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Solve for y(x) using the Fundamental Theorem of Calculus

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the integral equation for y(x):
    y(x) = 1 + ∫ { [y(t)]^2 / (1 + t^2) } dt
    (integral from 0 to x)

    See attached image for the equation in a nicer format.

    2. Relevant equations
    Fundamental Theorem of Calculus

    3. The attempt at a solution
    dy/dx = y(x)^2 / (1 + x^2)
    ∫ dy/y^2 = ∫ dx/(1 + x^2)
    -1/y = arctan(x) + C
    y = -1/arctan(x) + C

    y(0) = 1 = -1/arctan(0) + C
    C = 1 + 1/arctan(0)
    Since arctan(0) = 0, the fraction is indeterminate.

    In lectures we only did one practice example, so if anyone could give me a hand and point out where I am going wrong, it'll be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Jan 20, 2015 #2

    Dick

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    Use parentheses! -1/arctan(x)+C is not the same thing as -1/(arctan(x)+C)! One is indeterminant, one is -1/C. Which one do you want?
     
  4. Jan 20, 2015 #3

    LCKurtz

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    This is a good example of why you need to not be sloppy about using parentheses.

    [Edit]: Dick wins again.
     
  5. Jan 21, 2015 #4
    I definitely want -1/C. Thank you for helping me figure it out. I knew it was something small but I couldn't see it.
     
  6. Jan 21, 2015 #5
    It wasn't that I was sloppy. I commonly leave the C on the outer edge of my equations because C cannot loss its generality. But it definitely reminded me to use the generality to my advantage. Thank you to Dick and yourself for helping students with their problems.
     
  7. Jan 21, 2015 #6

    Dick

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    Once you've put in the C after you integrate, you have to leave it where it is. You can't just move it someplace else later on without a good algebraic reason. It doesn't have that much 'generality'. -1/(arctan(x)+C) is a general solution to your ode. -1/arctan(x)+C is NOT.
     
  8. Jan 21, 2015 #7
    Thanks. I didn't know that.
     
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