Hello,
Consider the following differential equation
y=function('y')(x)
eq=6*diff(y,x)-2*y==x*y^4
sol=desolve(eq,y,show_method=True)
This code gives the following solution
[e^(1/3*x)/(-1/2*(x - 1)*e^x + _C)^(1/3), 'bernoulli']
When I solve this with Mathematica it gives me 3 solutions for y(x) (for each root, as expected).
Question #1: Why Sage gives one solution? When I solve this differential equation analytically, I would multiply both sides by y^(-4) (assuming y!=0) and then change the variable y^3=v. The differential equation becomes a 1st order linear differential equation in the dependent variable v and independent variable x. After solving for v, the cube-root brings the 3 solutions Mathematica is giving.
Question #2: Ok, this is really bugging me the most: Consider the same differential equation but this time let's also give the initial condition y(0)=-2
y=function('y')(x)
eq=6*diff(y,x)-2*y==x*y^4
sol=desolve(eq,y,show_method=True,ics=[0,-2])
This gives the following
[e^(1/3*x)/(-1/2*(x - 1)*e^x - 1/2)^(1/3), 'bernoulli']
However, this function blows up at x=0! How can I have a solution by calling desolve with the initial condition y(0)=-2 and end up with a solution that doesn't satisfy the initial condition ?!
Question #3: I tried to figure out the constant by myself with the following code:
y=function('y')(x)
eq=6*diff(y,x)-2*y==x*y^4
sol=desolve(eq,y,show_method=True)
c=sol[0].variables()[0] #<--- this refs _C
icsol=solve(sol[0].subs(x==0)==-2,c)
icsol gives me
[(_C + 1/2)^(1/3) == (-1/2)]
Why doesn't it give as [_C==...] ?
Question #4: For the code given in Question #3, if I try to substitute icsol back in the solution as
sol[0].subs(icsol)
I got sol[0] thrown back at me unchanged since there is no (_C + 1/2)^(1/3) in sol[0].
How can I even substitute back the constant C in the solution?
Thanks in advance for your help.
