \right ) {\mathrm e}^{x}}{12}-3 x \\ -\frac {31}{4}+\frac {15 \,{\mathrm e}^{2 x}}{2}-\frac {8 \,{\mathrm e}^{4 x}}{3}+\frac {5 \left (7-15 x \right ) {\mathrm e}^{x}}{12}-2 x^{2}-4 x \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (3 c_{3} -8\right ) {\mathrm e}^{4 x}}{48}+\frac {9 \,{\mathrm e}^{2 x}}{4}+\frac {\left (\left (48 c_{2} -300\right ) x +48 c_{1} -48 c_{2} +440\right ) {\mathrm e}^{x}}{48}-\frac {3 x^{2}}{2}-\frac {27 x}{4}-\frac {45}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=\frac {c_{3}}{16}+c_{1} -c_{2} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (3 c_{3} -8\right ) {\mathrm e}^{4 x}}{12}+\frac {9 \,{\mathrm e}^{2 x}}{2}+\frac {\left (48 c_{2} -300\right ) {\mathrm e}^{x}}{48}+\frac {\left (\left (48 c_{2} -300\right ) x +48 c_{1} -48 c_{2} +440\right ) {\mathrm e}^{x}}{48}-3 x -\frac {27}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=7 \\ {} & {} & 7=\frac {c_{3}}{4}+c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (3 c_{3} -8\right ) {\mathrm e}^{4 x}}{3}+9 \,{\mathrm e}^{2 x}+\frac {\left (48 c_{2} -300\right ) {\mathrm e}^{x}}{24}+\frac {\left (\left (48 c_{2} -300\right ) x +48 c_{1} -48 c_{2} +440\right ) {\mathrm e}^{x}}{48}-3 \\ \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 {115}{9}, c_{2} =\frac {31}{3}, c_{3} =-\frac {208}{9}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {9 \,{\mathrm e}^{2 x}}{4}-\frac {29 \,{\mathrm e}^{4 x}}{18}+\frac {\left (147 x +418\right ) {\mathrm e}^{x}}{36}-\frac {3 x^{2}}{2}-\frac {27 x}{4}-\frac {45}{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: 0.015 (sec). Leaf size: 35
dsolve([diff(y(x),x$3)-6*diff(y(x),x$2)+9*diff(y(x),x)-4*y(x)=8*x^2+3-6*exp(2*x),y(0) = 1, D(y)(0) = 7, (D@@2)(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = -2 x^{2}-9 x +3 \,{\mathrm e}^{2 x}-15+\frac {44 \,{\mathrm e}^{x}}{3}-\frac {5 \,{\mathrm e}^{4 x}}{3}+2 \,{\mathrm e}^{x} x \]
✓ Solution by Mathematica
Time used: 0.21 (sec). Leaf size: 42
DSolve[{y'''[x]-6*y''[x]+9*y'[x]-4*y[x]==8*x^2+3-6*Exp[2*x],{y[0]==1,y'[0]==7,y''[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -2 x^2-9 x+3 e^{2 x}-\frac {5 e^{4 x}}{3}+e^x \left (2 x+\frac {44}{3}\right )-15 \]