Internal problem ID [15374]
Internal file name [OUTPUT/15375_Wednesday_May_08_2024_03_57_23_PM_34810919/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.3 Nonhomogeneous linear equations with
constant coefficients. Initial value problem. Exercises page 140
Problem number: 606.
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, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime \prime }-y=2 x} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 2] \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 \[ y^{\prime \prime \prime }-y = 0 \] The characteristic equation is \[ \lambda ^{3}-1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -\frac {1}{2}-\frac {i \sqrt {3}}{2}\\ \lambda _3 &= -\frac {1}{2}+\frac {i \sqrt {3}}{2} \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} c_{2} +{\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} \\ y_3 &= {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-y = 2 x \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{x}, {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x}, {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x}\right \} \] 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. \[ y_p = A_{2} x +A_{1} \] 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 \[ -A_{2} x -A_{1} = 2 x \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 0, A_{2} = -2] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -2 x \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} c_{2} +{\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} c_{3}\right ) + \left (-2 x\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}+{\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} c_{2} +{\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} c_{3} -2 x \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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{2} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} {\mathrm e}^{x}+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} c_{2} +\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} c_{3} -2 \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = -2+\frac {i \left (c_{2} -c_{3} \right ) \sqrt {3}}{2}+c_{1} -\frac {c_{2}}{2}-\frac {c_{3}}{2}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{x}+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} c_{2} +\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} c_{3} \end {align*}
substituting \(y^{\prime \prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = \frac {i \left (-c_{2} +c_{3} \right ) \sqrt {3}}{2}+c_{1} -\frac {c_{2}}{2}-\frac {c_{3}}{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 {4}{3}}\\ c_{2}&=-{\frac {2}{3}}\\ c_{3}&=-{\frac {2}{3}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {4 \,{\mathrm e}^{x}}{3}-\frac {2 \,{\mathrm e}^{\frac {\left (i \sqrt {3}-1\right ) x}{2}}}{3}-\frac {2 \,{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}}}{3}-2 x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {4 \,{\mathrm e}^{x}}{3}-\frac {2 \,{\mathrm e}^{\frac {\left (i \sqrt {3}-1\right ) x}{2}}}{3}-\frac {2 \,{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}}}{3}-2 x \\ \end{align*}
Verification of solutions
\[ y = \frac {4 \,{\mathrm e}^{x}}{3}-\frac {2 \,{\mathrm e}^{\frac {\left (i \sqrt {3}-1\right ) x}{2}}}{3}-\frac {2 \,{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}}}{3}-2 x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-y=2 x , y \left (0\right )=0, 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 \}}}}=2\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 )=2 x +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 )=2 x +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 \\ 1 & 0 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 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 \\ 2 x \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \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 [-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}} \\ 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 complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & {\mathrm e}^{-\frac {x}{2}}\cdot \left (\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )\right )\cdot \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ \cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2} \\ -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2} \\ \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ -\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}\right ) & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}\right ) & {\mathrm e}^{-\frac {x}{2}} \left (\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) & -{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\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}{ccc} {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}\right ) & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \left (-\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}\right ) & {\mathrm e}^{-\frac {x}{2}} \left (\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) & -{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & -\frac {1}{2} & -\frac {\sqrt {3}}{2} \\ 1 & -\frac {1}{2} & \frac {\sqrt {3}}{2} \\ 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} \frac {{\mathrm e}^{x}}{3}+\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} & \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{3} & \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{3} \\ \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{3} & \frac {{\mathrm e}^{x}}{3}+\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} & \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{3} \\ \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}+\frac {{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{3} & \frac {{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}-\frac {{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{3} & \frac {{\mathrm e}^{x}}{3}+\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} \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 {2 \,{\mathrm e}^{x}}{3}+\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{3}-2 x -\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} \\ \frac {2 \,{\mathrm e}^{x}}{3}-2+\frac {4 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} \\ -\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \left (-{\mathrm e}^{\frac {3 x}{2}}+\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}+\cos \left (\frac {\sqrt {3}\, x}{2}\right )\right )}{3} \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 {2 \,{\mathrm e}^{x}}{3}+\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{3}-2 x -\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} \\ \frac {2 \,{\mathrm e}^{x}}{3}-2+\frac {4 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3} \\ -\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \left (-{\mathrm e}^{\frac {3 x}{2}}+\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}+\cos \left (\frac {\sqrt {3}\, x}{2}\right )\right )}{3} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} +\frac {4}{3}\right ) \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\left (\left (c_{2} -\frac {4}{3}\right ) \sqrt {3}-c_{3} \right ) {\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\left (6 c_{1} +4\right ) {\mathrm e}^{x}}{6}-2 x \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-\frac {c_{3} \sqrt {3}}{2}-\frac {c_{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} +\frac {4}{3}\right ) \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} +\frac {4}{3}\right ) \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{4}+\frac {\left (\left (c_{2} -\frac {4}{3}\right ) \sqrt {3}-c_{3} \right ) {\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{4}-\frac {\left (\left (c_{2} -\frac {4}{3}\right ) \sqrt {3}-c_{3} \right ) {\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{4}+\frac {\left (6 c_{1} +4\right ) {\mathrm e}^{x}}{6}-2 \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {c_{3} \sqrt {3}}{4}+\frac {c_{2}}{4}-1-\frac {\left (\left (c_{2} -\frac {4}{3}\right ) \sqrt {3}-c_{3} \right ) \sqrt {3}}{4}+c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} +\frac {4}{3}\right ) \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{4}-\frac {{\mathrm e}^{-\frac {x}{2}} \left (c_{3} \sqrt {3}+c_{2} +\frac {4}{3}\right ) \sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{4}+\frac {\left (\left (c_{2} -\frac {4}{3}\right ) \sqrt {3}-c_{3} \right ) {\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{4}+\frac {\left (\left (c_{2} -\frac {4}{3}\right ) \sqrt {3}-c_{3} \right ) {\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{4}+\frac {\left (6 c_{1} +4\right ) {\mathrm e}^{x}}{6} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=\frac {c_{3} \sqrt {3}}{4}+\frac {c_{2}}{4}+1+\frac {\left (\left (c_{2} -\frac {4}{3}\right ) \sqrt {3}-c_{3} \right ) \sqrt {3}}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {2}{3}, c_{2} =\frac {4}{3}, c_{3} =0\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {4 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {\sqrt {3}\, x}{2}\right )}{3}-2 x +\frac {4 \,{\mathrm e}^{x}}{3} \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: 0.032 (sec). Leaf size: 25
dsolve([diff(y(x),x$3)-y(x)=2*x,y(0) = 0, D(y)(0) = 0, (D@@2)(y)(0) = 2],y(x), singsol=all)
\[ y \left (x \right ) = -2 x +\frac {4 \,{\mathrm e}^{x}}{3}-\frac {4 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {x \sqrt {3}}{2}\right )}{3} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 38
DSolve[{y'''[x]-y[x]==2*x,{y[0]==0,y'[0]==0,y''[0]==2}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{3} \left (-6 x+4 e^x-4 e^{-x/2} \cos \left (\frac {\sqrt {3} x}{2}\right )\right ) \]