Internal problem ID [14651]
Internal file name [OUTPUT/14331_Wednesday_April_03_2024_02_17_20_PM_64728628/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: 33.
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]]
\[ \boxed {y^{\left (5\right )}+4 y^{\prime \prime \prime \prime }=0} \] The characteristic equation is \[ \lambda ^{5}+4 \lambda ^{4} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -4\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 0\\ \lambda _5 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=t^{3} c_{4} +t^{2} c_{3} +c_{2} t +c_{1} +{\mathrm e}^{-4 t} c_{5} \] 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 &= t^{3}\\ y_5 &= {\mathrm e}^{-4 t} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= t^{3} c_{4} +t^{2} c_{3} +c_{2} t +c_{1} +{\mathrm e}^{-4 t} c_{5} \\ \end{align*}
Verification of solutions
\[ y = t^{3} c_{4} +t^{2} c_{3} +c_{2} t +c_{1} +{\mathrm e}^{-4 t} c_{5} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\left (5\right )}+4 y^{\prime \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_{5}\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_{5}\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 & 0 & -4 \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 & 0 & -4 \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 [-4, \left [\begin {array}{c} \frac {1}{256} \\ -\frac {1}{64} \\ \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]\right ], \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 [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-4, \left [\begin {array}{c} \frac {1}{256} \\ -\frac {1}{64} \\ \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-4 t}\cdot \left [\begin {array}{c} \frac {1}{256} \\ -\frac {1}{64} \\ \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\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}}_{2}=\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}}_{3}=\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}}_{4}=\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}}_{5}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\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}+c_{5} {\moverset {\rightarrow }{y}}_{5} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-4 t}\cdot \left [\begin {array}{c} \frac {1}{256} \\ -\frac {1}{64} \\ \frac {1}{16} \\ -\frac {1}{4} \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{2} \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {c_{1} {\mathrm e}^{-4 t}}{256}+c_{2} \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: 25
dsolve(diff(y(t),t$5)+4*diff(y(t),t$4)=0,y(t), singsol=all)
\[ y \left (t \right ) = c_{1} +t c_{2} +c_{3} t^{2}+c_{4} t^{3}+c_{5} {\mathrm e}^{-4 t} \]
✓ Solution by Mathematica
Time used: 0.04 (sec). Leaf size: 33
DSolve[y'''''[t]+4*y''''[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{256} c_1 e^{-4 t}+t (t (c_5 t+c_4)+c_3)+c_2 \]