problem 434

Internal problem ID [15223]
Internal ﬁle name [OUTPUT/15224_Wednesday_May_08_2024_03_53_53_PM_54640770/index.tex]

Book: A book of problems in ordinary diﬀerential equations. M.L. KRASNOV, A.L. KISELYOV, G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.2 Homogeneous diﬀerential equations with constant coeﬃcients. Exercises page 121
Problem number: 434.
ODE order: 3.
ODE degree: 1.

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

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

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

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

Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{x}+x \,{\mathrm e}^{x} c_{2} +x^{2} {\mathrm e}^{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 } = c_{1} {\mathrm e}^{x}+{\mathrm e}^{x} c_{2} +x \,{\mathrm e}^{x} c_{2} +2 x \,{\mathrm e}^{x} c_{3} +x^{2} {\mathrm e}^{x} c_{3} \end {align*}

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

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

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

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

Summary

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

Veriﬁcation of solutions

\[ y = {\mathrm e}^{x} \left (x +1\right ) \] 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`

15.3 problem 434