7.5 problem 4(f)

Internal problem ID [5976]
Internal file name [OUTPUT/5224_Sunday_June_05_2022_03_27_53_PM_58330443/index.tex]

Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY 1961
Section: Chapter 2. Linear equations with constant coefficients. Page 74
Problem number: 4(f).
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]]

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

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

Summary

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

Verification of solutions

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

7.5.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+5 y^{\prime \prime }+4 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 )=-5 y_{3}\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_{4}^{\prime }\left (x \right )=-5 y_{3}\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 ) \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 \\ -4 & 0 & -5 & 0 \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 \\ -4 & 0 & -5 & 0 \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 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {-I}, \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I}, \left [\begin {array}{c} \mathrm {I} \\ -1 \\ \mathrm {-I} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \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 \\ {} & {} & \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ -\frac {\cos \left (2 x \right )}{4}+\frac {\mathrm {I} \sin \left (2 x \right )}{4} \\ \frac {\mathrm {I}}{2} \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}}_{1}\left (x \right )=\left [\begin {array}{c} -\frac {\sin \left (2 x \right )}{8} \\ -\frac {\cos \left (2 x \right )}{4} \\ \frac {\sin \left (2 x \right )}{2} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{2}\left (x \right )=\left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{8} \\ \frac {\sin \left (2 x \right )}{4} \\ \frac {\cos \left (2 x \right )}{2} \\ -\sin \left (2 x \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I}, \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} x}\cdot \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 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 (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} \mathrm {-I} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ -\cos \left (x \right )+\mathrm {I} \sin \left (x \right ) \\ \mathrm {I} \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}}_{3}\left (x \right )=\left [\begin {array}{c} -\sin \left (x \right ) \\ -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )=\left [\begin {array}{c} -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \\ -\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}\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}}=\left [\begin {array}{c} -c_{4} \cos \left (x \right )-c_{3} \sin \left (x \right )-\frac {c_{2} \cos \left (2 x \right )}{8}-\frac {\sin \left (2 x \right ) c_{1}}{8} \\ c_{4} \sin \left (x \right )-c_{3} \cos \left (x \right )+\frac {c_{2} \sin \left (2 x \right )}{4}-\frac {c_{1} \cos \left (2 x \right )}{4} \\ c_{4} \cos \left (x \right )+c_{3} \sin \left (x \right )+\frac {c_{2} \cos \left (2 x \right )}{2}+\frac {\sin \left (2 x \right ) c_{1}}{2} \\ -c_{4} \sin \left (x \right )+c_{3} \cos \left (x \right )-c_{2} \sin \left (2 x \right )+c_{1} \cos \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-c_{4} \cos \left (x \right )-c_{3} \sin \left (x \right )-\frac {c_{2} \cos \left (2 x \right )}{8}-\frac {\sin \left (2 x \right ) c_{1}}{8} \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: 29

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

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

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 30

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

\[ y(x)\to c_1 \cos (2 x)+c_4 \sin (x)+\cos (x) (2 c_2 \sin (x)+c_3) \]