problem section 9.2, problem 19

Internal problem ID [1483]
Internal ﬁle name [OUTPUT/1484_Sunday_June_05_2022_02_19_21_AM_92724626/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 19.
ODE order: 3.
ODE degree: 1.

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

\[ \boxed {3 y^{\prime \prime \prime }-y^{\prime \prime }-7 y^{\prime }+5 y=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {14}{5}}, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 10\right ] \end {align*}

The characteristic equation is \[ 3 \lambda ^{3}-\lambda ^{2}-7 \lambda +5 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -{\frac {5}{3}}\\ \lambda _2 &= 1\\ \lambda _3 &= 1 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+{\mathrm e}^{x} c_{2} x +{\mathrm e}^{-\frac {5 x}{3}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{x}\\ y_2 &= x \,{\mathrm e}^{x}\\ y_3 &= {\mathrm e}^{-\frac {5 x}{3}} \end {align*}

Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{x}+{\mathrm e}^{x} c_{2} x +{\mathrm e}^{-\frac {5 x}{3}} 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 \(y = {\frac {14}{5}}\) and \(x = 0\) in the above gives \begin {align*} {\frac {14}{5}} = c_{1} +c_{3}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} {\mathrm e}^{x}+{\mathrm e}^{x} c_{2} x +c_{2} {\mathrm e}^{x}-\frac {5 \,{\mathrm e}^{-\frac {5 x}{3}} c_{3}}{3} \end {align*}

substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{2} -\frac {5 c_{3}}{3}\tag {2A} \end {align*}

Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{x}+{\mathrm e}^{x} c_{2} x +2 c_{2} {\mathrm e}^{x}+\frac {25 \,{\mathrm e}^{-\frac {5 x}{3}} c_{3}}{9} \end {align*}

substituting \(y^{\prime \prime } = 10\) and \(x = 0\) in the above gives \begin {align*} 10 = c_{1} +2 c_{2} +\frac {25 c_{3}}{9}\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}&=1\\ c_{2}&=2\\ c_{3}&={\frac {9}{5}} \end {align*}

Substituting these values back in above solution results in \begin {align*} y = 2 x \,{\mathrm e}^{x}+\frac {9 \,{\mathrm e}^{-\frac {5 x}{3}}}{5}+{\mathrm e}^{x} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 2 x \,{\mathrm e}^{x}+\frac {9 \,{\mathrm e}^{-\frac {5 x}{3}}}{5}+{\mathrm e}^{x} \\ \end{align*}

Veriﬁcation of solutions

\[ y = 2 x \,{\mathrm e}^{x}+\frac {9 \,{\mathrm e}^{-\frac {5 x}{3}}}{5}+{\mathrm e}^{x} \] Veriﬁed OK.

18.19.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [3 y^{\prime \prime \prime }-y^{\prime \prime }-7 y^{\prime }+5 y=0, y \left (0\right )=\frac {14}{5}, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=10\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 }}{3}+\frac {7 y^{\prime }}{3}-\frac {5 y}{3} \\ \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 }}{3}-\frac {7 y^{\prime }}{3}+\frac {5 y}{3}=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 )}{3}+\frac {7 y_{2}\left (x \right )}{3}-\frac {5 y_{1}\left (x \right )}{3} \\ & {} & \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 )}{3}+\frac {7 y_{2}\left (x \right )}{3}-\frac {5 y_{1}\left (x \right )}{3}\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 {5}{3} & \frac {7}{3} & \frac {1}{3} \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 {5}{3} & \frac {7}{3} & \frac {1}{3} \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 {5}{3}, \left [\begin {array}{c} \frac {9}{25} \\ -\frac {3}{5} \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {5}{3}, \left [\begin {array}{c} \frac {9}{25} \\ -\frac {3}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-\frac {5 x}{3}}\cdot \left [\begin {array}{c} \frac {9}{25} \\ -\frac {3}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} 1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 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 =1\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 =1 \\ \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} 1 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -\frac {5}{3} & \frac {7}{3} & \frac {1}{3} \end {array}\right ]-1\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} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} 1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{x}\cdot \left (x \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 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 {5 x}{3}}\cdot \left [\begin {array}{c} \frac {9}{25} \\ -\frac {3}{5} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+{\mathrm e}^{x} c_{3} \cdot \left (x \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -1 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\left (\left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{\frac {8 x}{3}}+\frac {9 c_{1}}{25}\right ) {\mathrm e}^{-\frac {5 x}{3}} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=\frac {14}{5} \\ {} & {} & \frac {14}{5}=c_{2} -c_{3} +\frac {9 c_{1}}{25} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\left ({\mathrm e}^{\frac {8 x}{3}} c_{3} +\frac {8 \left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{\frac {8 x}{3}}}{3}\right ) {\mathrm e}^{-\frac {5 x}{3}}-\frac {5 \left (\left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{\frac {8 x}{3}}+\frac {9 c_{1}}{25}\right ) {\mathrm e}^{-\frac {5 x}{3}}}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{2} -\frac {3 c_{1}}{5} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\left (\frac {16 \,{\mathrm e}^{\frac {8 x}{3}} c_{3}}{3}+\frac {64 \left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{\frac {8 x}{3}}}{9}\right ) {\mathrm e}^{-\frac {5 x}{3}}-\frac {10 \left ({\mathrm e}^{\frac {8 x}{3}} c_{3} +\frac {8 \left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{\frac {8 x}{3}}}{3}\right ) {\mathrm e}^{-\frac {5 x}{3}}}{3}+\frac {25 \left (\left (\left (x -1\right ) c_{3} +c_{2} \right ) {\mathrm e}^{\frac {8 x}{3}}+\frac {9 c_{1}}{25}\right ) {\mathrm e}^{-\frac {5 x}{3}}}{9} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=10 \\ {} & {} & 10=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =5, c_{2} =3, c_{3} =2\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-\frac {5 x}{3}} \left (\frac {9}{5}+\left (1+2 x \right ) {\mathrm e}^{\frac {8 x}{3}}\right ) \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`

\[ y \left (x \right ) = {\mathrm e}^{-\frac {5 x}{3}} \left (\frac {9}{5}+\left (2 x +1\right ) {\mathrm e}^{\frac {8 x}{3}}\right ) \]

18.19 problem section 9.2, problem 19

18.19.1 Maple step by step solution