2.3.26 Problem 26

Internal problem ID [10158]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 26
Date solved : Friday, April 10, 2026 at 01:41:29 AM
CAS classification : [[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

2.3.26.1 second order ode missing x

0.256 (sec)

\begin{align*} y^{\prime \prime }+\sin \left (y\right ) {y^{\prime }}^{2}&=0 \\ \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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\sin \left (y \right ) p \left (y \right )^{2} = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

2.3.26.2 Solved by factoring the differential equation

Time used: 0.050 (sec)

\begin{align*} p p^{\prime }+\sin \left (y \right ) p^{2}&=0 \\ \end{align*}

Writing the ode as

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

Therefore we need to solve the following equations

\begin{align*} \tag{1} p &= 0 \\ \tag{2} \sin \left (y \right ) p+p^{\prime } &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

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

\begin{align*} p = 0 \end{align*}

Solving gives

\begin{align*} p &= 0 \\ \end{align*}

Solving equation (2)

Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} p^{\prime } + q(y)p &= p(y) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(y) &=\sin \left (y \right )\\ p(y) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dy}}\\ &= {\mathrm e}^{\int \sin \left (y \right )d y}\\ &= {\mathrm e}^{-\cos \left (y \right )} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \mu p &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (p \,{\mathrm e}^{-\cos \left (y \right )}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} p \,{\mathrm e}^{-\cos \left (y \right )}&= \int {0 \,dy} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-\cos \left (y \right )}\) gives the final solution

\[ p = {\mathrm e}^{\cos \left (y \right )} c_1 \]

Simplifying the above gives

\begin{align*} p &= {\mathrm e}^{\cos \left (y \right )} c_1 \\ \end{align*}

Solving for constants of integration from initial conditions gives

\begin{align*} c_1 &= c_1 \\ \end{align*}

Substituting the above in the solution gives

\begin{align*} p &= {\mathrm e}^{\cos \left (y \right )} c_1 \\ \end{align*}

Summary of solutions found

\begin{align*} p &= 0 \\ p &= {\mathrm e}^{\cos \left (y \right )} c_1 \\ \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

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

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\). Integrating gives

\begin{align*} y &= c_2 \end{align*}

Solving for constants of integration from initial conditions gives

\begin{align*} c_2 &= c_2 \\ \end{align*}

Substituting the above in the solution gives

\begin{align*} y &= c_2 \\ \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = {\mathrm e}^{\cos \left (y\right )} 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

\[ \int _{}^{y}\frac {{\mathrm e}^{-\cos \left (\tau \right )}}{c_1}d \tau = x +c_3 \]

Solving for constants of integration from initial conditions gives

\begin{align*} c_1 &= c_1 \\ c_3 &= c_3 \\ \end{align*}

Substituting the above in the solution gives

\begin{align*} \int _{}^{y}\frac {{\mathrm e}^{-\cos \left (\tau \right )}}{c_1}d \tau &= x +c_3 \\ \end{align*}

Summary of solutions found

\begin{align*} \int _{}^{y}\frac {{\mathrm e}^{-\cos \left (\tau \right )}}{c_1}d \tau &= x +c_3 \\ y &= c_2 \\ \end{align*}

2.3.26.3 ✓ Maple. Time used: 0.003 (sec). Leaf size: 21

ode:=diff(diff(y(x),x),x)+sin(y(x))*diff(y(x),x)^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ \int _{}^{y}{\mathrm e}^{-\cos \left (\textit {\_a} \right )}d \textit {\_a} -c_1 x -c_2 = 0 \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
<- 2nd_order Liouville successful
 

2.3.26.4 ✓ Mathematica. Time used: 0.154 (sec). Leaf size: 129

ode=D[y[x],{x,2}]+y[x]*Sin[y[x]](D[y[x],x])^2==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\exp \left (-\int _1^{K[2]}-K[1] \sin (K[1])dK[1]\right )}{c_1}dK[2]\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {\exp \left (-\int _1^{K[2]}-K[1] \sin (K[1])dK[1]\right )}{c_1}dK[2]\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\exp \left (-\int _1^{K[2]}-K[1] \sin (K[1])dK[1]\right )}{c_1}dK[2]\&\right ][x+c_2] \end{align*}

2.3.26.5 ✗ Sympy

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

NotImplementedError : The given ODE -sqrt(-Derivative(y(x), (x, 2))/sin(y(x))) + Derivative(y(x), x)
 

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', 'Liouville', 'Liouville_Integral')