Internal problem ID [1569]
Internal file name [OUTPUT/1570_Sunday_June_05_2022_02_22_52_AM_59473195/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 72.
ODE order: 4.
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
[[_high_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }+2 y^{\prime \prime }+2 y^{\prime }+y={\mathrm e}^{-x} \left (20-12 x \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = -4, y^{\prime \prime }\left (0\right ) = 7, y^{\prime \prime \prime }\left (0\right ) = -22] \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 \[ y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }+2 y^{\prime \prime }+2 y^{\prime }+y = 0 \] The characteristic equation is \[ \lambda ^{4}+2 \lambda ^{3}+2 \lambda ^{2}+2 \lambda +1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= i\\ \lambda _2 &= -i\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} x \,{\mathrm e}^{-x}+{\mathrm e}^{i x} c_{3} +{\mathrm e}^{-i x} c_{4} \] 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}^{i x} \\ y_4 &= {\mathrm e}^{-i x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }+2 y^{\prime \prime }+2 y^{\prime }+y = {\mathrm e}^{-x} \left (20-12 x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{-x} \left (20-12 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{-x}, {\mathrm e}^{-x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{-x}, {\mathrm e}^{i x}, {\mathrm e}^{-x}, {\mathrm e}^{-i x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{-x}, x^{2} {\mathrm e}^{-x}\}] \] Since \(x \,{\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-x}, {\mathrm e}^{-x} x^{3}\}] \] 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. \[ y_p = A_{1} x^{2} {\mathrm e}^{-x}+A_{2} {\mathrm e}^{-x} x^{3} \] 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 \[ 12 A_{2} {\mathrm e}^{-x} x +4 A_{1} {\mathrm e}^{-x}-12 A_{2} {\mathrm e}^{-x} = {\mathrm e}^{-x} \left (20-12 x \right ) \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 2, A_{2} = -1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = 2 x^{2} {\mathrm e}^{-x}-{\mathrm e}^{-x} x^{3} \] 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}^{i x} c_{3} +{\mathrm e}^{-i x} c_{4}\right ) + \left (2 x^{2} {\mathrm e}^{-x}-{\mathrm e}^{-x} x^{3}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{-i x} c_{4} +{\mathrm e}^{i x} c_{3} +\left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+2 x^{2} {\mathrm e}^{-x}-{\mathrm e}^{-x} x^{3} \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = {\mathrm e}^{-i x} c_{4} +{\mathrm e}^{i x} c_{3} +\left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+2 x^{2} {\mathrm e}^{-x}-{\mathrm e}^{-x} x^{3} \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 \(y = 3\) and \(x = 0\) in the above gives \begin {align*} 3 = c_{4} +c_{3} +c_{1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -i {\mathrm e}^{-i x} c_{4} +i {\mathrm e}^{i x} c_{3} +c_{2} {\mathrm e}^{-x}-\left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+4 x \,{\mathrm e}^{-x}-5 x^{2} {\mathrm e}^{-x}+{\mathrm e}^{-x} x^{3} \end {align*}
substituting \(y^{\prime } = -4\) and \(x = 0\) in the above gives \begin {align*} -4 = c_{3} i-c_{4} i-c_{1} +c_{2}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = -{\mathrm e}^{-i x} c_{4} -{\mathrm e}^{i x} c_{3} -2 c_{2} {\mathrm e}^{-x}+\left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+4 \,{\mathrm e}^{-x}-14 x \,{\mathrm e}^{-x}+8 x^{2} {\mathrm e}^{-x}-{\mathrm e}^{-x} x^{3} \end {align*}
substituting \(y^{\prime \prime } = 7\) and \(x = 0\) in the above gives \begin {align*} 7 = -c_{4} -c_{3} -2 c_{2} +c_{1} +4\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = i {\mathrm e}^{-i x} c_{4} -i {\mathrm e}^{i x} c_{3} +3 c_{2} {\mathrm e}^{-x}-\left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}-18 \,{\mathrm e}^{-x}+30 x \,{\mathrm e}^{-x}-11 x^{2} {\mathrm e}^{-x}+{\mathrm e}^{-x} x^{3} \end {align*}
substituting \(y^{\prime \prime \prime } = -22\) and \(x = 0\) in the above gives \begin {align*} -22 = -c_{3} i+c_{4} i-c_{1} +3 c_{2} -18\tag {4A} \end {align*}
Equations {1A,2A,3A,4A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}\}\). Solving for the constants gives \begin {align*} c_{1}&=2\\ c_{2}&=-1\\ c_{3}&=\frac {1}{2}+\frac {i}{2}\\ c_{4}&=\frac {1}{2}-\frac {i}{2} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -{\mathrm e}^{-x} x^{3}+2 x^{2} {\mathrm e}^{-x}-x \,{\mathrm e}^{-x}+2 \,{\mathrm e}^{-x}+\cos \left (x \right )-\sin \left (x \right ) \end {align*}
Which simplifies to \[ y = \left (-x^{3}+2 x^{2}-x +2\right ) {\mathrm e}^{-x}+\cos \left (x \right )-\sin \left (x \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (-x^{3}+2 x^{2}-x +2\right ) {\mathrm e}^{-x}+\cos \left (x \right )-\sin \left (x \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (-x^{3}+2 x^{2}-x +2\right ) {\mathrm e}^{-x}+\cos \left (x \right )-\sin \left (x \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }+2 y^{\prime \prime }+2 y^{\prime }+y={\mathrm e}^{-x} \left (20-12 x \right ), y \left (0\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-4, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=7, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-22\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=-y-12 x \,{\mathrm e}^{-x}+20 \,{\mathrm e}^{-x}-2 y^{\prime \prime \prime }-2 y^{\prime \prime }-2 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 \prime }+2 y^{\prime \prime \prime }+2 y^{\prime \prime }+2 y^{\prime }+y=-4 \,{\mathrm e}^{-x} \left (3 x -5\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 {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=-12 x \,{\mathrm e}^{-x}+20 \,{\mathrm e}^{-x}-2 y_{4}\left (x \right )-2 y_{3}\left (x \right )-2 y_{2}\left (x \right )-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_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{4}^{\prime }\left (x \right )=-12 x \,{\mathrm e}^{-x}+20 \,{\mathrm e}^{-x}-2 y_{4}\left (x \right )-2 y_{3}\left (x \right )-2 y_{2}\left (x \right )-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 ) \\ y_{4}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & -2 & -2 & -2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ -12 x \,{\mathrm e}^{-x}+20 \,{\mathrm e}^{-x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ -12 x \,{\mathrm e}^{-x}+20 \,{\mathrm e}^{-x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & -2 & -2 & -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 [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [\mathrm {-I}, \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I}, \left [\begin {array}{c} \mathrm {I} \\ -1 \\ \mathrm {-I} \\ 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 \\ 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 \\ 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}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & -2 & -2 & -2 \end {array}\right ]-\left (-1\right )\cdot \left [\begin {array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 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 \\ 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 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I}, \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} x}\cdot \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 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 (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} \mathrm {-I} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ -\cos \left (x \right )+\mathrm {I} \sin \left (x \right ) \\ \mathrm {I} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \cos \left (x \right )-\mathrm {I} \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} -\sin \left (x \right ) \\ -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )=\left [\begin {array}{c} -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \\ -\sin \left (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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\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}{cccc} -{\mathrm e}^{-x} & \left (-x -1\right ) {\mathrm e}^{-x} & -\sin \left (x \right ) & -\cos \left (x \right ) \\ {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & -\cos \left (x \right ) & \sin \left (x \right ) \\ -{\mathrm e}^{-x} & -x \,{\mathrm e}^{-x} & \sin \left (x \right ) & \cos \left (x \right ) \\ {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & \cos \left (x \right ) & -\sin \left (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 \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}{cccc} -{\mathrm e}^{-x} & \left (-x -1\right ) {\mathrm e}^{-x} & -\sin \left (x \right ) & -\cos \left (x \right ) \\ {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & -\cos \left (x \right ) & \sin \left (x \right ) \\ -{\mathrm e}^{-x} & -x \,{\mathrm e}^{-x} & \sin \left (x \right ) & \cos \left (x \right ) \\ {\mathrm e}^{-x} & x \,{\mathrm e}^{-x} & \cos \left (x \right ) & -\sin \left (x \right ) \end {array}\right ]\cdot \left (\left [\begin {array}{cccc} -1 & -1 & 0 & -1 \\ 1 & 0 & -1 & 0 \\ -1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \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}{cccc} \left (x +1\right ) {\mathrm e}^{-x} & x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{-x}}{2}-\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} & x \,{\mathrm e}^{-x}+{\mathrm e}^{-x}-\cos \left (x \right ) & x \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{-x}}{2}-\frac {\cos \left (x \right )}{2}-\frac {\sin \left (x \right )}{2} \\ -x \,{\mathrm e}^{-x} & \frac {{\mathrm e}^{-x}}{2}-x \,{\mathrm e}^{-x}+\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} & -x \,{\mathrm e}^{-x}+\sin \left (x \right ) & \frac {{\mathrm e}^{-x}}{2}-x \,{\mathrm e}^{-x}-\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ x \,{\mathrm e}^{-x} & -\frac {{\mathrm e}^{-x}}{2}+x \,{\mathrm e}^{-x}-\frac {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2} & x \,{\mathrm e}^{-x}+\cos \left (x \right ) & -\frac {{\mathrm e}^{-x}}{2}+x \,{\mathrm e}^{-x}+\frac {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2} \\ -x \,{\mathrm e}^{-x} & \frac {{\mathrm e}^{-x}}{2}-x \,{\mathrm e}^{-x}-\frac {\cos \left (x \right )}{2}-\frac {\sin \left (x \right )}{2} & -x \,{\mathrm e}^{-x}-\sin \left (x \right ) & \frac {{\mathrm e}^{-x}}{2}-x \,{\mathrm e}^{-x}+\frac {\cos \left (x \right )}{2}-\frac {\sin \left (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 )=\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} \left (-2 x^{3}+7 x^{2}+10 x +3\right ) {\mathrm e}^{-x}-3 \cos \left (x \right )-7 \sin \left (x \right ) \\ \left (2 x^{3}-13 x^{2}+4 x +7\right ) {\mathrm e}^{-x}-7 \cos \left (x \right )+3 \sin \left (x \right ) \\ \left (-2 x^{3}+13 x^{2}-10 x -3\right ) {\mathrm e}^{-x}+3 \cos \left (x \right )+7 \sin \left (x \right ) \\ \left (2 x^{3}-13 x^{2}+16 x -7\right ) {\mathrm e}^{-x}+7 \cos \left (x \right )-3 \sin \left (x \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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right )+\left [\begin {array}{c} \left (-2 x^{3}+7 x^{2}+10 x +3\right ) {\mathrm e}^{-x}-3 \cos \left (x \right )-7 \sin \left (x \right ) \\ \left (2 x^{3}-13 x^{2}+4 x +7\right ) {\mathrm e}^{-x}-7 \cos \left (x \right )+3 \sin \left (x \right ) \\ \left (-2 x^{3}+13 x^{2}-10 x -3\right ) {\mathrm e}^{-x}+3 \cos \left (x \right )+7 \sin \left (x \right ) \\ \left (2 x^{3}-13 x^{2}+16 x -7\right ) {\mathrm e}^{-x}+7 \cos \left (x \right )-3 \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\left (-2 x^{3}+7 x^{2}+\left (-c_{2} +10\right ) x -c_{1} -c_{2} +3\right ) {\mathrm e}^{-x}+\left (-c_{4} -3\right ) \cos \left (x \right )-\sin \left (x \right ) \left (c_{3} +7\right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=-c_{1} -c_{2} -c_{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (-6 x^{2}-c_{2} +14 x +10\right ) {\mathrm e}^{-x}-\left (-2 x^{3}+7 x^{2}+\left (-c_{2} +10\right ) x -c_{1} -c_{2} +3\right ) {\mathrm e}^{-x}-\left (-c_{4} -3\right ) \sin \left (x \right )-\left (c_{3} +7\right ) \cos \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-4 \\ {} & {} & -4=c_{1} -c_{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\left (-12 x +14\right ) {\mathrm e}^{-x}-2 \left (-6 x^{2}-c_{2} +14 x +10\right ) {\mathrm e}^{-x}+\left (-2 x^{3}+7 x^{2}+\left (-c_{2} +10\right ) x -c_{1} -c_{2} +3\right ) {\mathrm e}^{-x}-\left (-c_{4} -3\right ) \cos \left (x \right )+\sin \left (x \right ) \left (c_{3} +7\right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=7 \\ {} & {} & 7=c_{2} -c_{1} +c_{4} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-12 \,{\mathrm e}^{-x}-3 \left (-12 x +14\right ) {\mathrm e}^{-x}+3 \left (-6 x^{2}-c_{2} +14 x +10\right ) {\mathrm e}^{-x}-\left (-2 x^{3}+7 x^{2}+\left (-c_{2} +10\right ) x -c_{1} -c_{2} +3\right ) {\mathrm e}^{-x}+\left (-c_{4} -3\right ) \sin \left (x \right )+\left (c_{3} +7\right ) \cos \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-22 \\ {} & {} & -22=-20-2 c_{2} +c_{1} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-5, c_{2} =-2, c_{3} =-1, c_{4} =4\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (-2 x^{3}+7 x^{2}+12 x +10\right ) {\mathrm e}^{-x}-7 \cos \left (x \right )-6 \sin \left (x \right ) \end {array} \]
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 4; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 31
dsolve([diff(y(x),x$4)+2*diff(y(x),x$3)+2*diff(y(x),x$2)+2*diff(y(x),x)+1*y(x)=exp(-x)*(20-12*x),y(0) = 3, D(y)(0) = -4, (D@@2)(y)(0) = 7, (D@@3)(y)(0) = -22],y(x), singsol=all)
\[ y \left (x \right ) = \left (-x^{3}+2 x^{2}-x +2\right ) {\mathrm e}^{-x}+\cos \left (x \right )-\sin \left (x \right ) \]
✓ Solution by Mathematica
Time used: 0.085 (sec). Leaf size: 39
DSolve[{y''''[x]+2*y'''[x]+2*y''[x]+2*y'[x]+1*y[x]==Exp[-x]*(20-12*x),{y[0]==3,y'[0]==-4,y''[0]==7,y'''[0]==-22}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x} \left (-x^3+2 x^2-x-e^x \sin (x)+e^x \cos (x)+2\right ) \]