15.48 problem 51

Internal problem ID [14756]
Internal file name [OUTPUT/14436_Monday_April_08_2024_06_24_56_AM_75093460/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.7, page 195
Problem number: 51.
ODE order: 3.
ODE degree: 1.

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

Maple gives the following as the ode type

[[_3rd_order, _with_linear_symmetries]]

\[ \boxed {x^{3} y^{\prime \prime \prime }+y^{\prime } x -y=0} \] This is Euler ODE of higher order. Let \(y = x^{\lambda }\). Hence \begin {align*} y^{\prime } &= \lambda \,x^{\lambda -1}\\ y^{\prime \prime } &= \lambda \left (\lambda -1\right ) x^{\lambda -2}\\ y^{\prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3} \end {align*}

Substituting these back into \[ x^{3} y^{\prime \prime \prime }+y^{\prime } x -y = 0 \] gives \[ x \lambda \,x^{\lambda -1}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}-x^{\lambda } = 0 \] Which simplifies to \[ \lambda \,x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }-x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes

\[ \lambda +\lambda \left (\lambda -1\right ) \left (\lambda -2\right )-1 = 0 \] Simplifying gives the characteristic equation as \[ \left (\lambda -1\right )^{3} = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 1\\ \lambda _3 &= 1 \end {align*}

This table summarises the result

root multiplicity type of root
\(1\) \(3\) real root

The solution is generated by going over the above table. For each real root \(\lambda \) of multiplicity one generates a \(c_1x^{\lambda }\) basis solution. Each real root of multiplicty two, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) basis solutions. Each real root of multiplicty three, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) and \(c_3x^{\lambda } \ln \left (x \right )^{2}\) basis solutions, and so on. Each complex root \(\alpha \pm i \beta \) of multiplicity one generates \(x^{\alpha } \left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity two generates \(\ln \left (x \right ) x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity three generates \(\ln \left (x \right )^{2} x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And so on. Using the above show that the solution is

\[ y = c_{1} x +c_{2} x \ln \left (x \right )+c_{3} \ln \left (x \right )^{2} x \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= x\\ y_2 &= x \ln \left (x \right )\\ y_3 &= \ln \left (x \right )^{2} x \end {align*}

Summary

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

Verification of solutions

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

Maple trace

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

Solution by Maple

Time used: 0.0 (sec). Leaf size: 18

dsolve(x^3*diff(y(x),x$3)+x*diff(y(x),x)-y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = x \left (c_{1} +c_{2} \ln \left (x \right )+c_{3} \ln \left (x \right )^{2}\right ) \]

Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 22

DSolve[x^3*y'''[x]+x*y'[x]-y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to x \left (c_3 \log ^2(x)+c_2 \log (x)+c_1\right ) \]