Internal problem ID [15235]
Internal file name [OUTPUT/15236_Wednesday_May_08_2024_03_53_59_PM_23615901/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.2 Homogeneous differential equations with
constant coefficients. Exercises page 121
Problem number: 446.
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]]
\[ \boxed {y^{\left (5\right )}+4 y^{\prime \prime \prime \prime }+5 y^{\prime \prime \prime }-6 y^{\prime }-4 y=0} \] The characteristic equation is \[ \lambda ^{5}+4 \lambda ^{4}+5 \lambda ^{3}-6 \lambda -4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -2\\ \lambda _3 &= -1\\ \lambda _4 &= -1-i\\ \lambda _5 &= -1+i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+{\mathrm e}^{-2 x} c_{2} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{\left (-1-i\right ) x} c_{4} +{\mathrm e}^{\left (-1+i\right ) x} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{-2 x}\\ y_3 &= {\mathrm e}^{x}\\ y_4 &= {\mathrm e}^{\left (-1-i\right ) x}\\ y_5 &= {\mathrm e}^{\left (-1+i\right ) x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+{\mathrm e}^{-2 x} c_{2} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{\left (-1-i\right ) x} c_{4} +{\mathrm e}^{\left (-1+i\right ) x} c_{5} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x}+{\mathrm e}^{-2 x} c_{2} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{\left (-1-i\right ) x} c_{4} +{\mathrm e}^{\left (-1+i\right ) x} c_{5} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\left (5\right )}+4 y^{\prime \prime \prime \prime }+5 y^{\prime \prime \prime }-6 y^{\prime }-4 y=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 (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 )=y^{\prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{5}\left (x \right ) \\ {} & {} & y_{5}\left (x \right )=y^{\prime \prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{5}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{5}^{\prime }\left (x \right )=-4 y_{5}\left (x \right )-5 y_{4}\left (x \right )+6 y_{2}\left (x \right )+4 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_{5}\left (x \right )=y_{4}^{\prime }\left (x \right ), y_{5}^{\prime }\left (x \right )=-4 y_{5}\left (x \right )-5 y_{4}\left (x \right )+6 y_{2}\left (x \right )+4 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 ) \\ y_{5}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \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 \\ 4 & 6 & 0 & -5 & -4 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \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 \\ 4 & 6 & 0 & -5 & -4 \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}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [-1-\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [-1+\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {1}{4}-\frac {\mathrm {I}}{4} \\ \frac {\mathrm {I}}{2} \\ -\frac {1}{2}-\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} \frac {1}{16} \\ -\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}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-1-\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-1-\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\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 \\ {} & {} & {\mathrm e}^{-x}\cdot \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {1}{4}+\frac {\mathrm {I}}{4} \\ -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {\cos \left (x \right )}{4}+\frac {\mathrm {I} \sin \left (x \right )}{4} \\ \left (\frac {1}{4}+\frac {\mathrm {I}}{4}\right ) \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ -\frac {\mathrm {I}}{2} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \left (-\frac {1}{2}+\frac {\mathrm {I}}{2}\right ) \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \cos \left (x \right )-\mathrm {I} \sin \left (x \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 (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {\cos \left (x \right )}{4} \\ \frac {\cos \left (x \right )}{4}+\frac {\sin \left (x \right )}{4} \\ -\frac {\sin \left (x \right )}{2} \\ -\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{5}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {\sin \left (x \right )}{4} \\ \frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \\ -\frac {\cos \left (x \right )}{2} \\ \frac {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2} \\ -\sin \left (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}+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right )+c_{5} {\moverset {\rightarrow }{y}}_{5}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {1}{16} \\ -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+c_{4} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {\cos \left (x \right )}{4} \\ \frac {\cos \left (x \right )}{4}+\frac {\sin \left (x \right )}{4} \\ -\frac {\sin \left (x \right )}{2} \\ -\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ \cos \left (x \right ) \end {array}\right ]+c_{5} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {\sin \left (x \right )}{4} \\ \frac {\cos \left (x \right )}{4}-\frac {\sin \left (x \right )}{4} \\ -\frac {\cos \left (x \right )}{2} \\ \frac {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2} \\ -\sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-2 x} \left (c_{3} {\mathrm e}^{3 x}+\left (-\frac {\cos \left (x \right ) c_{4}}{4}+\frac {\sin \left (x \right ) c_{5}}{4}+c_{2} \right ) {\mathrm e}^{x}+\frac {c_{1}}{16}\right ) \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: 30
dsolve(diff(y(x),x$5)+4*diff(y(x),x$4)+5*diff(y(x),x$3)-6*diff(y(x),x)-4*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-2 x} \left (c_{2} {\mathrm e}^{3 x}+\left (\sin \left (x \right ) c_{4} +\cos \left (x \right ) c_{5} +c_{1} \right ) {\mathrm e}^{x}+c_{3} \right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 44
DSolve[y'''''[x]+4*y''''[x]+5*y'''[x]-6*y'[x]-4*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-2 x} \left (c_4 e^x+c_5 e^{3 x}+c_2 e^x \cos (x)+c_1 e^x \sin (x)+c_3\right ) \]