problem Problem 39

Internal problem ID [2617]
Internal ﬁle name [OUTPUT/2109_Sunday_June_05_2022_02_48_53_AM_70023368/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 1, First-Order Diﬀerential Equations. Section 1.2, Basic Ideas and Terminology. page 21
Problem number: Problem 39.
ODE order: 3.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime }=6 x} \] 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*}

This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{3} x^{2}+c_{2} x +c_{1} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= x^{2} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime } = 6 x \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ x \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, x^{2}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2}, x^{3}\}] \] Since \(x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3}, x^{4}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{2} x^{4}+A_{1} x^{3} \] The unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coeﬃcients. Substituting the trial solution into the ODE and simplifying gives \[ 24 x A_{2}+6 A_{1} = 6 x \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = 0, A_{2} = {\frac {1}{4}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{4}}{4} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{3} x^{2}+c_{2} x +c_{1}\right ) + \left (\frac {x^{4}}{4}\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = c_{3} x^{2}+c_{2} x +c_{1} +\frac {1}{4} x^{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 \(y = 1\) 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 } = x^{3}+2 c_{3} x +c_{2} \end {align*}

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

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

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

Substituting these values back in above solution results in \begin {align*} y = -2 x^{2}-x +1+\frac {1}{4} x^{4} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -2 x^{2}-x +1+\frac {1}{4} x^{4} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -2 x^{2}-x +1+\frac {1}{4} x^{4} \] Veriﬁed OK.

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`

1.31 problem Problem 39