2.1.69 Problem 69
Internal
problem
ID
[10055]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
69
Date
solved
:
Wednesday, April 29, 2026 at 06:35:00 AM
CAS
classification
:
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
2.1.69.1 second order ode missing x
11.217 (sec)
\begin{align*}
a y^{2} y^{\prime \prime }+b y^{2}&=c \\
\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*} a \,y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+b \,y^{2} = c \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 {b \,y^{2}-c}{a \,y^{2} p}
\end{equation}
is separable as it can be written as
\begin{align*} p^{\prime }&= -\frac {b \,y^{2}-c}{a \,y^{2} p}\\ &= f(y) g(p) \end{align*}
Where
\begin{align*} f(y) &= -\frac {b \,y^{2}-c}{a \,y^{2}}\\ 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 {b \,y^{2}-c}{a \,y^{2}} \,dy} \\
\end{align*}
\[
\frac {p^{2}}{2}=\frac {-b \,y^{2}-c}{a y}+c_1
\]
Simplifying the above gives
\begin{align*}
\frac {p^{2}}{2} &= \frac {c_1 a y -b \,y^{2}-c}{a y} \\
\end{align*}
Solving for \(p\) gives
\begin{align*}
p &= \frac {\sqrt {2}\, \sqrt {a y \left (c_1 a y -b \,y^{2}-c \right )}}{a y} \\
p &= -\frac {\sqrt {2}\, \sqrt {a y \left (c_1 a y -b \,y^{2}-c \right )}}{a y} \\
\end{align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is
\begin{align*} y^{\prime } = \frac {\sqrt {2}\, \sqrt {a y \left (y c_1 a -b y^{2}-c \right )}}{a y} \end{align*}
Entering first order ode autonomous solverIntegration could not be done or too complicated. Since
initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as
\begin{align*} \int _{}^{y}-\frac {a \tau \sqrt {2}}{2 \sqrt {a \tau \left (c_1 a \tau -b \,\tau ^{2}-c \right )}}d \tau +x +c_2 = 0\end{align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is
\begin{align*} y^{\prime } = -\frac {\sqrt {2}\, \sqrt {a y \left (y c_1 a -b y^{2}-c \right )}}{a y} \end{align*}
Entering first order ode autonomous solverIntegration could not be done or too complicated. Since
initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as
\begin{align*} \int _{}^{y}\frac {a \tau \sqrt {2}}{2 \sqrt {a \tau \left (c_1 a \tau -b \,\tau ^{2}-c \right )}}d \tau +x +c_3 = 0\end{align*}
Summary of solutions found
\begin{align*}
\int _{}^{y}\frac {a \tau \sqrt {2}}{2 \sqrt {a \tau \left (c_1 a \tau -b \,\tau ^{2}-c \right )}}d \tau +x +c_3 &= 0 \\
\int _{}^{y}-\frac {a \tau \sqrt {2}}{2 \sqrt {a \tau \left (c_1 a \tau -b \,\tau ^{2}-c \right )}}d \tau +x +c_2 &= 0 \\
\end{align*}
2.1.69.2 ✓ Maple. Time used: 0.013 (sec). Leaf size: 76
ode:=a*y(x)^2*diff(diff(y(x),x),x)+b*y(x)^2 = c;
dsolve(ode,y(x), singsol=all);
\begin{align*}
a \int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} a \left (c_1 \textit {\_a} a -2 b \,\textit {\_a}^{2}-2 c \right )}}d \textit {\_a} -x -c_2 &= 0 \\
-a \int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} a \left (c_1 \textit {\_a} a -2 b \,\textit {\_a}^{2}-2 c \right )}}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
-> Computing symmetries using: way = exp_sym
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+(_a^2*b-c)/_a^2/a = 0,
_b(_a)
*** Sublevel 2 ***
Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
<- Bernoulli successful
<- differential order: 2; canonical coordinates successful
<- differential order 2; missing variables successful
2.1.69.3 ✓ Mathematica. Time used: 0.495 (sec). Leaf size: 346
ode=a*y[x]^2*D[y[x],{x,2}]+b*y[x]^2==c;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\[
\text {Solve}\left [-\frac {\left (\sqrt {-16 b c+a^2 c_1{}^2}-a c_1\right ) \left (\sqrt {-16 b c+a^2 c_1{}^2}+a c_1\right ){}^2 \left (1+\frac {4 b y(x)}{\sqrt {-16 b c+a^2 c_1{}^2}-a c_1}\right ) \left (1-\frac {4 b y(x)}{\sqrt {-16 b c+a^2 c_1{}^2}+a c_1}\right ) \left (E\left (i \text {arcsinh}\left (2 \sqrt {\frac {b}{\sqrt {a^2 c_1{}^2-16 b c}-a c_1}} \sqrt {y(x)}\right )|\frac {a c_1-\sqrt {a^2 c_1{}^2-16 b c}}{a c_1+\sqrt {a^2 c_1{}^2-16 b c}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (2 \sqrt {\frac {b}{\sqrt {a^2 c_1{}^2-16 b c}-a c_1}} \sqrt {y(x)}\right ),\frac {a c_1-\sqrt {a^2 c_1{}^2-16 b c}}{a c_1+\sqrt {a^2 c_1{}^2-16 b c}}\right )\right ){}^2}{16 b^3 y(x) \left (-\frac {2 \left (b y(x)^2+c\right )}{a y(x)}+c_1\right )}=(x+c_2){}^2,y(x)\right ]
\]
2.1.69.4 ✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
b = symbols("b")
c = symbols("c")
y = Function("y")
ode = Eq(a*y(x)**2*Derivative(y(x), (x, 2)) + b*y(x)**2 - c,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
Timed Out
Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0]
Sympy version 1.14.0