2.1.65 Problem 65

2.1.65.1 Solved by factoring the differential equation
2.1.65.2 Maple
2.1.65.3 Mathematica
2.1.65.4 Sympy

Internal problem ID [10051]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 65
Date solved : Monday, December 08, 2025 at 07:12:25 PM
CAS classification : [[_2nd_order, _quadrature]]

2.1.65.1 Solved by factoring the differential equation

Time used: 0.010 (sec)

\begin{align*} y^{2} y^{\prime \prime }&=0 \\ \end{align*}
Writing the ode as
\begin{align*} \left (y^{2}\right )\left (y^{\prime \prime }\right )&=0 \end{align*}

Therefore we need to solve the following equations

\begin{align*} \tag{1} y^{2} &= 0 \\ \tag{2} y^{\prime \prime } &= 0 \\ \end{align*}
Now each of the above equations is solved in turn.

Solving equation (1)

Entering zero order ode solverSolving for \(y\) from

\begin{align*} y^{2} = 0 \end{align*}

Solving gives

\begin{align*} y &= 0 \\ \end{align*}
Solving equation (2)

Entering second order ode quadrature solverIntegrating twice gives the solution

\[ y= c_1 x + c_2 \]

Summary of solutions found

\begin{align*} y &= 0 \\ y &= c_1 x +c_2 \\ \end{align*}
2.1.65.2 Maple. Time used: 0.001 (sec). Leaf size: 13
ode:=y(x)^2*diff(diff(y(x),x),x) = 0; 
dsolve(ode,y(x), singsol=all);
 
\begin{align*} y &= 0 \\ y &= c_1 x +c_2 \\ \end{align*}

Maple trace

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

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right )^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\mathit {C1} \end {array} \]
2.1.65.3 Mathematica. Time used: 0.002 (sec). Leaf size: 17
ode=y[x]^2*D[y[x],{x,2}]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
\begin{align*} y(x)&\to 0\\ y(x)&\to c_2 x+c_1 \end{align*}
2.1.65.4 Sympy. Time used: 0.099 (sec). Leaf size: 7
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)**2*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
\[ y{\left (x \right )} = C_{1} + C_{2} x \]