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Differential equations [arctan(x) to tan(x)]

  1. Jun 17, 2015 #1
    1. The problem statement, all variables and given/known data
    dx/dt = (1+x2)et ; x(0) = 1

    1/1+x2 dx/dt = et
    ∫ 1/1+x2 (dx/dt) * dt = ∫ et dt
    ∫ 1/(1+x2) dx = ∫ et dt
    arctan(x) = et + c
    so x(t) = tan (et+c)

    2. Relevant equations


    3. The attempt at a solution

    dx/dt = (1+x2)et ; x(0) = 1

    1/1+x2 dx/dt = et
    ∫ 1/1+x2 (dx/dt) * dt = ∫ et dt
    ∫ 1/(1+x2) dx = ∫ et dt

    arctan(x) = et + c
    so x(t) = tan (et+c)

    I'm a bit confused by the last step. How does arctan(x) = et + c become x(t) = tan (et+c). Cheers!
     
  2. jcsd
  3. Jun 17, 2015 #2

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    For any function f, if x is in the domain of f and x=y, then f(x)=f(y). Also, we have tan(arctan z)=z for all z in the domain of arctan.

    By the way, you shouldn't write 1/(1+x2) as 1/1+x2. The latter is equal to 1+x2.
     
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