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 29, 2025 at 09:01:38 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

2.1.62.1 second order ode missing x

1.516 (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

\[ \int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\tau \right )+2 c_1}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \sqrt {2 \ln \left (y \right )+2 c_1}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{-c_1} \end{align*}

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

\[ \int _{}^{y}-\frac {1}{\sqrt {2 \ln \left (\tau \right )+2 c_1}}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} -\sqrt {2 \ln \left (y \right )+2 c_1}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{-c_1} \end{align*}

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

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