2.2.11 Problem 12

Internal problem ID [9134]
Book : Second order enumerated odes
Section : section 2
Problem number : 12
Date solved : Sunday, March 30, 2025 at 02:23:04 PM
CAS classification : [_Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

Solved as second nonlinear ode solved by Mainardi Lioville method

Time used: 0.440 (sec)

Solve

$$\begin{array}{rl}3y^{\prime \prime }+\cos \left(x\right)y^{\prime }+\sin \left(y\right){y^{\prime }}^2& =0\end{array}$$

The ode has the Liouville form given by

$$\begin{array}{}\text{(1A)}& y^{\prime \prime }+f(x)y^{\prime }+g(y){y^{\prime }}^2& =0\end{array}$$

Where in this problem

$$\begin{array}{rl}f(x)& =\frac{\cos \left(x\right)}{3}\\ g(y)& =\frac{\sin \left(y\right)}{3}\end{array}$$

Dividing through by $y^{\prime }$ then Eq (1A) becomes

$$\begin{array}{}\text{(2A)}& \frac{y^{\prime \prime }}{y^{\prime }}+f+g{y}^{\prime }& =0\end{array}$$

But the first term in Eq (2A) can be written as

$$\begin{array}{}\text{(3A)}& \frac{y^{\prime \prime }}{y^{\prime }}& =\frac{d}{dx}\ln \left(y^{\prime }\right)\end{array}$$

And the last term in Eq (2A) can be written as

$$\begin{array}{rl}g\frac{dy}{dx}& =(\frac{d}{dy}\int gdy)\frac{dy}{dx}\\ \text{(4A)}& & =\frac{d}{dx}\int gdy\end{array}$$

Substituting (3A,4A) back into (2A) gives

$$\begin{array}{}\text{(5A)}& \frac{d}{dx}\ln \left(y^{\prime }\right)+\frac{d}{dx}\int gdy& =-f\end{array}$$

Integrating the above w.r.t. $x$ gives

$$\begin{array}{rl}\ln \left(y^{\prime }\right)+\int gdy& =-\int fdx+c_1\end{array}$$

Where $c_1$ is arbitrary constant. Taking the exponential of the above gives

$$\begin{array}{}\text{(6A)}& y^{\prime }& =c_2{e}^{\int -gdy}{e}^{\int -fdx}\end{array}$$

Where $c_2$ is a new arbitrary constant. But since $g=\frac{\sin \left(y\right)}{3}$ and $f=\frac{\cos \left(x\right)}{3}$, then

$$\begin{array}{rl}\int -gdy& =\int -\frac{\sin \left(y\right)}{3}dy\\ & =\frac{\cos \left(y\right)}{3}\\ \int -fdx& =\int -\frac{\cos \left(x\right)}{3}dx\\ & =-\frac{\sin \left(x\right)}{3}\end{array}$$

Substituting the above into Eq(6A) gives

$$y^{\prime }=c_2{\mathrm{e}}^{\frac{\cos \left(y\right)}{3}}{\mathrm{e}}^{-\frac{\sin \left(x\right)}{3}}$$

Which is now solved as first order separable ode. The ode

$$\begin{array}{}\text{(1)}& y^{\prime }=c_2{\mathrm{e}}^{\frac{\cos \left(y\right)}{3}}{\mathrm{e}}^{-\frac{\sin \left(x\right)}{3}}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{y}^{\prime }& =c_2{\mathrm{e}}^{\frac{\cos \left(y\right)}{3}}{\mathrm{e}}^{-\frac{\sin \left(x\right)}{3}}\\ & =f(x)g(y)\end{array}$$

Where

$$\begin{array}{rl}f(x)& ={\mathrm{e}}^{-\frac{\sin \left(x\right)}{3}}{c}_2\\ g(y)& ={\mathrm{e}}^{\frac{\cos \left(y\right)}{3}}\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(y)}dy& =\int f(x)dx\\ \int {\mathrm{e}}^{-\frac{\cos \left(y\right)}{3}}dy& =\int {\mathrm{e}}^{-\frac{\sin \left(x\right)}{3}}{c}_{2}dx\end{array}$$

$${\int }_{}^y{\mathrm{e}}^{-\frac{\cos \left(\tau \right)}{3}}d\tau =\int {\mathrm{e}}^{-\frac{\sin \left(x\right)}{3}}{c}_{2}dx+2c_3$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}{\int }_{}^y{\mathrm{e}}^{-\frac{\cos \left(\tau \right)}{3}}d\tau & =\int {\mathrm{e}}^{-\frac{\sin \left(x\right)}{3}}{c}_{2}dx+2c_3\end{array}$$

✓ Maple. Time used: 0.004 (sec). Leaf size: 27

ode:=3*diff(diff(y(x),x),x)+cos(x)*diff(y(x),x)+sin(y(x))*diff(y(x),x)^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

$${\int }_{}^y{\mathrm{e}}^{-\frac{\cos \left(\mathit{\text{\_a}}\right)}{3}}d\mathit{\text{\_a}}-c_1\int {\mathrm{e}}^{-\frac{\sin \left(x\right)}{3}}dx-c_2=0$$

Maple trace

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

✓ Mathematica. Time used: 1.388 (sec). Leaf size: 67

ode=3*D[y[x],{x,2}]+Cos[x]*D[y[x],x]+Sin[y[x]]*(D[y[x],x])^2==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \text{InverseFunction}[{\int }_1^{\mathrm{\#}\text{1}}\exp (-{\int }_1^{K[3]}-\frac{1}{3}\sin (K[1])dK[1])dK[3]\mathrm{\&}][{\int }_1^x-\exp (-{\int }_1^{K[4]}\frac{1}{3}\cos (K[2])dK[2])c_{1}dK[4]+c_2]$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(sin(y(x))*Derivative(y(x), x)**2 + cos(x)*Derivative(y(x), x) + 3*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

ZeroDivisionError : polynomial division