2.1.62 Problem 62
Internal
problem
ID
[10048]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
62
Date
solved
:
Monday, December 08, 2025 at 07:12:18 PM
CAS
classification
:
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
2.1.62.1 second order ode missing x
2.287 (sec)
\begin{align*}
y y^{\prime \prime }&=1 \\
\end{align*}
Entering second order ode missing \(x\) solverThis is missing independent variable second order ode.
Solved by reduction of order by using substitution which makes the dependent variable \(y\) an
independent variable. Using \begin{align*} y' &= p \end{align*}
Then
\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}
Hence the ode becomes
\begin{align*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = 1 \end{align*}
Which is now solved as first order ode for \(p(y)\).
Entering first order ode separable solverThe ode
\begin{equation}
p^{\prime } = \frac {1}{y p}
\end{equation}
is separable as it can be written as
\begin{align*} p^{\prime }&= \frac {1}{y p}\\ &= f(y) g(p) \end{align*}
Where
\begin{align*} f(y) &= \frac {1}{y}\\ g(p) &= \frac {1}{p} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\
\int { p\,dp} &= \int { \frac {1}{y} \,dy} \\
\end{align*}
\[
\frac {p^{2}}{2}=\ln \left (y \right )+c_1
\]
Solving for \(p\) gives \begin{align*}
p &= \sqrt {2 \ln \left (y \right )+2 c_1} \\
p &= -\sqrt {2 \ln \left (y \right )+2 c_1} \\
\end{align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a
new first order ode to solve which is \begin{align*} y^{\prime } = \sqrt {2 \ln \left (y\right )+2 c_1} \end{align*}
Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and
since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\tau \right )+2 c_1}}d \tau = x +c_2 \]
For solution (2) found
earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin{align*} y^{\prime } = -\sqrt {2 \ln \left (y\right )+2 c_1} \end{align*}
Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and
since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {1}{\sqrt {2 \ln \left (\tau \right )+2 c_1}}d \tau = x +c_3 \]
Summary of solutions found
\begin{align*}
\int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\tau \right )+2 c_1}}d \tau &= x +c_2 \\
\int _{}^{y}-\frac {1}{\sqrt {2 \ln \left (\tau \right )+2 c_1}}d \tau &= x +c_3 \\
\end{align*}
2.1.62.2 ✓ Maple. Time used: 0.017 (sec). Leaf size: 51
ode:=y(x)*diff(diff(y(x),x),x) = 1;
dsolve(ode,y(x), singsol=all);
\begin{align*}
\int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )-c_1}}d \textit {\_a} -x -c_2 &= 0 \\
-\int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )-c_1}}d \textit {\_a} -x -c_2 &= 0 \\
\end{align*}
Maple trace
Methods for second order ODEs:
--- Trying classification methods ---
trying 2nd order Liouville
trying 2nd order WeierstrassP
trying 2nd order JacobiSN
differential order: 2; trying a linearization to 3rd order
trying 2nd order ODE linearizable_by_differentiation
trying 2nd order, 2 integrating factors of the form mu(x,y)
trying differential order: 2; missing variables
-> Computing symmetries using: way = 3
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)-1/_a = 0, _b(_a), HINT
= [[_a, 0]]
*** Sublevel 2 ***
symmetry methods on request
1st order, trying reduction of order with given symmetries:
[_a, 0]
1st order, trying the canonical coordinates of the invariance group
<- 1st order, canonical coordinates successful
<- differential order: 2; canonical coordinates successful
<- differential order 2; missing variables successful
2.1.62.3 ✓ Mathematica. Time used: 60.04 (sec). Leaf size: 93
ode=y[x]*D[y[x],{x,2}]==1;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to \exp \left (-\text {erf}^{-1}\left (-i \sqrt {\frac {2}{\pi }} \sqrt {e^{c_1} (x+c_2){}^2}\right ){}^2-\frac {c_1}{2}\right )\\ y(x)&\to \exp \left (-\text {erf}^{-1}\left (i \sqrt {\frac {2}{\pi }} \sqrt {e^{c_1} (x+c_2){}^2}\right ){}^2-\frac {c_1}{2}\right ) \end{align*}
2.1.62.4 ✗ Sympy
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(y(x)*Derivative(y(x), (x, 2)) - 1,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
Timed Out