12.24 problem 19.4 (h)

12.24.1 Maple step by step solution

Internal problem ID [13639]
Internal file name [OUTPUT/12811_Saturday_February_17_2024_08_44_55_AM_37692248/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 19. Arbitrary Homogeneous linear equations with constant coefficients. Additional exercises page 369
Problem number: 19.4 (h).
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]]

\[ \boxed {4 y^{\prime \prime \prime \prime }+15 y^{\prime \prime }-4 y=0} \] The characteristic equation is \[ 4 \lambda ^{4}+15 \lambda ^{2}-4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2 i\\ \lambda _2 &= -2 i\\ \lambda _3 &= -{\frac {1}{2}}\\ \lambda _4 &= {\frac {1}{2}} \end {align*}

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

Summary

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

Verification of solutions

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

12.24.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime \prime \prime }+15 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 } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=-\frac {15 y^{\prime \prime }}{4}+y \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+\frac {15 y^{\prime \prime }}{4}-y=0 \\ \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 )=-\frac {15 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 )=-\frac {15 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 \\ 1 & 0 & -\frac {15}{4} & 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 \\ 1 & 0 & -\frac {15}{4} & 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 [-\frac {1}{2}, \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 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 ], \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 eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {1}{2}, \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\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}}_{3}\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}}_{4}\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 {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}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} -8 \\ 4 \\ -2 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {c_{3} \sin \left (2 x \right )}{8}-\frac {c_{4} \cos \left (2 x \right )}{8} \\ \frac {c_{4} \sin \left (2 x \right )}{4}-\frac {c_{3} \cos \left (2 x \right )}{4} \\ \frac {c_{4} \cos \left (2 x \right )}{2}+\frac {c_{3} \sin \left (2 x \right )}{2} \\ c_{3} \cos \left (2 x \right )-c_{4} \sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-8 c_{1} {\mathrm e}^{-\frac {x}{2}}+8 c_{2} {\mathrm e}^{\frac {x}{2}}-\frac {c_{4} \cos \left (2 x \right )}{8}-\frac {c_{3} \sin \left (2 x \right )}{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: 29

dsolve(4*diff(y(x),x$4)+15*diff(y(x),x$2)-4*y(x)=0,y(x), singsol=all)
 

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

Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 37

DSolve[4*y''''[x]+15*y''[x]-4*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

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