10.15 problem 15

Internal problem ID [2131]
Internal file name [OUTPUT/2131_Monday_February_26_2024_09_17_46_AM_71634107/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 18, page 82
Problem number: 15.
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, _missing_x]]

\[ \boxed {y^{\prime \prime \prime \prime }+5 y^{\prime \prime }+6 y=0} \] The characteristic equation is \[ \lambda ^{4}+5 \lambda ^{2}+6 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= i \sqrt {2}\\ \lambda _2 &= -i \sqrt {2}\\ \lambda _3 &= i \sqrt {3}\\ \lambda _4 &= -i \sqrt {3} \end {align*}

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

Summary

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

Verification of solutions

\[ y = {\mathrm e}^{-i \sqrt {2}\, x} c_{1} +{\mathrm e}^{i \sqrt {3}\, x} c_{2} +{\mathrm e}^{-i \sqrt {3}\, x} c_{3} +{\mathrm e}^{i \sqrt {2}\, x} c_{4} \] Verified OK.

10.15.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }+5 \frac {d}{d x}y^{\prime }+6 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\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 )=\frac {d}{d x}y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=\frac {d}{d x}y^{\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 )=-5 y_{3}\left (x \right )-6 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 )=-5 y_{3}\left (x \right )-6 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 \\ -6 & 0 & -5 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \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 \\ -6 & 0 & -5 & 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 ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {\mathrm {I}}{4} \sqrt {2} \\ -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {-I} \sqrt {3}, \left [\begin {array}{c} -\frac {\mathrm {I}}{9} \sqrt {3} \\ -\frac {1}{3} \\ \frac {\mathrm {I}}{3} \sqrt {3} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I} \sqrt {2}, \left [\begin {array}{c} \frac {\mathrm {I}}{4} \sqrt {2} \\ -\frac {1}{2} \\ -\frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I} \sqrt {3}, \left [\begin {array}{c} \frac {\mathrm {I}}{9} \sqrt {3} \\ -\frac {1}{3} \\ -\frac {\mathrm {I}}{3} \sqrt {3} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {\mathrm {I}}{4} \sqrt {2} \\ -\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 {\mathrm {I}}{4} \sqrt {2} \\ -\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 {\mathrm {I}}{4} \sqrt {2} \\ -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{4} \left (\cos \left (\sqrt {2}\, x \right )-\mathrm {I} \sin \left (\sqrt {2}\, x \right )\right ) \sqrt {2} \\ -\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}}_{1}\left (x \right )=\left [\begin {array}{c} -\frac {\sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{4} \\ -\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}}_{2}\left (x \right )=\left [\begin {array}{c} -\frac {\sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{4} \\ \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 {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I} \sqrt {3}, \left [\begin {array}{c} -\frac {\mathrm {I}}{9} \sqrt {3} \\ -\frac {1}{3} \\ \frac {\mathrm {I}}{3} \sqrt {3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} \sqrt {3}\, x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{9} \sqrt {3} \\ -\frac {1}{3} \\ \frac {\mathrm {I}}{3} \sqrt {3} \\ 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 {3}\, x \right )-\mathrm {I} \sin \left (\sqrt {3}\, x \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{9} \sqrt {3} \\ -\frac {1}{3} \\ \frac {\mathrm {I}}{3} \sqrt {3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{9} \left (\cos \left (\sqrt {3}\, x \right )-\mathrm {I} \sin \left (\sqrt {3}\, x \right )\right ) \sqrt {3} \\ -\frac {\cos \left (\sqrt {3}\, x \right )}{3}+\frac {\mathrm {I} \sin \left (\sqrt {3}\, x \right )}{3} \\ \frac {\mathrm {I}}{3} \left (\cos \left (\sqrt {3}\, x \right )-\mathrm {I} \sin \left (\sqrt {3}\, x \right )\right ) \sqrt {3} \\ \cos \left (\sqrt {3}\, x \right )-\mathrm {I} \sin \left (\sqrt {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}}_{3}\left (x \right )=\left [\begin {array}{c} -\frac {\sin \left (\sqrt {3}\, x \right ) \sqrt {3}}{9} \\ -\frac {\cos \left (\sqrt {3}\, x \right )}{3} \\ \frac {\sin \left (\sqrt {3}\, x \right ) \sqrt {3}}{3} \\ \cos \left (\sqrt {3}\, x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )=\left [\begin {array}{c} -\frac {\cos \left (\sqrt {3}\, x \right ) \sqrt {3}}{9} \\ \frac {\sin \left (\sqrt {3}\, x \right )}{3} \\ \frac {\cos \left (\sqrt {3}\, x \right ) \sqrt {3}}{3} \\ -\sin \left (\sqrt {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}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\left [\begin {array}{c} -\frac {c_{4} \cos \left (\sqrt {3}\, x \right ) \sqrt {3}}{9}-\frac {c_{3} \sin \left (\sqrt {3}\, x \right ) \sqrt {3}}{9}-\frac {c_{2} \sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{4}-\frac {c_{1} \sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{4} \\ \frac {c_{4} \sin \left (\sqrt {3}\, x \right )}{3}-\frac {c_{3} \cos \left (\sqrt {3}\, x \right )}{3}+\frac {c_{2} \sin \left (\sqrt {2}\, x \right )}{2}-\frac {c_{1} \cos \left (\sqrt {2}\, x \right )}{2} \\ \frac {c_{4} \cos \left (\sqrt {3}\, x \right ) \sqrt {3}}{3}+\frac {c_{3} \sin \left (\sqrt {3}\, x \right ) \sqrt {3}}{3}+\frac {c_{2} \sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{2}+\frac {c_{1} \sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{2} \\ -c_{4} \sin \left (\sqrt {3}\, x \right )+c_{3} \cos \left (\sqrt {3}\, x \right )-c_{2} \sin \left (\sqrt {2}\, x \right )+c_{1} \cos \left (\sqrt {2}\, x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {c_{4} \cos \left (\sqrt {3}\, x \right ) \sqrt {3}}{9}-\frac {c_{3} \sin \left (\sqrt {3}\, x \right ) \sqrt {3}}{9}-\frac {c_{2} \sqrt {2}\, \cos \left (\sqrt {2}\, x \right )}{4}-\frac {c_{1} \sqrt {2}\, \sin \left (\sqrt {2}\, x \right )}{4} \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: 37

dsolve(diff(y(x),x$4)+5*diff(y(x),x$2)+6*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} \sin \left (\sqrt {2}\, x \right )+c_{2} \cos \left (\sqrt {2}\, x \right )+c_{3} \sin \left (\sqrt {3}\, x \right )+c_{4} \cos \left (\sqrt {3}\, x \right ) \]

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 50

DSolve[y''''[x]+5*y''[x]+6*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_3 \cos \left (\sqrt {2} x\right )+c_1 \cos \left (\sqrt {3} x\right )+c_4 \sin \left (\sqrt {2} x\right )+c_2 \sin \left (\sqrt {3} x\right ) \]