problem section 9.2, problem 18

Internal problem ID [1482]
Internal ﬁle name [OUTPUT/1483_Sunday_June_05_2022_02_19_19_AM_45254044/index.tex]

Book: Elementary diﬀerential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.2. constant coeﬃcient. Page 483
Problem number: section 9.2, problem 18.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime }-4 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 6, y^{\prime }\left (0\right ) = 3, y^{\prime \prime }\left (0\right ) = 22] \end {align*}

The characteristic equation is \[ \lambda ^{3}-2 \lambda -4 = 0 \] 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*}

Looking at the above solution \begin {align*} 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} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 6\) and \(x = 0\) in the above gives \begin {align*} 6 = c_{1} +c_{2} +c_{3}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 c_{1} {\mathrm e}^{2 x}+\left (-1-i\right ) {\mathrm e}^{\left (-1-i\right ) x} c_{2} +\left (-1+i\right ) {\mathrm e}^{\left (-1+i\right ) x} c_{3} \end {align*}

substituting \(y^{\prime } = 3\) and \(x = 0\) in the above gives \begin {align*} 3 = 2 c_{1} +\left (-1-i\right ) c_{2} +\left (-1+i\right ) c_{3}\tag {2A} \end {align*}

Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 4 c_{1} {\mathrm e}^{2 x}+2 i {\mathrm e}^{\left (-1-i\right ) x} c_{2} -2 i {\mathrm e}^{\left (-1+i\right ) x} c_{3} \end {align*}

substituting \(y^{\prime \prime } = 22\) and \(x = 0\) in the above gives \begin {align*} 22 = 2 c_{2} i-2 c_{3} i+4 c_{1}\tag {3A} \end {align*}

Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=4\\ c_{2}&=1-\frac {3 i}{2}\\ c_{3}&=1+\frac {3 i}{2} \end {align*}

Substituting these values back in above solution results in \begin {align*} y = 4 \,{\mathrm e}^{2 x}+{\mathrm e}^{\left (-1-i\right ) x}-\frac {3 i {\mathrm e}^{\left (-1-i\right ) x}}{2}+{\mathrm e}^{\left (-1+i\right ) x}+\frac {3 i {\mathrm e}^{\left (-1+i\right ) x}}{2} \end {align*}

Summary

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

Veriﬁcation of solutions

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

18.18.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-2 y^{\prime }-4 y=0, y \left (0\right )=6, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=3, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=22\right ] \\ \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 )=2 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 )=2 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 & 2 & 0 \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 & 2 & 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, \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 (\cos \left (x \right )-\mathrm {I} \sin \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 (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \left (-\frac {1}{2}+\frac {\mathrm {I}}{2}\right ) \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}}_{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 {\cos \left (x \right )}{2}+\frac {\sin \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 {\cos \left (x \right )}{2}+\frac {\sin \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 {\left (-2 c_{2} \sin \left (x \right )-2 c_{3} \cos \left (x \right )\right ) {\mathrm e}^{-x}}{4}+\frac {c_{1} {\mathrm e}^{2 x}}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=6 \\ {} & {} & 6=-\frac {c_{3}}{2}+\frac {c_{1}}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (-2 c_{2} \cos \left (x \right )+2 c_{3} \sin \left (x \right )\right ) {\mathrm e}^{-x}}{4}-\frac {\left (-2 c_{2} \sin \left (x \right )-2 c_{3} \cos \left (x \right )\right ) {\mathrm e}^{-x}}{4}+\frac {c_{1} {\mathrm e}^{2 x}}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=3 \\ {} & {} & 3=-\frac {c_{2}}{2}+\frac {c_{3}}{2}+\frac {c_{1}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (2 c_{2} \sin \left (x \right )+2 c_{3} \cos \left (x \right )\right ) {\mathrm e}^{-x}}{4}-\frac {\left (-2 c_{2} \cos \left (x \right )+2 c_{3} \sin \left (x \right )\right ) {\mathrm e}^{-x}}{2}+\frac {\left (-2 c_{2} \sin \left (x \right )-2 c_{3} \cos \left (x \right )\right ) {\mathrm e}^{-x}}{4}+c_{1} {\mathrm e}^{2 x} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=22 \\ {} & {} & 22=c_{1} +c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =16, c_{2} =6, c_{3} =-4\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (-3 \sin \left (x \right )+2 \cos \left (x \right )\right ) {\mathrm e}^{-x}+4 \,{\mathrm e}^{2 x} \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`

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

18.18 problem section 9.2, problem 18

18.18.1 Maple step by step solution