2.1.8 Problem 8

Internal problem ID [10367]
Book : Second order enumerated odes
Section : section 1
Problem number : 8
Date solved : Monday, January 26, 2026 at 09:46:38 PM
CAS classification : [[_2nd_order, _quadrature]]

2.1.8.1 Solved by factoring the differential equation

Time used: 0.033 (sec)

\begin{align*} {y^{\prime \prime }}^{2}&=1 \\ \end{align*}

Writing the ode as

\begin{align*} \left (y^{\prime \prime }-1\right )\left (y^{\prime \prime }+1\right )&=0 \end{align*}

Therefore we need to solve the following equations

\begin{align*} \tag{1} y^{\prime \prime }-1 &= 0 \\ \tag{2} y^{\prime \prime }+1 &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Entering second order ode quadrature solverThe ODE can be written as

\[ y^{\prime \prime } = 1 \]

Integrating once gives

\[ y^{\prime }= x + c_1 \]

Integrating again gives

\[ y= \frac {x^{2}}{2} + c_1 x + c_2 \]

Solving equation (2)

Entering second order ode quadrature solverThe ODE can be written as

\[ y^{\prime \prime } = -1 \]

Integrating once gives

\[ y^{\prime }= -x + c_3 \]

Integrating again gives

\[ y= -\frac {x^{2}}{2} + c_3 x + c_4 \]

Summary of solutions found

\begin{align*} y &= -\frac {1}{2} x^{2}+c_3 x +c_4 \\ y &= \frac {1}{2} x^{2}+c_1 x +c_2 \\ \end{align*}

2.1.8.2 ✓ Maple. Time used: 0.001 (sec). Leaf size: 27

ode:=diff(diff(y(x),x),x)^2 = 1; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= \frac {1}{2} x^{2}+c_1 x +c_2 \\ y &= -\frac {1}{2} x^{2}+c_1 x +c_2 \\ \end{align*}

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful 
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} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )^{2}=1 \\ \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 {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}u \left (x \right )\right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}u \left (x \right )=-1, \frac {d}{d x}u \left (x \right )=1\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}u \left (x \right )=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}u \left (x \right )\right )d x =\int \left (-1\right )d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (x \right )=-x +\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}u \left (x \right )=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}u \left (x \right )\right )d x =\int 1d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (x \right )=x +\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{u \left (x \right )=-x +\mathit {C1} , u \left (x \right )=x +\mathit {C1} \right \} \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=\frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} \frac {d}{d x}y \left (x \right )=u \left (y \right ),\frac {d^{2}}{d x^{2}}y \left (x \right )=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (y \right )^{2} \left (\frac {d}{d y}u \left (y \right )\right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d y}u \left (y \right )=\frac {1}{u \left (y \right )}, \frac {d}{d y}u \left (y \right )=-\frac {1}{u \left (y \right )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d y}u \left (y \right )=\frac {1}{u \left (y \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int 1d y +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=y +\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {2 \textit {\_C1} +2 y}, u \left (y \right )=-\sqrt {2 \textit {\_C1} +2 y}\right \} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{u \left (y \right )=\sqrt {\textit {\_C1} +2 y}, u \left (y \right )=-\sqrt {\textit {\_C1} +2 y}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d y}u \left (y \right )=-\frac {1}{u \left (y \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int \left (-1\right )d y +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=-y +\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {2 \textit {\_C1} -2 y}, u \left (y \right )=-\sqrt {2 \textit {\_C1} -2 y}\right \} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{u \left (y \right )=\sqrt {\textit {\_C1} -2 y}, u \left (y \right )=-\sqrt {\textit {\_C1} -2 y}\right \} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{u \left (y \right )=\sqrt {\mathit {C1} -2 y}, u \left (y \right )=\sqrt {\mathit {C1} +2 y}, u \left (y \right )=-\sqrt {\mathit {C1} -2 y}, u \left (y \right )=-\sqrt {\mathit {C1} +2 y}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {\mathit {C1} -2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {\mathit {C1} -2 y \left (x \right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {\mathit {C1} -2 y \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} -2 y \left (x \right )}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} -2 y \left (x \right )}}d x =\int 1d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\sqrt {\mathit {C1} -2 y \left (x \right )}=x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=-\frac {1}{2} \mathit {C2}^{2}-\mathit {C2} x -\frac {1}{2} x^{2}+\frac {1}{2} \mathit {C1} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x -\frac {1}{2} x^{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {\mathit {C1} +2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {\mathit {C1} +2 y \left (x \right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {\mathit {C1} +2 y \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} +2 y \left (x \right )}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} +2 y \left (x \right )}}d x =\int 1d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \sqrt {\mathit {C1} +2 y \left (x \right )}=x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {1}{2} \mathit {C2}^{2}+\mathit {C2} x +\frac {1}{2} x^{2}-\frac {1}{2} \mathit {C1} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\frac {1}{2} x^{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve 3rd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {\mathit {C1} -2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sqrt {\mathit {C1} -2 y \left (x \right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sqrt {\mathit {C1} -2 y \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} -2 y \left (x \right )}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} -2 y \left (x \right )}}d x =\int \left (-1\right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\sqrt {\mathit {C1} -2 y \left (x \right )}=-x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=-\frac {1}{2} \mathit {C2}^{2}+\mathit {C2} x -\frac {1}{2} x^{2}+\frac {1}{2} \mathit {C1} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x -\frac {1}{2} x^{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve 4th ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {\mathit {C1} +2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sqrt {\mathit {C1} +2 y \left (x \right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sqrt {\mathit {C1} +2 y \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} +2 y \left (x \right )}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} +2 y \left (x \right )}}d x =\int \left (-1\right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \sqrt {\mathit {C1} +2 y \left (x \right )}=-x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {1}{2} \mathit {C2}^{2}-\mathit {C2} x +\frac {1}{2} x^{2}-\frac {1}{2} \mathit {C1} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\frac {1}{2} x^{2}+\mathit {C1} \end {array} \]

2.1.8.3 ✓ Mathematica. Time used: 0.002 (sec). Leaf size: 37

ode=(D[y[x],{x,2}])^2==1; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to -\frac {x^2}{2}+c_2 x+c_1\\ y(x)&\to \frac {x^2}{2}+c_2 x+c_1 \end{align*}

2.1.8.4 ✓ Sympy. Time used: 0.217 (sec). Leaf size: 24

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), (x, 2))**2 - 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ \left [ y{\left (x \right )} = C_{1} + C_{2} x - \frac {x^{2}}{2}, \ y{\left (x \right )} = C_{1} + C_{2} x + \frac {x^{2}}{2}\right ] \]

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('factorable', 'nth_algebraic', 'nth_algebraic_Integral')