10.18 problem 18

Internal problem ID [2134]
Internal file name [OUTPUT/2134_Monday_February_26_2024_09_17_47_AM_94047950/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 18, page 82
Problem number: 18.
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]]

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

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

Summary

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

Verification of solutions

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

10.18.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }+\frac {d}{d x}y^{\prime \prime }-3 \frac {d}{d x}y^{\prime }-4 y^{\prime }-4 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\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 )=\frac {d}{d x}y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=\frac {d}{d x}y^{\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 )=-y_{4}\left (x \right )+3 y_{3}\left (x \right )+4 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_{4}^{\prime }\left (x \right )=-y_{4}\left (x \right )+3 y_{3}\left (x \right )+4 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 ) \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 & 4 & 3 & -1 \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 & 4 & 3 & -1 \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}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{3}} \\ \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{3}} \\ \frac {1}{\left (-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} -\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}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \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}}_{2}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{3}} \\ \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{3}} \\ \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{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}^{-\frac {x}{2}}\cdot \left (\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )\right )\cdot \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{3}} \\ \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{3}} \\ \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\right )}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ \cos \left (\frac {\sqrt {3}\, x}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, x}{2}\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}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \\ -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2} \\ -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2} \\ \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \\ -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ -\sin \left (\frac {\sqrt {3}\, x}{2}\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}}={\mathrm e}^{-2 x} c_{1} \cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+{\mathrm e}^{2 x} c_{2} \cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \\ -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}-\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2} \\ -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right )}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2} \\ \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ]+{\mathrm e}^{-\frac {x}{2}} c_{4} \cdot \left [\begin {array}{c} -\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \\ -\frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ \frac {\cos \left (\frac {\sqrt {3}\, x}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, x}{2}\right )}{2} \\ -\sin \left (\frac {\sqrt {3}\, x}{2}\right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{4 x} c_{2} +8 c_{3} \cos \left (\frac {\sqrt {3}\, x}{2}\right ) {\mathrm e}^{\frac {3 x}{2}}-8 c_{4} \sin \left (\frac {\sqrt {3}\, x}{2}\right ) {\mathrm e}^{\frac {3 x}{2}}-c_{1} \right ) {\mathrm e}^{-2 x}}{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: 0.016 (sec). Leaf size: 43

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

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

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 64

DSolve[y''''[x]+y'''[x]-3*y''[x]-4*y'[x]-4*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-2 x} \left (c_4 e^{4 x}+c_2 e^{3 x/2} \cos \left (\frac {\sqrt {3} x}{2}\right )+c_1 e^{3 x/2} \sin \left (\frac {\sqrt {3} x}{2}\right )+c_3\right ) \]