- #1
nonequilibrium
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Is there an insightful reason for the fact x*y''(x) + y'(x) + y(x) = 0 can go "bad" in x = 0 and y''(x) + x*y'(x) + y(x) = 0 doesn't?
Why xy''+y=0 a problem but y''+xy=0 not?
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Discussion Overview
The discussion explores the differences in behavior between the differential equations \( xy'' + y' + y = 0 \) and \( y'' + xy = 0 \), particularly focusing on the implications of their coefficients at \( x = 0 \). The scope includes theoretical considerations and potential implications in mathematical contexts.
Discussion Character
- Technical explanation
- Debate/contested
Main Points Raised
- One participant questions why \( xy'' + y' + y = 0 \) exhibits problematic behavior at \( x = 0 \) while \( y'' + xy = 0 \) does not.
- Another participant suggests that if a function misbehaves, its higher derivatives will misbehave even more strongly, implying a relationship between the behavior of the function and its derivatives.
- A different perspective notes that the "lead coefficient" \( x \) in the first equation is not a unit, drawing a parallel to algebraic geometry where similar behaviors are observed in equations like \( xy^2 + xy + 1 = 0 \).
- One participant raises a question about the possibility of having an arbitrary value for \( y'(0) \) if \( y(0) \) is specified, indicating uncertainty about initial conditions.
- Another participant points out that the first equation has two families of solutions, including Bessel functions, and expresses confusion about the characterization of the logarithmic behavior of one of the solutions as "bad."
Areas of Agreement / Disagreement
Participants express differing views on the implications of the equations' behaviors at \( x = 0 \), and there is no consensus on the nature of the problems associated with the first equation compared to the second.
Contextual Notes
Limitations include the dependence on the definitions of "bad" behavior and the unresolved nature of the implications of initial conditions on the solutions of the equations.
- #2
arildno
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Because, if a function misbehaves, its higher derivatives will misbehave even more strongly.
- #3
mathwonk
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because the "lead coefficient" X in the first one is not a unit.
the same thing happens in alkgebraic geometry with equations like xy^2 + xy + 1 = 0.
i.e. as a function of y, it is quadratic for all x except x = 0. so projection on the x-axis is two to one except over x=0 where the graph of the curve goes off to infinity.
the same thing happens in alkgebraic geometry with equations like xy^2 + xy + 1 = 0.
i.e. as a function of y, it is quadratic for all x except x = 0. so projection on the x-axis is two to one except over x=0 where the graph of the curve goes off to infinity.
- #4
Dickfore
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Is the possible to have an arbitraray value for y'(0) if y(0) is given?
- #5
JJacquelin
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x*y''(x) + y'(x) + y(x) = 0 has two families of solutions :
One is the Bessel function of the first kind which is equal to 1 in x=0.
The other is the Bessel function of second kind whiich is equivalent to a logarithmic function in x close to 0.
It's strange to say that the logarithm go "bad" ! It behaves just like a logarithm usually behaves.
I cannot understand where is the problem.
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