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Seperable diff eq, differing intial-conditions

  1. Jun 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve:

    [tex] \frac {dy}{dx} = 2x \sqrt{1-y^{2}} [/tex]

    then find a solution for:

    [tex] y(0)=0 [/tex]

    and can you find a solution for:

    [tex] y(0)=2 [/tex]

    2. Relevant equations



    3. The attempt at a solution

    Just want to know if this is right.

    First the equation can be rearranged to:

    [tex] \frac {dy}{\sqrt{1-y^{2}}}= 2x dx [/tex]

    [tex] \int \frac {1}{\sqrt{1-y^{2}}} dy = \int 2x dx [/tex]

    [tex] sin^{-1}(y) = x^{2}+c [/tex]

    [tex] y=sin(x^{2}+c) [/tex]

    for y(0)=0

    [tex] 0=sin(c) [/tex]

    which is valid when:

    [tex] c = 0 \;\; or \;\; c=k \pi, \;\; \forall \; k \; \in \; Z [/tex]

    and for y(0)=2

    [tex] 2=sin(c) [/tex]

    [tex] c = sin^{-1}(2) = \frac {\pi}{2}-\frac {ln(4\sqrt{3}+7)}{2} i[/tex]

    which is non-real. I assume that they are looking for me to realize that the answer isn't real? I am use to dealing with complex numbers from electrical engineering so this isn't that strange to me. Although I am not sure what this really means in this context. (the context being abstract math-land.)
     
  2. jcsd
  3. Jun 27, 2011 #2

    rock.freak667

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    Homework Helper

    I think they just wanted you to realize that for sinc>1 yields no real solutions for c.
     
  4. Jun 27, 2011 #3

    micromass

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    Staff Emeritus
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    2016 Award

    Hi Asphyxiated! :smile:

    That seems very good.

    Regarding the y(0)=2 case. When solve ODE's like this, it is assumed that we work in the real numbers (unless specified otherwise). So only the real solutions count. So fro y(0)=2 there are no solutions.
     
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