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An integral equation

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Find [itex]y(\pi/3)[/itex] in the following integral equation:

    $$y(x)=e^{-\int_{1}^{x}\frac{y(t)}{\sin^2(t)}dt}$$
    2. Relevant equations

    3. The attempt at a solution

    Differentiating both sides gives a differential equation with the general solution [itex]y(x)=(-\cot(x)+c)^{-1}[/itex], and since y(1)=1, we have [itex]c=1+\cot(1)[/itex].

    Here is two questions:

    1. The hint says the answer is [itex]y(\pi/3)=3/(6-\sqrt{3})[/itex]. Is this correct?

    2. Is there any shortcut way to solve this problem without differentiating?

    Thanks in advance.
     
  2. jcsd
  3. Aug 1, 2012 #2
    No,no. You can check it you know. Here's how in Mathematica:

    Code (Text):

    In[20]:=
    y[x_] := 1/(-Cot[x] + Cot[1] + 1)
    tval = Pi/3;
    Exp[-NIntegrate[y[t]/Sin[t]^2, {t, 1, tval}]]
    N[3/(6 - Sqrt[3])]
    N[y[\[Pi]/3]]

    Out[22]= 0.939194

    Out[23]= 0.702914

    Out[24]= 0.939194
     
    at least that's what I'm turin' in.
     
    Last edited: Aug 1, 2012
  4. Aug 1, 2012 #3

    uart

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    You wont get the given answer with that initial condition. Are you sure it's not [itex]y(\pi/4) = 1\,[/itex]?
     
  5. Aug 1, 2012 #4

    HallsofIvy

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    Differentiating both sides of the integral equation gives
    [tex]\frac{dy}{dt}= \frac{y}{sin^2(t)}[/tex]
    which separates as
    [tex]\frac{dy}{y}= \frac{dt}{sin^2(t)}= cosec^2(t)dt[/tex]

    That integrates as
    [tex]ln(y)= - cot(t)+ C[/tex]
    Did you accidently write [itex]y^2[/itex] instead of y?
     
  6. Aug 1, 2012 #5
    I don't think so Hall. I belive it's

    [tex]y'=-\frac{y^2}{\sin^2(x)}[/tex]

    (forgot the negative sign)
     
    Last edited: Aug 1, 2012
  7. Aug 1, 2012 #6

    uart

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    I took logs of both sides first and then took derivatives, (using the chain rule on the LHS and the FTC on the RHS), to verify the OPs proposed solution was indeed correct.
     
  8. Aug 1, 2012 #7

    HallsofIvy

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    Oh, blast! I completely neglected the exponential part!

    Thanks.
     
  9. Aug 2, 2012 #8

    gabbagabbahey

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    The initial condition [itex]y(1)=1[/itex] comes from the original form of the question. So either there is a typo in the question, or the given answer:

    [tex]y(1)=\text{e}^{-\int_{1}^{1} \frac{ y(t) }{\sin^2 (t)} dt } = \text{e}^{0} = 1[/tex]

    @asmani, your solution looks fine to me :approve:
     
  10. Aug 3, 2012 #9

    uart

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    Arh yes I see. It's a definite integral so y(1)=1 is not an arbitrary initial condition.

    In that case then the answer in the book is definitely wrong.
     
  11. Aug 4, 2012 #10
    Thanks anybody. :wink:
     
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