Internal
problem
ID
[10964]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
Additional
non-linear
first
order
Problem
number
:
968
Date
solved
:
Sunday, March 30, 2025 at 07:33:46 PM
CAS
classification
:
[[_homogeneous, `class D`]]
ode:=diff(y(x),x) = 1/2*(-sin(y(x)/x)*y(x)*x-y(x)*sin(y(x)/x)+y(x)*sin(3/2*y(x)/x)*cos(1/2*y(x)/x)*x+y(x)*sin(3/2*y(x)/x)*cos(1/2*y(x)/x)+y(x)*cos(1/2*y(x)/x)*sin(1/2*y(x)/x)*x+y(x)*cos(1/2*y(x)/x)*sin(1/2*y(x)/x)+2*sin(y(x)/x)*x^4*cos(1/2*y(x)/x)*sin(1/2*y(x)/x))/cos(y(x)/x)/cos(1/2*y(x)/x)/sin(1/2*y(x)/x)/x/(1+x); dsolve(ode,y(x), singsol=all);
ode=D[y[x],x] == (Csc[y[x]/(2*x)]*Sec[y[x]/(2*x)]*Sec[y[x]/x]*(x^4*Cos[y[x]/(2*x)]*Sin[y[x]/(2*x)]*Sin[y[x]/x] + (Cos[y[x]/(2*x)]*Sin[y[x]/(2*x)]*y[x])/2 + (x*Cos[y[x]/(2*x)]*Sin[y[x]/(2*x)]*y[x])/2 - (Sin[y[x]/x]*y[x])/2 - (x*Sin[y[x]/x]*y[x])/2 + (Cos[y[x]/(2*x)]*Sin[(3*y[x])/(2*x)]*y[x])/2 + (x*Cos[y[x]/(2*x)]*Sin[(3*y[x])/(2*x)]*y[x])/2))/(x*(1 + x)); ic={}; DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
from sympy import * x = symbols("x") y = Function("y") ode = Eq(Derivative(y(x), x) - (2*x**4*sin(y(x)/(2*x))*sin(y(x)/x)*cos(y(x)/(2*x)) + x*y(x)*sin(y(x)/(2*x))*cos(y(x)/(2*x)) - x*y(x)*sin(y(x)/x) + x*y(x)*sin(3*y(x)/(2*x))*cos(y(x)/(2*x)) + y(x)*sin(y(x)/(2*x))*cos(y(x)/(2*x)) - y(x)*sin(y(x)/x) + y(x)*sin(3*y(x)/(2*x))*cos(y(x)/(2*x)))/(2*x*(x + 1)*sin(y(x)/(2*x))*cos(y(x)/(2*x))*cos(y(x)/x)),0) ics = {} dsolve(ode,func=y(x),ics=ics)
PolynomialDivisionFailed : couldnt reduce degree in a polynomial division algorithm when dividing [[], []] by [[ANP([mpq(1,1)], [mpq(1,1), mpq(0,1), mpq(1,1)], QQ), ANP([mpq(-2,1), mpq(0,1)], [mpq(1,1), mpq(0,1), mpq(1,1)], QQ), ANP([mpq(-2,1)], [mpq(1,1), mpq(0,1), mpq(1,1)], QQ)]]. This can happen when its not possible to detect zero in the coefficient domain. The domain of computation is QQ<I>. Zero detection is guaranteed in this coefficient domain. This may indicate a bug in SymPy or the domain is user defined and doesnt implement zero detection properly.