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IVP problem

  1. Apr 14, 2007 #1


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    1. The problem statement, all variables and given/known data
    Please take a look at my work and help me figure out where I went wrong. Thanks!
    Use separation of variables to solve the IVP
    [tex]\[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0 [/tex]

    2. Relevant equations

    3. The attempt at a solution
    Use separation of variables to solve the IVP
    [tex] \[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0 [/tex]

    [tex]dy/dt = \sqrt{1-y^2}/t [/tex]

    [tex]1/\sqrt{1-y^2} dy = 1/t dt[/tex]

    [tex] \int 1/\sqrt{1-y^2} dy = \int 1/t dt[/tex]

    [tex] arcsin(y) = ln (abs (t)) + C [/tex]

    [tex] C = 1 [/tex]

    therefore the formula should be

    [tex]y = sin(ln(abs(t))+1) [/tex]

    Last edited: Apr 14, 2007
  2. jcsd
  3. Apr 15, 2007 #2

    Gib Z

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    Homework Helper

    I don't understand what you think is the problem.
  4. Apr 15, 2007 #3
    I don't see a problem except that your value of C is incorrect, because with that choice y(1) is not equal to 1.
  5. Apr 15, 2007 #4


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    arcsin(1) is [itex]\pi/2[/itex], not 1.

    I will also point out that this is of the form [itex]y'= \sqrt{1- y^2}/t[/itex] and the function is NOT Lipshchitz in y on any interval containing y= 1. Therefore, the solution is not unique. y= 1 for all t is another obvious solution and, in fact, there are an infinite number of solutions.
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