13.4 problem 21

13.4.1 Maple step by step solution

Internal problem ID [14639]
Internal file name [OUTPUT/14319_Wednesday_April_03_2024_02_17_17_PM_97534733/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: 21.
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]]

\[ \boxed {y^{\prime \prime \prime \prime }+16 y^{\prime \prime }=0} \] The characteristic equation is \[ \lambda ^{4}+16 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 4 i\\ \lambda _4 &= -4 i \end {align*}

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

Summary

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

Verification of solutions

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

13.4.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+16 y^{\prime \prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \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 (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 {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (t \right )=-16 y_{3}\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_{4}^{\prime }\left (t \right )=-16 y_{3}\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 ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -16 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \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 \\ 0 & 0 & -16 & 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 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [-4 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{64} \\ -\frac {1}{16} \\ \frac {\mathrm {I}}{4} \\ 1 \end {array}\right ]\right ], \left [4 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{64} \\ -\frac {1}{16} \\ -\frac {\mathrm {I}}{4} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 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 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 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 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-4 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{64} \\ -\frac {1}{16} \\ \frac {\mathrm {I}}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-4 \,\mathrm {I} t}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{64} \\ -\frac {1}{16} \\ \frac {\mathrm {I}}{4} \\ 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 (4 t \right )-\mathrm {I} \sin \left (4 t \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{64} \\ -\frac {1}{16} \\ \frac {\mathrm {I}}{4} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{64} \left (\cos \left (4 t \right )-\mathrm {I} \sin \left (4 t \right )\right ) \\ -\frac {\cos \left (4 t \right )}{16}+\frac {\mathrm {I} \sin \left (4 t \right )}{16} \\ \frac {\mathrm {I}}{4} \left (\cos \left (4 t \right )-\mathrm {I} \sin \left (4 t \right )\right ) \\ \cos \left (4 t \right )-\mathrm {I} \sin \left (4 t \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 (t \right )=\left [\begin {array}{c} -\frac {\sin \left (4 t \right )}{64} \\ -\frac {\cos \left (4 t \right )}{16} \\ \frac {\sin \left (4 t \right )}{4} \\ \cos \left (4 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (t \right )=\left [\begin {array}{c} -\frac {\cos \left (4 t \right )}{64} \\ \frac {\sin \left (4 t \right )}{16} \\ \frac {\cos \left (4 t \right )}{4} \\ -\sin \left (4 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}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\left [\begin {array}{c} -\frac {c_{4} \cos \left (4 t \right )}{64}-\frac {c_{3} \sin \left (4 t \right )}{64}+c_{1} \\ \frac {c_{4} \sin \left (4 t \right )}{16}-\frac {c_{3} \cos \left (4 t \right )}{16} \\ \frac {c_{4} \cos \left (4 t \right )}{4}+\frac {c_{3} \sin \left (4 t \right )}{4} \\ -c_{4} \sin \left (4 t \right )+c_{3} \cos \left (4 t \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 (4 t \right )}{64}-\frac {c_{3} \sin \left (4 t \right )}{64}+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: 21

dsolve(diff(y(t),t$4)+16*diff(y(t),t$2)=0,y(t), singsol=all)
 

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

Solution by Mathematica

Time used: 0.17 (sec). Leaf size: 32

DSolve[y''''[t]+16*y''[t]==0,y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to c_4 t-\frac {1}{16} c_1 \cos (4 t)-\frac {1}{16} c_2 \sin (4 t)+c_3 \]