15.14 problem 445

Internal problem ID [15234]
Internal file name [OUTPUT/15235_Wednesday_May_08_2024_03_53_59_PM_56013244/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: 445.
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]]

\[ \boxed {y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }+4 y^{\prime \prime }-2 y^{\prime }-5 y=0} \] The characteristic equation is \[ \lambda ^{4}+2 \lambda ^{3}+4 \lambda ^{2}-2 \lambda -5 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -1\\ \lambda _3 &= -1-2 i\\ \lambda _4 &= -1+2 i \end {align*}

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

Verification of solutions

\[ y = c_{1} {\mathrm e}^{-x}+{\mathrm e}^{x} c_{2} +{\mathrm e}^{\left (-1+2 i\right ) x} c_{3} +{\mathrm e}^{\left (-1-2 i\right ) x} c_{4} \] Verified OK.

15.14.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }+4 y^{\prime \prime }-2 y^{\prime }-5 y=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 (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 {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=-2 y_{4}\left (x \right )-4 y_{3}\left (x \right )+2 y_{2}\left (x \right )+5 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 )=-2 y_{4}\left (x \right )-4 y_{3}\left (x \right )+2 y_{2}\left (x \right )+5 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 \\ 5 & 2 & -4 & -2 \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 \\ 5 & 2 & -4 & -2 \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 [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [-1-2 \,\mathrm {I}, \left [\begin {array}{c} \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ], \left [-1+2 \,\mathrm {I}, \left [\begin {array}{c} \frac {11}{125}+\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}-\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 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 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-1-2 \,\mathrm {I}, \left [\begin {array}{c} \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-1-2 \,\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{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 \\ {} & {} & {\mathrm e}^{-x}\cdot \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right )\cdot \left [\begin {array}{c} \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-x}\cdot \left [\begin {array}{c} \left (\frac {11}{125}-\frac {2 \,\mathrm {I}}{125}\right ) \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \left (-\frac {3}{25}-\frac {4 \,\mathrm {I}}{25}\right ) \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \left (-\frac {1}{5}+\frac {2 \,\mathrm {I}}{5}\right ) \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \cos \left (2 x \right )-\mathrm {I} \sin \left (2 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 )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {11 \cos \left (2 x \right )}{125}-\frac {2 \sin \left (2 x \right )}{125} \\ -\frac {3 \cos \left (2 x \right )}{25}-\frac {4 \sin \left (2 x \right )}{25} \\ -\frac {\cos \left (2 x \right )}{5}+\frac {2 \sin \left (2 x \right )}{5} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {11 \sin \left (2 x \right )}{125}-\frac {2 \cos \left (2 x \right )}{125} \\ \frac {3 \sin \left (2 x \right )}{25}-\frac {4 \cos \left (2 x \right )}{25} \\ \frac {\sin \left (2 x \right )}{5}+\frac {2 \cos \left (2 x \right )}{5} \\ -\sin \left (2 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}}=c_{1} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+{\mathrm e}^{x} c_{2} \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} \frac {11 \cos \left (2 x \right )}{125}-\frac {2 \sin \left (2 x \right )}{125} \\ -\frac {3 \cos \left (2 x \right )}{25}-\frac {4 \sin \left (2 x \right )}{25} \\ -\frac {\cos \left (2 x \right )}{5}+\frac {2 \sin \left (2 x \right )}{5} \\ \cos \left (2 x \right ) \end {array}\right ]+c_{4} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {11 \sin \left (2 x \right )}{125}-\frac {2 \cos \left (2 x \right )}{125} \\ \frac {3 \sin \left (2 x \right )}{25}-\frac {4 \cos \left (2 x \right )}{25} \\ \frac {\sin \left (2 x \right )}{5}+\frac {2 \cos \left (2 x \right )}{5} \\ -\sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (11 c_{3} -2 c_{4} \right ) \cos \left (2 x \right )+\left (-2 c_{3} -11 c_{4} \right ) \sin \left (2 x \right )-125 c_{1} \right ) {\mathrm e}^{-x}}{125}+{\mathrm e}^{x} 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: 28

dsolve(diff(y(x),x$4)+2*diff(y(x),x$3)+4*diff(y(x),x$2)-2*diff(y(x),x)-5*y(x)=0,y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 36

DSolve[y''''[x]+2*y'''[x]+4*y''[x]-2*y'[x]-5*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

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