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Kreyszig ODE

  1. Sep 11, 2011 #1
    Hi! I think I have to ask this since I'm having health problems-

    from Kreyszig, for xy'=-y how do you verify the solution y=h(x)=clnx by differentiating
    y'=h'(x)=-clnx^2? I don't see how you get the x^2 term
    also for ODEs the solution is on an open interval a<x<b but how does it include special cases of the intervals -inf<x<b, a<x<inf, -inf<x<inf; wouldn't the open interval a<x<b exclude -inf<x<b?
    thanks very much!
  2. jcsd
  3. Sep 11, 2011 #2


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    Homework Helper

    That is not a solution.
    It is a separable differential equation.
  4. Sep 11, 2011 #3
    thanks very much!
    I separated leading to-
    so then you just find the substitution c/x? (I misread the text: l is /)
  5. Sep 11, 2011 #4


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    Well, you don't just "find" a substitution. You always solve an equation of the form f(y)= F(x) for y by taking [itex]f^{-1}[/itex] of both sides. The inverse function to ln(x) is, of course, [itex]e^x[/itex] so take the exponential of both sides:
    [tex]e^{-ln|y|}= e^{ln|x|+ C}[/tex]

    [itex]-ln|y|= ln|y^{-1}|[/itex] so the left side is [itex]y^{-1}= 1/y[/itex]. Because [itex]e^{a+ b}= e^ae^b[/itex] the right side is [itex]e^{ln|x|}e^C= C' |x|[/itex] where C'= e^C.
    That is, [itex]1/|y|= C'|x|[/itex] or [itex]|y|= C'/|x|[/itex]. We can then "absorb" the absolute values into C' by allowing it to be positive or negative.
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