problem section 9.2, problem 22

Internal problem ID [1486]
Internal ﬁle name [OUTPUT/1487_Sunday_June_05_2022_02_19_25_AM_18782849/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 22.
ODE order: 3.
ODE degree: 1.

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

\[ \boxed {8 y^{\prime \prime \prime }-4 y^{\prime \prime }-2 y^{\prime }+y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 4, y^{\prime }\left (0\right ) = -3, y^{\prime \prime }\left (0\right ) = -1] \end {align*}

The characteristic equation is \[ 8 \lambda ^{3}-4 \lambda ^{2}-2 \lambda +1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -{\frac {1}{2}}\\ \lambda _2 &= {\frac {1}{2}}\\ \lambda _3 &= {\frac {1}{2}} \end {align*}

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

Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{-\frac {x}{2}}+c_{2} {\mathrm e}^{\frac {x}{2}}+x \,{\mathrm e}^{\frac {x}{2}} 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 \(y = 4\) and \(x = 0\) in the above gives \begin {align*} 4 = c_{1} +c_{2}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {c_{1} {\mathrm e}^{-\frac {x}{2}}}{2}+\frac {c_{2} {\mathrm e}^{\frac {x}{2}}}{2}+{\mathrm e}^{\frac {x}{2}} c_{3} +\frac {x \,{\mathrm e}^{\frac {x}{2}} c_{3}}{2} \end {align*}

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

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

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

Substituting these values back in above solution results in \begin {align*} y = -2 \,{\mathrm e}^{\frac {x}{2}} x +3 \,{\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}} \end {align*}

Summary

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

Veriﬁcation of solutions

\[ y = -2 \,{\mathrm e}^{\frac {x}{2}} x +3 \,{\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}} \] Veriﬁed OK.

18.22.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [8 y^{\prime \prime \prime }-4 y^{\prime \prime }-2 y^{\prime }+y=0, y \left (0\right )=4, 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 \}}}}=-1\right ] \\ \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 }=\frac {y^{\prime \prime }}{2}+\frac {y^{\prime }}{4}-\frac {y}{8} \\ \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 }-\frac {y^{\prime \prime }}{2}-\frac {y^{\prime }}{4}+\frac {y}{8}=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 )=\frac {y_{3}\left (x \right )}{2}+\frac {y_{2}\left (x \right )}{4}-\frac {y_{1}\left (x \right )}{8} \\ & {} & \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 )=\frac {y_{3}\left (x \right )}{2}+\frac {y_{2}\left (x \right )}{4}-\frac {y_{1}\left (x \right )}{8}\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 \\ -\frac {1}{8} & \frac {1}{4} & \frac {1}{2} \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 \\ -\frac {1}{8} & \frac {1}{4} & \frac {1}{2} \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} 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} 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} 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =\frac {1}{2}\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -\frac {1}{8} & \frac {1}{4} & \frac {1}{2} \end {array}\right ]-\frac {1}{2}\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -8 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{\frac {x}{2}}\cdot \left (x \cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -8 \\ 0 \\ 0 \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}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]+{\mathrm e}^{\frac {x}{2}} c_{3} \cdot \left (x \cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -8 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\left (\left (4 x -8\right ) c_{3} +4 c_{2} \right ) {\mathrm e}^{\frac {x}{2}}+4 c_{1} {\mathrm e}^{-\frac {x}{2}} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=4 \\ {} & {} & 4=-8 c_{3} +4 c_{2} +4 c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=4 \,{\mathrm e}^{\frac {x}{2}} c_{3} +\frac {\left (\left (4 x -8\right ) c_{3} +4 c_{2} \right ) {\mathrm e}^{\frac {x}{2}}}{2}-2 c_{1} {\mathrm e}^{-\frac {x}{2}} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-3 \\ {} & {} & -3=2 c_{2} -2 c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=4 \,{\mathrm e}^{\frac {x}{2}} c_{3} +\frac {\left (\left (4 x -8\right ) c_{3} +4 c_{2} \right ) {\mathrm e}^{\frac {x}{2}}}{4}+c_{1} {\mathrm e}^{-\frac {x}{2}} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-1 \\ {} & {} & -1=c_{1} +c_{2} +2 c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {3}{4}, c_{2} =-\frac {3}{4}, c_{3} =-\frac {1}{2}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-2 \,{\mathrm e}^{\frac {x}{2}} x +3 \,{\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{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`

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

\[ y(x)\to -\frac {2}{21} e^{-x/4} \left (9 e^{5 x/4}+13 \sqrt {3} \sin \left (\frac {\sqrt {3} x}{4}\right )-51 \cos \left (\frac {\sqrt {3} x}{4}\right )\right ) \]

18.22 problem section 9.2, problem 22

18.22.1 Maple step by step solution