Use sympy solve differential equation with initial condition and get an implicit result

Question

Here is the question and what I got so far, but I do not know how I should go for the next to get C1. Solve the IVP (1/x + 2y^2x)dx + (2yx^2 - cos(y))dy = 0, y(1) = pi. Give an implicit solution

from sympy import *

x = symbols('x')

y = Function('y')

deq = diff(y(x),x) + (1/x + 2*y(x)**2*x)/(2*y(x)*x**2 - cos(y(x)))

ysoln = dsolve(deq, y(x))

Answer 1

The following should work:

from sympy import *

x = symbols('x')
y = Function('y')
deq = diff(y(x),x)*(2*y(x)*x**2 - cos(y(x))) + (1/x + 2*y(x)**2*x)

# this leads to an error
# ysoln = dsolve(deq, y(x), ics={y(0): pi})

# so we do it our own way
ysoln = dsolve(deq, y(x))
C1 = solve(ysoln.subs(x, 1).subs(y(1), pi), 'C1')[0]
ysoln = ysoln.subs('C1', C1)
print(ysoln)
# Eq(x**2*y(x)**2 + log(x) - sin(y(x)), pi**2)

My version couldn't solve the equation in the form that you have so I had to restructure deq a bit. It probably just a little problem with all the division.

Note that this is likely will not work for every ODE. It worked now because one can solve for the unique solution of C1 given the initial conditions. Also, in the future, SymPy might not use C1 as the name of the arbitrary constant and functions such as .subs('C1', C1) will not work in that case.

As an interactive session though, the above method will work just fine.