Why xy''+y=0 a problem but y''+xy=0 not?

  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #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.
 
  • #4
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Is the possible to have an arbitraray value for y'(0) if y(0) is given?
 
  • #5
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x*y''(x) + y'(x) + y(x) = 0 can go "bad" in x = 0
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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