2.1.3 Problem 3

Internal problem ID [9074]
Book : Second order enumerated odes
Section : section 1
Problem number : 3
Date solved : Sunday, March 30, 2025 at 02:06:16 PM
CAS classification : [[_2nd_order, _quadrature]]

✓ Maple. Time used: 0.003 (sec). Leaf size: 9

ode:=diff(diff(y(x),x),x)^n = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_{1}x+c_2$$

Maple trace

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

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & {\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)}^n=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=0\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{0\pm \left(\sqrt{0}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=0\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(x\right)=1\\ \bullet & & \text{Repeated root, multiply}{y}_1\left(x\right)\text{by}x\text{to ensure linear independence}\\ & & y_2\left(x\right)=x\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{y}_1\left(x\right)+\mathit{C}\mathit{2}{y}_2\left(x\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(x\right)=\mathit{C}\mathit{2}x+\mathit{C}\mathit{1}\end{array}$$

✓ Mathematica. Time used: 0.004 (sec). Leaf size: 24

ode=(D[y[x],{x,2}])^n==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{1}{2}{0}^{\frac{1}{n}}{x}^2+c_{2}x+c_1$$

✗ Sympy

from sympy import * 
x = symbols("x") 
n = symbols("n") 
y = Function("y") 
ode = Eq(Derivative(y(x), (x, 2))**n,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

TypeError : cannot determine truth value of Relational