10.14 problem 14

Internal problem ID [2130]
Internal file name [OUTPUT/2130_Monday_February_26_2024_09_17_46_AM_14734510/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: 14.
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]]

\[ \boxed {y^{\prime \prime \prime \prime }+y^{\prime \prime }-20 y=0} \] The characteristic equation is \[ \lambda ^{4}+\lambda ^{2}-20 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2\\ \lambda _3 &= i \sqrt {5}\\ \lambda _4 &= -i \sqrt {5} \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{-i \sqrt {5}\, x} c_{3} +{\mathrm e}^{i \sqrt {5}\, x} c_{4} \] 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}\\ y_3 &= {\mathrm e}^{-i \sqrt {5}\, x}\\ y_4 &= {\mathrm e}^{i \sqrt {5}\, x} \end {align*}

Summary

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

Verification of solutions

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

10.14.1 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 }+\frac {d}{d x}y^{\prime }-20 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 )=-y_{3}\left (x \right )+20 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 )=-y_{3}\left (x \right )+20 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 \\ 20 & 0 & -1 & 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 \\ 20 & 0 & -1 & 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 [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {-I} \sqrt {5}, \left [\begin {array}{c} -\frac {\mathrm {I}}{25} \sqrt {5} \\ -\frac {1}{5} \\ \frac {\mathrm {I}}{5} \sqrt {5} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I} \sqrt {5}, \left [\begin {array}{c} \frac {\mathrm {I}}{25} \sqrt {5} \\ -\frac {1}{5} \\ -\frac {\mathrm {I}}{5} \sqrt {5} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I} \sqrt {5}, \left [\begin {array}{c} -\frac {\mathrm {I}}{25} \sqrt {5} \\ -\frac {1}{5} \\ \frac {\mathrm {I}}{5} \sqrt {5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} \sqrt {5}\, x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{25} \sqrt {5} \\ -\frac {1}{5} \\ \frac {\mathrm {I}}{5} \sqrt {5} \\ 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 {5}\, x \right )-\mathrm {I} \sin \left (\sqrt {5}\, x \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{25} \sqrt {5} \\ -\frac {1}{5} \\ \frac {\mathrm {I}}{5} \sqrt {5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{25} \left (\cos \left (\sqrt {5}\, x \right )-\mathrm {I} \sin \left (\sqrt {5}\, x \right )\right ) \sqrt {5} \\ -\frac {\cos \left (\sqrt {5}\, x \right )}{5}+\frac {\mathrm {I} \sin \left (\sqrt {5}\, x \right )}{5} \\ \frac {\mathrm {I}}{5} \left (\cos \left (\sqrt {5}\, x \right )-\mathrm {I} \sin \left (\sqrt {5}\, x \right )\right ) \sqrt {5} \\ \cos \left (\sqrt {5}\, x \right )-\mathrm {I} \sin \left (\sqrt {5}\, 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 {\sqrt {5}\, \sin \left (\sqrt {5}\, x \right )}{25} \\ -\frac {\cos \left (\sqrt {5}\, x \right )}{5} \\ \frac {\sqrt {5}\, \sin \left (\sqrt {5}\, x \right )}{5} \\ \cos \left (\sqrt {5}\, x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )=\left [\begin {array}{c} -\frac {\sqrt {5}\, \cos \left (\sqrt {5}\, x \right )}{25} \\ \frac {\sin \left (\sqrt {5}\, x \right )}{5} \\ \frac {\sqrt {5}\, \cos \left (\sqrt {5}\, x \right )}{5} \\ -\sin \left (\sqrt {5}\, 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}+c_{2} {\moverset {\rightarrow }{y}}_{2}+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}}={\mathrm e}^{-2 x} c_{1} \cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+{\mathrm e}^{2 x} c_{2} \cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {c_{4} \sqrt {5}\, \cos \left (\sqrt {5}\, x \right )}{25}-\frac {c_{3} \sqrt {5}\, \sin \left (\sqrt {5}\, x \right )}{25} \\ \frac {c_{4} \sin \left (\sqrt {5}\, x \right )}{5}-\frac {c_{3} \cos \left (\sqrt {5}\, x \right )}{5} \\ \frac {c_{4} \sqrt {5}\, \cos \left (\sqrt {5}\, x \right )}{5}+\frac {c_{3} \sqrt {5}\, \sin \left (\sqrt {5}\, x \right )}{5} \\ -c_{4} \sin \left (\sqrt {5}\, x \right )+c_{3} \cos \left (\sqrt {5}\, 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} c_{1}}{8}+\frac {{\mathrm e}^{2 x} c_{2}}{8}-\frac {c_{4} \sqrt {5}\, \cos \left (\sqrt {5}\, x \right )}{25}-\frac {c_{3} \sqrt {5}\, \sin \left (\sqrt {5}\, x \right )}{25} \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: 33

dsolve(diff(y(x),x$4)+diff(y(x),x$2)-20*y(x)=0,y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 44

DSolve[y''''[x]+y''[x]-20*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_3 e^{-2 x}+c_4 e^{2 x}+c_1 \cos \left (\sqrt {5} x\right )+c_2 \sin \left (\sqrt {5} x\right ) \]