Comment #3 on issue 3993 by [email protected]: Cannot solve x*y'' - xy' + y
= 0
http://code.google.com/p/sympy/issues/detail?id=3993
This is a 2nd order linear equation with variable coefficients. sympy
currently correctly finds 1 solution. The second can be found using
reduction of order, or directly through series methods. By steps in
sympy, ...
In [18]: -x*Derivative(y(x), x) + x*Derivative(y(x), x, x) + y(x)
Out[18]:
2
d d
- x⋅──(y(x)) + x⋅───(y(x)) + y(x)
dx 2
dx
In [19]: _.subs(y(x),x).doit()
Out[19]: 0
In [20]: dsolve(__)
Out[20]:
⎛ 6⎞
y(x) = C₀⋅x + O⎝x ⎠
In [21]: Out[18].subs(y(x),x*f(x))
Out[21]:
2
d d
x⋅f(x) - x⋅──(x⋅f(x)) + x⋅───(x⋅f(x))
dx 2
dx
In [22]: _.doit()
Out[22]:
⎛ 2 ⎞
⎛ d ⎞ ⎜ d d ⎟
- x⋅⎜x⋅──(f(x)) + f(x)⎟ + x⋅⎜x⋅───(f(x)) + 2⋅──(f(x))⎟ + x⋅f(x)
⎝ dx ⎠ ⎜ 2 dx ⎟
⎝ dx ⎠
In [23]: _.expand()
Out[23]:
2
2 d 2 d d
- x ⋅──(f(x)) + x ⋅───(f(x)) + 2⋅x⋅──(f(x))
dx 2 dx
dx
In [25]: __.subs(f(x).diff(x),g(x))
Out[25]:
2 2 d
- x ⋅g(x) + x ⋅──(g(x)) + 2⋅x⋅g(x)
dx
In [28]: dsolve(_,g(x))
Out[28]:
x
C₁⋅ℯ
g(x) = ─────
2
x
In [30]: _
Out[30]:
x
C₁⋅ℯ
g(x) = ─────
2
x
In [31]: Integral(_.rhs,x)
Out[31]:
⌠
⎮ x
⎮ C₁⋅ℯ
⎮ ───── dx
⎮ 2
⎮ x
⌡
In [32]: _.doit()
Out[32]:
⎛ x⎞
⎜ ℯ ⎟
C₁⋅⎜Ei(x) - ──⎟
⎝ x ⎠
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