18.16 problem 22

18.16.1 Maple step by step solution

Internal problem ID [14842]
Internal file name [OUTPUT/14522_Monday_April_08_2024_06_26_28_AM_34098867/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Chapter 4 review exercises, page 219
Problem number: 22.
ODE order: 3.
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

[[_3rd_order, _missing_x]]

\[ \boxed {9 y^{\prime \prime \prime }+36 y^{\prime \prime }+40 y^{\prime }=0} \] The characteristic equation is \[ 9 \lambda ^{3}+36 \lambda ^{2}+40 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= -2+\frac {2 i}{3}\\ \lambda _3 &= -2-\frac {2 i}{3} \end {align*}

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

Summary

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

Verification of solutions

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

18.16.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 y^{\prime \prime \prime }+36 y^{\prime \prime }+40 y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-4 y^{\prime \prime }-\frac {40 y^{\prime }}{9} \\ \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 }+4 y^{\prime \prime }+\frac {40 y^{\prime }}{9}=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 {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=-4 y_{3}\left (x \right )-\frac {40 y_{2}\left (x \right )}{9} \\ & {} & \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_{3}^{\prime }\left (x \right )=-4 y_{3}\left (x \right )-\frac {40 y_{2}\left (x \right )}{9}\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 ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -\frac {40}{9} & -4 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -\frac {40}{9} & -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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [-2-\frac {2 \,\mathrm {I}}{3}, \left [\begin {array}{c} \frac {9}{50}-\frac {27 \,\mathrm {I}}{200} \\ -\frac {9}{20}+\frac {3 \,\mathrm {I}}{20} \\ 1 \end {array}\right ]\right ], \left [-2+\frac {2 \,\mathrm {I}}{3}, \left [\begin {array}{c} \frac {9}{50}+\frac {27 \,\mathrm {I}}{200} \\ -\frac {9}{20}-\frac {3 \,\mathrm {I}}{20} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2-\frac {2 \,\mathrm {I}}{3}, \left [\begin {array}{c} \frac {9}{50}-\frac {27 \,\mathrm {I}}{200} \\ -\frac {9}{20}+\frac {3 \,\mathrm {I}}{20} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-2-\frac {2 \,\mathrm {I}}{3}\right ) x}\cdot \left [\begin {array}{c} \frac {9}{50}-\frac {27 \,\mathrm {I}}{200} \\ -\frac {9}{20}+\frac {3 \,\mathrm {I}}{20} \\ 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}^{-2 x}\cdot \left (\cos \left (\frac {2 x}{3}\right )-\mathrm {I} \sin \left (\frac {2 x}{3}\right )\right )\cdot \left [\begin {array}{c} \frac {9}{50}-\frac {27 \,\mathrm {I}}{200} \\ -\frac {9}{20}+\frac {3 \,\mathrm {I}}{20} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \left (\frac {9}{50}-\frac {27 \,\mathrm {I}}{200}\right ) \left (\cos \left (\frac {2 x}{3}\right )-\mathrm {I} \sin \left (\frac {2 x}{3}\right )\right ) \\ \left (-\frac {9}{20}+\frac {3 \,\mathrm {I}}{20}\right ) \left (\cos \left (\frac {2 x}{3}\right )-\mathrm {I} \sin \left (\frac {2 x}{3}\right )\right ) \\ \cos \left (\frac {2 x}{3}\right )-\mathrm {I} \sin \left (\frac {2 x}{3}\right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {9 \cos \left (\frac {2 x}{3}\right )}{50}-\frac {27 \sin \left (\frac {2 x}{3}\right )}{200} \\ -\frac {9 \cos \left (\frac {2 x}{3}\right )}{20}+\frac {3 \sin \left (\frac {2 x}{3}\right )}{20} \\ \cos \left (\frac {2 x}{3}\right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} -\frac {9 \sin \left (\frac {2 x}{3}\right )}{50}-\frac {27 \cos \left (\frac {2 x}{3}\right )}{200} \\ \frac {9 \sin \left (\frac {2 x}{3}\right )}{20}+\frac {3 \cos \left (\frac {2 x}{3}\right )}{20} \\ -\sin \left (\frac {2 x}{3}\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}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}={\mathrm e}^{-2 x} c_{2} \cdot \left [\begin {array}{c} \frac {9 \cos \left (\frac {2 x}{3}\right )}{50}-\frac {27 \sin \left (\frac {2 x}{3}\right )}{200} \\ -\frac {9 \cos \left (\frac {2 x}{3}\right )}{20}+\frac {3 \sin \left (\frac {2 x}{3}\right )}{20} \\ \cos \left (\frac {2 x}{3}\right ) \end {array}\right ]+c_{3} {\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} -\frac {9 \sin \left (\frac {2 x}{3}\right )}{50}-\frac {27 \cos \left (\frac {2 x}{3}\right )}{200} \\ \frac {9 \sin \left (\frac {2 x}{3}\right )}{20}+\frac {3 \cos \left (\frac {2 x}{3}\right )}{20} \\ -\sin \left (\frac {2 x}{3}\right ) \end {array}\right ]+\left [\begin {array}{c} c_{1} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {9 \left (c_{2} -\frac {3 c_{3}}{4}\right ) {\mathrm e}^{-2 x} \cos \left (\frac {2 x}{3}\right )}{50}-\frac {27 \,{\mathrm e}^{-2 x} \left (c_{2} +\frac {4 c_{3}}{3}\right ) \sin \left (\frac {2 x}{3}\right )}{200}+c_{1} \end {array} \]

Maple trace

`Methods for third 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: 26

dsolve(9*diff(y(x),x$3)+36*diff(y(x),x$2)+40*diff(y(x),x)=0,y(x), singsol=all)
 

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

Solution by Mathematica

Time used: 0.242 (sec). Leaf size: 48

DSolve[9*y'''[x]+36*y''[x]+40*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_3-\frac {3}{20} e^{-2 x} \left ((c_1+3 c_2) \cos \left (\frac {2 x}{3}\right )+(3 c_1-c_2) \sin \left (\frac {2 x}{3}\right )\right ) \]