15.11 problem 442

Internal problem ID [15231]
Internal file name [OUTPUT/15232_Wednesday_May_08_2024_03_53_57_PM_58150107/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: 442.
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]]

\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+10 y^{\prime \prime }+12 y^{\prime }+5 y=0} \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{3}+10 \lambda ^{2}+12 \lambda +5 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1-2 i\\ \lambda _2 &= -1+2 i\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end {align*}

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

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

15.11.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+10 y^{\prime \prime }+12 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 )=-4 y_{4}\left (x \right )-10 y_{3}\left (x \right )-12 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 )=-4 y_{4}\left (x \right )-10 y_{3}\left (x \right )-12 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 & -12 & -10 & -4 \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 & -12 & -10 & -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 [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \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, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =-1\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =-1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} -1 \\ {} & {} & \left (\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -5 & -12 & -10 & -4 \end {array}\right ]-\left (-1\right )\cdot \left [\begin {array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-x}\cdot \left (x \cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+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 ]+c_{2} {\mathrm e}^{-x}\cdot \left (x \cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\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=-\left (\frac {\left (-11 c_{3} +2 c_{4} \right ) \cos \left (2 x \right )}{125}+\frac {\left (2 c_{3} +11 c_{4} \right ) \sin \left (2 x \right )}{125}+\left (x +1\right ) c_{2} +c_{1} \right ) {\mathrm e}^{-x} \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(x),x$4)+4*diff(y(x),x$3)+10*diff(y(x),x$2)+12*diff(y(x),x)+5*y(x)=0,y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 32

DSolve[y''''[x]+4*y'''[x]+10*y''[x]+12*y'[x]+5*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

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