problem section 9.2, problem 20

Internal problem ID [1484]
Internal ﬁle name [OUTPUT/1485_Sunday_June_05_2022_02_19_22_AM_88452520/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 20.
ODE order: 3.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime }-6 y^{\prime \prime }+12 y^{\prime }-8 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = -1, y^{\prime \prime }\left (0\right ) = -4] \end {align*}

The characteristic equation is \[ \lambda ^{3}-6 \lambda ^{2}+12 \lambda -8 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 2\\ \lambda _3 &= 2 \end {align*}

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

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

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

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

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

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

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

Summary

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

Veriﬁcation of solutions

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

Maple trace

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

18.20 problem section 9.2, problem 20