14.31 problem 33

Internal problem ID [2232]
Internal file name [OUTPUT/2232_Monday_February_26_2024_09_18_35_AM_92822207/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 23, page 106
Problem number: 33.
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, _missing_y]]

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

Therefore the homogeneous solution is \[ y_h(x)=c_{1} +{\mathrm e}^{-i \sqrt {2}\, x} c_{2} +{\mathrm e}^{i \sqrt {2}\, x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{-i \sqrt {2}\, x} \\ y_3 &= {\mathrm e}^{i \sqrt {2}\, x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+2 y^{\prime } = \sin \left (x \right ) x^{2} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ \sin \left (x \right ) x^{2} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \cos \left (x \right ), x^{2} \cos \left (x \right ), \sin \left (x \right ) x, \sin \left (x \right ) x^{2}, \cos \left (x \right ), \sin \left (x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{1, {\mathrm e}^{i \sqrt {2}\, x}, {\mathrm e}^{-i \sqrt {2}\, x}\right \} \] 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. \[ y_p = A_{1} x \cos \left (x \right )+A_{2} x^{2} \cos \left (x \right )+A_{3} \sin \left (x \right ) x +A_{4} \sin \left (x \right ) x^{2}+A_{5} \cos \left (x \right )+A_{6} \sin \left (x \right ) \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) 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 \[ -A_{1} \cos \left (x \right )-A_{1} x \sin \left (x \right )-6 A_{2} \sin \left (x \right )-2 A_{2} x \cos \left (x \right )-A_{2} x^{2} \sin \left (x \right )+A_{3} \cos \left (x \right ) x -A_{3} \sin \left (x \right )+A_{4} \cos \left (x \right ) x^{2}-2 A_{4} \sin \left (x \right ) x +6 A_{4} \cos \left (x \right )-A_{5} \sin \left (x \right )+A_{6} \cos \left (x \right ) = \sin \left (x \right ) x^{2} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 0, A_{2} = -1, A_{3} = -2, A_{4} = 0, A_{5} = 8, A_{6} = 0] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -x^{2} \cos \left (x \right )-2 \sin \left (x \right ) x +8 \cos \left (x \right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +{\mathrm e}^{-i \sqrt {2}\, x} c_{2} +{\mathrm e}^{i \sqrt {2}\, x} c_{3}\right ) + \left (-x^{2} \cos \left (x \right )-2 \sin \left (x \right ) x +8 \cos \left (x \right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +{\mathrm e}^{-i \sqrt {2}\, x} c_{2} +{\mathrm e}^{i \sqrt {2}\, x} c_{3} -x^{2} \cos \left (x \right )-2 \sin \left (x \right ) x +8 \cos \left (x \right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} +{\mathrm e}^{-i \sqrt {2}\, x} c_{2} +{\mathrm e}^{i \sqrt {2}\, x} c_{3} -x^{2} \cos \left (x \right )-2 \sin \left (x \right ) x +8 \cos \left (x \right ) \] Verified OK.

14.31.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+2 y^{\prime }=\sin \left (x \right ) x^{2} \\ \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 )=\sin \left (x \right ) x^{2}-2 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 )=\sin \left (x \right ) x^{2}-2 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 & -2 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ \sin \left (x \right ) x^{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ \sin \left (x \right ) x^{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & 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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ -\frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ]\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}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} \sqrt {2}\, x}\cdot \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {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 \\ {} & {} & \left (\cos \left (\sqrt {2}\, x \right )-\mathrm {I} \sin \left (\sqrt {2}\, x \right )\right )\cdot \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\cos \left (\sqrt {2}\, x \right )}{2}+\frac {\mathrm {I} \sin \left (\sqrt {2}\, x \right )}{2} \\ \frac {\mathrm {I}}{2} \left (\cos \left (\sqrt {2}\, x \right )-\mathrm {I} \sin \left (\sqrt {2}\, x \right )\right ) \sqrt {2} \\ \cos \left (\sqrt {2}\, x \right )-\mathrm {I} \sin \left (\sqrt {2}\, x \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 )=\left [\begin {array}{c} -\frac {\cos \left (\sqrt {2}\, x \right )}{2} \\ \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} \\ \cos \left (\sqrt {2}\, x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} \frac {\sin \left (\sqrt {2}\, x \right )}{2} \\ \frac {\sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{2} \\ -\sin \left (\sqrt {2}\, 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}+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} 1 & -\frac {\cos \left (\sqrt {2}\, x \right )}{2} & \frac {\sin \left (\sqrt {2}\, x \right )}{2} \\ 0 & \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} & \frac {\sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{2} \\ 0 & \cos \left (\sqrt {2}\, x \right ) & -\sin \left (\sqrt {2}\, 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}{ccc} 1 & -\frac {\cos \left (\sqrt {2}\, x \right )}{2} & \frac {\sin \left (\sqrt {2}\, x \right )}{2} \\ 0 & \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} & \frac {\sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{2} \\ 0 & \cos \left (\sqrt {2}\, x \right ) & -\sin \left (\sqrt {2}\, x \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & -\frac {1}{2} & 0 \\ 0 & 0 & \frac {\sqrt {2}}{2} \\ 0 & 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} 1 & \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} & \frac {1}{2}-\frac {\cos \left (\sqrt {2}\, x \right )}{2} \\ 0 & \cos \left (\sqrt {2}\, x \right ) & \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} \\ 0 & -\sqrt {2}\, \sin \left (\sqrt {2}\, x \right ) & \cos \left (\sqrt {2}\, x \right ) \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} -1-x^{2} \cos \left (x \right )-2 \sin \left (x \right ) x -7 \cos \left (\sqrt {2}\, x \right )+8 \cos \left (x \right ) \\ \frac {\sqrt {2}\, \left (\left (\sin \left (x \right ) x^{2}-4 x \cos \left (x \right )-10 \sin \left (x \right )\right ) \sqrt {2}+14 \sin \left (\sqrt {2}\, x \right )\right )}{2 \left (1+\sqrt {2}\right )^{3} \left (\sqrt {2}-1\right )^{3}} \\ x^{2} \cos \left (x \right )+6 \sin \left (x \right ) x +14 \cos \left (\sqrt {2}\, x \right )-14 \cos \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}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+\left [\begin {array}{c} -1-x^{2} \cos \left (x \right )-2 \sin \left (x \right ) x -7 \cos \left (\sqrt {2}\, x \right )+8 \cos \left (x \right ) \\ \frac {\sqrt {2}\, \left (\left (\sin \left (x \right ) x^{2}-4 x \cos \left (x \right )-10 \sin \left (x \right )\right ) \sqrt {2}+14 \sin \left (\sqrt {2}\, x \right )\right )}{2 \left (1+\sqrt {2}\right )^{3} \left (\sqrt {2}-1\right )^{3}} \\ x^{2} \cos \left (x \right )+6 \sin \left (x \right ) x +14 \cos \left (\sqrt {2}\, x \right )-14 \cos \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-1-x^{2} \cos \left (x \right )-2 \sin \left (x \right ) x -7 \cos \left (\sqrt {2}\, x \right )+8 \cos \left (x \right )+\frac {c_{3} \sin \left (\sqrt {2}\, x \right )}{2}-\frac {c_{2} \cos \left (\sqrt {2}\, x \right )}{2}+c_{1} \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) = sin(_a)*_a^2-2*_b(_a), _b(_a)`   *** Sublevel 2 *** 
   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`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 46

dsolve(diff(y(x),x$3)+2*diff(y(x),x)=x^2*sin(x),y(x), singsol=all)
 

\[ y \left (x \right ) = -\frac {c_{2} \sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{2}+\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right ) c_{1}}{2}-\cos \left (x \right ) x^{2}+8 \cos \left (x \right )-2 x \sin \left (x \right )+c_{3} \]

✓ Solution by Mathematica

Time used: 0.164 (sec). Leaf size: 55

DSolve[y'''[x]+2*y'[x]==x^2*Sin[x],y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to -\left (x^2-8\right ) \cos (x)-2 x \sin (x)-\frac {c_2 \cos \left (\sqrt {2} x\right )}{\sqrt {2}}+\frac {c_1 \sin \left (\sqrt {2} x\right )}{\sqrt {2}}+c_3 \]