problem Ex 6

Internal problem ID [11267]
Internal ﬁle name [OUTPUT/10253_Wednesday_December_21_2022_03_46_51_PM_39861667/index.tex]

Book: An elementary treatise on diﬀerential equations by Abraham Cohen. DC heath publishers. 1906
Section: Chapter VII, Linear diﬀerential equations with constant coeﬃcients. Article 50. Method of undetermined coeﬃcients. Page 107
Problem number: Ex 6.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime \prime }-3 y^{\prime }=3 x^{2}+\sin \left (x \right )} \] 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 \[ y^{\prime \prime \prime }-2 y^{\prime \prime }-3 y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{3}-2 \lambda ^{2}-3 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 3\\ \lambda _3 &= -1 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} +{\mathrm e}^{3 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= 1 \\ y_3 &= {\mathrm e}^{3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-2 y^{\prime \prime }-3 y^{\prime } = 3 x^{2}+\sin \left (x \right ) \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ 3 x^{2}+\sin \left (x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (x \right ), \sin \left (x \right )\}, \{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{-x}, {\mathrm e}^{3 x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{\cos \left (x \right ), \sin \left (x \right )\}, \{x, x^{2}, x^{3}\}] \] 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. \[ y_p = A_{1} \cos \left (x \right )+A_{2} \sin \left (x \right )+A_{3} x +A_{4} x^{2}+A_{5} x^{3} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coeﬃcients. Substituting the trial solution into the ODE and simplifying gives \[ 4 A_{1} \sin \left (x \right )-4 A_{2} \cos \left (x \right )+6 A_{5}+2 A_{1} \cos \left (x \right )+2 A_{2} \sin \left (x \right )-4 A_{4}-12 A_{5} x -3 A_{3}-6 A_{4} x -9 A_{5} x^{2} = 3 x^{2}+\sin \left (x \right ) \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = {\frac {1}{5}}, A_{2} = {\frac {1}{10}}, A_{3} = -{\frac {14}{9}}, A_{4} = {\frac {2}{3}}, A_{5} = -{\frac {1}{3}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {\cos \left (x \right )}{5}+\frac {\sin \left (x \right )}{10}-\frac {14 x}{9}+\frac {2 x^{2}}{3}-\frac {x^{3}}{3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} +{\mathrm e}^{3 x} c_{3}\right ) + \left (\frac {\cos \left (x \right )}{5}+\frac {\sin \left (x \right )}{10}-\frac {14 x}{9}+\frac {2 x^{2}}{3}-\frac {x^{3}}{3}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+c_{2} +{\mathrm e}^{3 x} c_{3} +\frac {\cos \left (x \right )}{5}+\frac {\sin \left (x \right )}{10}-\frac {14 x}{9}+\frac {2 x^{2}}{3}-\frac {x^{3}}{3} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{-x}+c_{2} +{\mathrm e}^{3 x} c_{3} +\frac {\cos \left (x \right )}{5}+\frac {\sin \left (x \right )}{10}-\frac {14 x}{9}+\frac {2 x^{2}}{3}-\frac {x^{3}}{3} \] Veriﬁed OK.

27.6.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-2 y^{\prime \prime }-3 y^{\prime }=3 x^{2}+\sin \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=3 x^{2}+\sin \left (x \right )+2 y_{3}\left (x \right )+3 y_{2}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=3 x^{2}+\sin \left (x \right )+2 y_{3}\left (x \right )+3 y_{2}\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 \\ 0 & 3 & 2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 3 x^{2}+\sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 3 x^{2}+\sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 3 & 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 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-x} & 1 & \frac {{\mathrm e}^{3 x}}{9} \\ -{\mathrm e}^{-x} & 0 & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-x} & 0 & {\mathrm e}^{3 x} \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\phi \left (x \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-x} & 1 & \frac {{\mathrm e}^{3 x}}{9} \\ -{\mathrm e}^{-x} & 0 & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-x} & 0 & {\mathrm e}^{3 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & 1 & \frac {1}{9} \\ -1 & 0 & \frac {1}{3} \\ 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} 1 & -\frac {3 \,{\mathrm e}^{-x}}{4}+\frac {2}{3}+\frac {{\mathrm e}^{3 x}}{12} & \frac {{\mathrm e}^{-x}}{4}-\frac {1}{3}+\frac {{\mathrm e}^{3 x}}{12} \\ 0 & \frac {3 \,{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{3 x}}{4} & -\frac {{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{3 x}}{4} \\ 0 & -\frac {3 \,{\mathrm e}^{-x}}{4}+\frac {3 \,{\mathrm e}^{3 x}}{4} & \frac {{\mathrm e}^{-x}}{4}+\frac {3 \,{\mathrm e}^{3 x}}{4} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left [\begin {array}{c} -\frac {x^{3}}{3}+\frac {2 x^{2}}{3}-\frac {14 x}{9}+\frac {\sin \left (x \right )}{10}+\frac {\cos \left (x \right )}{5}+\frac {29 \,{\mathrm e}^{3 x}}{1080}+\frac {31}{27}-\frac {11 \,{\mathrm e}^{-x}}{8} \\ \frac {29 \,{\mathrm e}^{3 x}}{360}-x^{2}+\frac {4 x}{3}-\frac {\sin \left (x \right )}{5}+\frac {\cos \left (x \right )}{10}-\frac {14}{9}+\frac {11 \,{\mathrm e}^{-x}}{8} \\ \frac {29 \,{\mathrm e}^{3 x}}{120}-2 x -\frac {\sin \left (x \right )}{10}-\frac {\cos \left (x \right )}{5}+\frac {4}{3}-\frac {11 \,{\mathrm e}^{-x}}{8} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} -\frac {x^{3}}{3}+\frac {2 x^{2}}{3}-\frac {14 x}{9}+\frac {\sin \left (x \right )}{10}+\frac {\cos \left (x \right )}{5}+\frac {29 \,{\mathrm e}^{3 x}}{1080}+\frac {31}{27}-\frac {11 \,{\mathrm e}^{-x}}{8} \\ \frac {29 \,{\mathrm e}^{3 x}}{360}-x^{2}+\frac {4 x}{3}-\frac {\sin \left (x \right )}{5}+\frac {\cos \left (x \right )}{10}-\frac {14}{9}+\frac {11 \,{\mathrm e}^{-x}}{8} \\ \frac {29 \,{\mathrm e}^{3 x}}{120}-2 x -\frac {\sin \left (x \right )}{10}-\frac {\cos \left (x \right )}{5}+\frac {4}{3}-\frac {11 \,{\mathrm e}^{-x}}{8} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (1080 c_{1} -1485\right ) {\mathrm e}^{-x}}{1080}+\frac {\left (120 c_{3} +29\right ) {\mathrm e}^{3 x}}{1080}-\frac {x^{3}}{3}+\frac {2 x^{2}}{3}-\frac {14 x}{9}+c_{2} +\frac {\cos \left (x \right )}{5}+\frac {\sin \left (x \right )}{10}+\frac {31}{27} \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] 
-> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = 3*_a^2+2*(diff(_b(_a), _a))+3*_b(_a)+sin(_a), _b(_a)`   *** Sublevel 
   Methods for second order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying high order exact linear fully integrable 
   trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
   trying a double symmetry of the form [xi=0, eta=F(x)] 
   -> Try solving first the homogeneous part of the ODE 
      checking if the LODE has constant coefficients 
      <- constant coefficients successful 
   <- solving first the homogeneous part of the ODE successful 
<- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`

\[ y \left (x \right ) = -\frac {x^{3}}{3}+\frac {2 x^{2}}{3}-c_{1} {\mathrm e}^{-x}+\frac {c_{2} {\mathrm e}^{3 x}}{3}+\frac {\sin \left (x \right )}{10}+\frac {\cos \left (x \right )}{5}-\frac {14 x}{9}+c_{3} \]

\[ y(x)\to \frac {1}{9} \left (-3 x^3+6 x^2-14 x-9 c_1 e^{-x}+3 c_2 e^{3 x}+9 c_3\right )+\frac {\sin (x)}{10}+\frac {\cos (x)}{5} \]

27.6 problem Ex 6

27.6.1 Maple step by step solution