13.17 problem 34

Internal problem ID [14652]
Internal file name [OUTPUT/14332_Wednesday_April_03_2024_02_17_21_PM_6249229/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.5, page 175
Problem number: 34.
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]]

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

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

Summary

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

Verification of solutions

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

13.17.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\left (5\right )}+4 y^{\prime \prime \prime }=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 (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (t \right ) \\ {} & {} & y_{4}\left (t \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{5}\left (t \right ) \\ {} & {} & y_{5}\left (t \right )=y^{\prime \prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{5}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{5}^{\prime }\left (t \right )=-4 y_{4}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=y_{1}^{\prime }\left (t \right ), y_{3}\left (t \right )=y_{2}^{\prime }\left (t \right ), y_{4}\left (t \right )=y_{3}^{\prime }\left (t \right ), y_{5}\left (t \right )=y_{4}^{\prime }\left (t \right ), y_{5}^{\prime }\left (t \right )=-4 y_{4}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \\ y_{4}\left (t \right ) \\ y_{5}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \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 \\ 0 & 0 & 0 & -4 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \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 \\ 0 & 0 & 0 & -4 & 0 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\left (t \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [-2 \,\mathrm {I}, \left [\begin {array}{c} \frac {1}{16} \\ -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} \frac {1}{16} \\ \frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 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 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2 \,\mathrm {I}, \left [\begin {array}{c} \frac {1}{16} \\ -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} t}\cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{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 (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right )\cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} \frac {\cos \left (2 t \right )}{16}-\frac {\mathrm {I} \sin \left (2 t \right )}{16} \\ -\frac {\mathrm {I}}{8} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ -\frac {\cos \left (2 t \right )}{4}+\frac {\mathrm {I} \sin \left (2 t \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \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 (t \right )=\left [\begin {array}{c} \frac {\cos \left (2 t \right )}{16} \\ -\frac {\sin \left (2 t \right )}{8} \\ -\frac {\cos \left (2 t \right )}{4} \\ \frac {\sin \left (2 t \right )}{2} \\ \cos \left (2 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{5}\left (t \right )=\left [\begin {array}{c} -\frac {\sin \left (2 t \right )}{16} \\ -\frac {\cos \left (2 t \right )}{8} \\ \frac {\sin \left (2 t \right )}{4} \\ \frac {\cos \left (2 t \right )}{2} \\ -\sin \left (2 t \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}+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right )+c_{5} {\moverset {\rightarrow }{y}}_{5}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\left [\begin {array}{c} -\frac {c_{5} \sin \left (2 t \right )}{16}+\frac {c_{4} \cos \left (2 t \right )}{16}+c_{1} \\ -\frac {c_{5} \cos \left (2 t \right )}{8}-\frac {c_{4} \sin \left (2 t \right )}{8} \\ \frac {c_{5} \sin \left (2 t \right )}{4}-\frac {c_{4} \cos \left (2 t \right )}{4} \\ \frac {c_{5} \cos \left (2 t \right )}{2}+\frac {c_{4} \sin \left (2 t \right )}{2} \\ -c_{5} \sin \left (2 t \right )+c_{4} \cos \left (2 t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {c_{5} \sin \left (2 t \right )}{16}+\frac {c_{4} \cos \left (2 t \right )}{16}+c_{1} \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: 26

dsolve(diff(y(t),t$5)+4*diff(y(t),t$3)=0,y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.204 (sec). Leaf size: 38

DSolve[y'''''[t]+4*y'''[t]==0,y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to c_5 t^2+c_4 t+\frac {1}{8} c_2 \cos (2 t)-\frac {1}{8} c_1 \sin (2 t)+c_3 \]