10.9 problem 9

Internal problem ID [12767]
Internal file name [OUTPUT/11420_Friday_November_03_2023_06_32_47_AM_78862228/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 4. N-th Order Linear Differential Equations. Exercises 4.3, page 210
Problem number: 9.
ODE order: 5.
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, _missing_x]]

\[ \boxed {y^{\left (5\right )}-y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }+35 y^{\prime \prime }+16 y^{\prime }-52 y=0} \] The characteristic equation is \[ \lambda ^{5}-\lambda ^{4}+\lambda ^{3}+35 \lambda ^{2}+16 \lambda -52 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 2-3 i\\ \lambda _3 &= 2+3 i\\ \lambda _4 &= -2\\ \lambda _5 &= -2 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+x \,{\mathrm e}^{-2 x} c_{2} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{\left (2-3 i\right ) x} c_{4} +{\mathrm e}^{\left (2+3 i\right ) x} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-2 x}\\ y_2 &= {\mathrm e}^{-2 x} x\\ y_3 &= {\mathrm e}^{x}\\ y_4 &= {\mathrm e}^{\left (2-3 i\right ) x}\\ y_5 &= {\mathrm e}^{\left (2+3 i\right ) x} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-2 x}+x \,{\mathrm e}^{-2 x} c_{2} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{\left (2-3 i\right ) x} c_{4} +{\mathrm e}^{\left (2+3 i\right ) x} c_{5} \\ \end{align*}

Verification of solutions

\[ y = c_{1} {\mathrm e}^{-2 x}+x \,{\mathrm e}^{-2 x} c_{2} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{\left (2-3 i\right ) x} c_{4} +{\mathrm e}^{\left (2+3 i\right ) x} c_{5} \] Verified OK.

10.9.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\left (5\right )}-y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }+35 y^{\prime \prime }+16 y^{\prime }-52 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 5 \\ {} & {} & y^{\left (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 {Define new variable}\hspace {3pt} y_{5}\left (x \right ) \\ {} & {} & y_{5}\left (x \right )=y^{\prime \prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{5}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{5}^{\prime }\left (x \right )=y_{5}\left (x \right )-y_{4}\left (x \right )-35 y_{3}\left (x \right )-16 y_{2}\left (x \right )+52 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_{5}\left (x \right )=y_{4}^{\prime }\left (x \right ), y_{5}^{\prime }\left (x \right )=y_{5}\left (x \right )-y_{4}\left (x \right )-35 y_{3}\left (x \right )-16 y_{2}\left (x \right )+52 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 ) \\ y_{5}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 52 & -16 & -35 & -1 & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 52 & -16 & -35 & -1 & 1 \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 ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-2, \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-2, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [2-3 \,\mathrm {I}, \left [\begin {array}{c} -\frac {119}{28561}-\frac {120 \,\mathrm {I}}{28561} \\ -\frac {46}{2197}+\frac {9 \,\mathrm {I}}{2197} \\ -\frac {5}{169}+\frac {12 \,\mathrm {I}}{169} \\ \frac {2}{13}+\frac {3 \,\mathrm {I}}{13} \\ 1 \end {array}\right ]\right ], \left [2+3 \,\mathrm {I}, \left [\begin {array}{c} -\frac {119}{28561}+\frac {120 \,\mathrm {I}}{28561} \\ -\frac {46}{2197}-\frac {9 \,\mathrm {I}}{2197} \\ -\frac {5}{169}-\frac {12 \,\mathrm {I}}{169} \\ \frac {2}{13}-\frac {3 \,\mathrm {I}}{13} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -2 \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 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 =-2\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 =-2 \\ \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} -2 \\ {} & {} & \left (\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 52 & -16 & -35 & -1 & 1 \end {array}\right ]-\left (-2\right )\cdot \left [\begin {array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{32} \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -2 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-2 x}\cdot \left (x \cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} \frac {1}{32} \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [2-3 \,\mathrm {I}, \left [\begin {array}{c} -\frac {119}{28561}-\frac {120 \,\mathrm {I}}{28561} \\ -\frac {46}{2197}+\frac {9 \,\mathrm {I}}{2197} \\ -\frac {5}{169}+\frac {12 \,\mathrm {I}}{169} \\ \frac {2}{13}+\frac {3 \,\mathrm {I}}{13} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (2-3 \,\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} -\frac {119}{28561}-\frac {120 \,\mathrm {I}}{28561} \\ -\frac {46}{2197}+\frac {9 \,\mathrm {I}}{2197} \\ -\frac {5}{169}+\frac {12 \,\mathrm {I}}{169} \\ \frac {2}{13}+\frac {3 \,\mathrm {I}}{13} \\ 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}^{2 x}\cdot \left (\cos \left (3 x \right )-\mathrm {I} \sin \left (3 x \right )\right )\cdot \left [\begin {array}{c} -\frac {119}{28561}-\frac {120 \,\mathrm {I}}{28561} \\ -\frac {46}{2197}+\frac {9 \,\mathrm {I}}{2197} \\ -\frac {5}{169}+\frac {12 \,\mathrm {I}}{169} \\ \frac {2}{13}+\frac {3 \,\mathrm {I}}{13} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \left (-\frac {119}{28561}-\frac {120 \,\mathrm {I}}{28561}\right ) \left (\cos \left (3 x \right )-\mathrm {I} \sin \left (3 x \right )\right ) \\ \left (-\frac {46}{2197}+\frac {9 \,\mathrm {I}}{2197}\right ) \left (\cos \left (3 x \right )-\mathrm {I} \sin \left (3 x \right )\right ) \\ \left (-\frac {5}{169}+\frac {12 \,\mathrm {I}}{169}\right ) \left (\cos \left (3 x \right )-\mathrm {I} \sin \left (3 x \right )\right ) \\ \left (\frac {2}{13}+\frac {3 \,\mathrm {I}}{13}\right ) \left (\cos \left (3 x \right )-\mathrm {I} \sin \left (3 x \right )\right ) \\ \cos \left (3 x \right )-\mathrm {I} \sin \left (3 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{4}\left (x \right )={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} -\frac {119 \cos \left (3 x \right )}{28561}-\frac {120 \sin \left (3 x \right )}{28561} \\ -\frac {46 \cos \left (3 x \right )}{2197}+\frac {9 \sin \left (3 x \right )}{2197} \\ -\frac {5 \cos \left (3 x \right )}{169}+\frac {12 \sin \left (3 x \right )}{169} \\ \frac {2 \cos \left (3 x \right )}{13}+\frac {3 \sin \left (3 x \right )}{13} \\ \cos \left (3 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{5}\left (x \right )={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {119 \sin \left (3 x \right )}{28561}-\frac {120 \cos \left (3 x \right )}{28561} \\ \frac {46 \sin \left (3 x \right )}{2197}+\frac {9 \cos \left (3 x \right )}{2197} \\ \frac {5 \sin \left (3 x \right )}{169}+\frac {12 \cos \left (3 x \right )}{169} \\ -\frac {2 \sin \left (3 x \right )}{13}+\frac {3 \cos \left (3 x \right )}{13} \\ -\sin \left (3 x \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right )+c_{5} {\moverset {\rightarrow }{y}}_{5}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+{\mathrm e}^{-2 x} c_{2} \cdot \left (x \cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} \frac {1}{32} \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right )+c_{3} {\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+{\mathrm e}^{2 x} c_{4} \cdot \left [\begin {array}{c} -\frac {119 \cos \left (3 x \right )}{28561}-\frac {120 \sin \left (3 x \right )}{28561} \\ -\frac {46 \cos \left (3 x \right )}{2197}+\frac {9 \sin \left (3 x \right )}{2197} \\ -\frac {5 \cos \left (3 x \right )}{169}+\frac {12 \sin \left (3 x \right )}{169} \\ \frac {2 \cos \left (3 x \right )}{13}+\frac {3 \sin \left (3 x \right )}{13} \\ \cos \left (3 x \right ) \end {array}\right ]+{\mathrm e}^{2 x} c_{5} \cdot \left [\begin {array}{c} \frac {119 \sin \left (3 x \right )}{28561}-\frac {120 \cos \left (3 x \right )}{28561} \\ \frac {46 \sin \left (3 x \right )}{2197}+\frac {9 \cos \left (3 x \right )}{2197} \\ \frac {5 \sin \left (3 x \right )}{169}+\frac {12 \cos \left (3 x \right )}{169} \\ -\frac {2 \sin \left (3 x \right )}{13}+\frac {3 \cos \left (3 x \right )}{13} \\ -\sin \left (3 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 x} \left (\frac {16 \left (\left (-119 c_{4} -120 c_{5} \right ) \cos \left (3 x \right )-120 \sin \left (3 x \right ) \left (c_{4} -\frac {119 c_{5}}{120}\right )\right ) {\mathrm e}^{4 x}}{28561}+16 c_{3} {\mathrm e}^{3 x}+\left (x +\frac {1}{2}\right ) c_{2} +c_{1} \right )}{16} \end {array} \]

Maple trace

`Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 40

dsolve(diff(y(x),x$5)-diff(y(x),x$4)+diff(y(x),x$3)+35*diff(y(x),x$2)+16*diff(y(x),x)-52*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = \left (c_{4} {\mathrm e}^{4 x} \sin \left (3 x \right )+c_{5} {\mathrm e}^{4 x} \cos \left (3 x \right )+c_{1} {\mathrm e}^{3 x}+c_{3} x +c_{2} \right ) {\mathrm e}^{-2 x} \]

✓ Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 50

DSolve[y'''''[x]-y''''[x]+y'''[x]+35*y''[x]+16*y'[x]-52*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-2 x} \left (c_4 x+c_5 e^{3 x}+c_2 e^{4 x} \cos (3 x)+c_1 e^{4 x} \sin (3 x)+c_3\right ) \]