problem section 9.1, problem 3

Internal problem ID [1458]
Internal ﬁle name [OUTPUT/1459_Sunday_June_05_2022_02_18_45_AM_77226135/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 9.1. Page 471
Problem number: section 9.1, problem 3.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }-7 y^{\prime \prime }-y^{\prime }+6 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 5, y^{\prime }\left (0\right ) = -6, y^{\prime \prime }\left (0\right ) = 10, y^{\prime \prime \prime }\left (0\right ) = -36] \end {align*}

The characteristic equation is \[ \lambda ^{4}+\lambda ^{3}-7 \lambda ^{2}-\lambda +6 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 2\\ \lambda _3 &= -3\\ \lambda _4 &= -1 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} +{\mathrm e}^{-3 x} c_{3} +{\mathrm e}^{2 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{x}\\ y_3 &= {\mathrm e}^{-3 x}\\ y_4 &= {\mathrm e}^{2 x} \end {align*}

Looking at the above solution \begin {align*} y = {\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} +{\mathrm e}^{-3 x} c_{3} +{\mathrm e}^{2 x} c_{4} \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 \(y = 5\) and \(x = 0\) in the above gives \begin {align*} 5 = c_{1} +c_{2} +c_{3} +c_{4}\tag {1A} \end {align*}

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

substituting \(y^{\prime } = -6\) and \(x = 0\) in the above gives \begin {align*} -6 = -c_{1} +c_{2} -3 c_{3} +2 c_{4}\tag {2A} \end {align*}

Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = {\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} +9 \,{\mathrm e}^{-3 x} c_{3} +4 \,{\mathrm e}^{2 x} c_{4} \end {align*}

substituting \(y^{\prime \prime } = 10\) and \(x = 0\) in the above gives \begin {align*} 10 = c_{1} +c_{2} +9 c_{3} +4 c_{4}\tag {3A} \end {align*}

Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -{\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} -27 \,{\mathrm e}^{-3 x} c_{3} +8 \,{\mathrm e}^{2 x} c_{4} \end {align*}

substituting \(y^{\prime \prime \prime } = -36\) and \(x = 0\) in the above gives \begin {align*} -36 = -c_{1} +c_{2} -27 c_{3} +8 c_{4}\tag {4A} \end {align*}

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

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

Summary

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

Veriﬁcation of solutions

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

Maple trace

`Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`

\[ y \left (x \right ) = \left (-{\mathrm e}^{5 x}+2 \,{\mathrm e}^{4 x}+3 \,{\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-3 x} \]

17.2 problem section 9.1, problem 3