10.2 problem 2

Internal problem ID [12760]
Internal file name [OUTPUT/11413_Friday_November_03_2023_06_32_46_AM_10799578/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 4. N-th Order Linear Differential Equations. Exercises 4.3, page 210
Problem number: 2.
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]]

\[ \boxed {y^{\prime \prime \prime }-4 y^{\prime \prime }+6 y^{\prime }-4 y=0} \] The characteristic equation is \[ \lambda ^{3}-4 \lambda ^{2}+6 \lambda -4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 1-i\\ \lambda _3 &= 1+i \end {align*}

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

Summary

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

Verification of solutions

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

10.2.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-4 y^{\prime \prime }+6 y^{\prime }-4 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\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 {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 )-6 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_{3}^{\prime }\left (x \right )=4 y_{3}\left (x \right )-6 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 ) \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 \\ 4 & -6 & 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 \\ 4 & -6 & 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 [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [1-\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{2} \\ \frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [1+\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ \frac {1}{2}-\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \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}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [1-\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{2} \\ \frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (1-\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} \frac {\mathrm {I}}{2} \\ \frac {1}{2}+\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 \\ {} & {} & {\mathrm e}^{x}\cdot \left (-\mathrm {I} \sin \left (x \right )+\cos \left (x \right )\right )\cdot \left [\begin {array}{c} \frac {\mathrm {I}}{2} \\ \frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{x}\cdot \left [\begin {array}{c} \frac {\mathrm {I}}{2} \left (-\mathrm {I} \sin \left (x \right )+\cos \left (x \right )\right ) \\ \left (\frac {1}{2}+\frac {\mathrm {I}}{2}\right ) \left (-\mathrm {I} \sin \left (x \right )+\cos \left (x \right )\right ) \\ -\mathrm {I} \sin \left (x \right )+\cos \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}}_{2}\left (x \right )={\mathrm e}^{x}\cdot \left [\begin {array}{c} \frac {\sin \left (x \right )}{2} \\ \frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{x}\cdot \left [\begin {array}{c} \frac {\cos \left (x \right )}{2} \\ -\frac {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2} \\ -\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}+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}}=c_{1} {\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{x}\cdot \left [\begin {array}{c} \frac {\sin \left (x \right )}{2} \\ \frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ \cos \left (x \right ) \end {array}\right ]+c_{3} {\mathrm e}^{x}\cdot \left [\begin {array}{c} \frac {\cos \left (x \right )}{2} \\ -\frac {\sin \left (x \right )}{2}+\frac {\cos \left (x \right )}{2} \\ -\sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {c_{1} {\mathrm e}^{2 x}}{4}+\frac {{\mathrm e}^{x} \left (c_{3} \cos \left (x \right )+c_{2} \sin \left (x \right )\right )}{2} \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: 0.0 (sec). Leaf size: 23

dsolve(diff(y(x),x$3)-4*diff(y(x),x$2)+6*diff(y(x),x)-4*y(x)=0,y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 26

DSolve[y'''[x]-4*y''[x]+6*y'[x]-4*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^x \left (c_3 e^x+c_2 \cos (x)+c_1 \sin (x)\right ) \]