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Separable Differential Equation

  1. Sep 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve the differential equation:
    (ex+1)cosy dy + ex(siny +1)dx=0 y(0)=3

    2. Relevant equations
    none

    3. The attempt at a solution
    (ex+1)cosy dy + ex(siny +1)dx=0
    (ex+1)cosy dy =- ex(siny +1)dx
    cosy/(siny+1)dy=-ex/(ex+1)dx
    ∫cosy/(siny+1)dy=-∫ex/(ex+1)dx
    using u sub on both the left and right for u=siny+1 on left and u=(ex+1) on right yeilds
    ∫1/u du=-∫ 1/u du
    ln|siny +1|= -ln|ex+1| +c
    exponentiating both sides ( this is where I think I messed up)
    siny +1= -x-1 +c
     
  2. jcsd
  3. Sep 25, 2016 #2

    Mark44

    Staff: Mentor

    Yes, it's messed up.
    Use the ideas that a * ln(b) = ln(ba) and that exp(m + n) = exp(m) * exp(n)
     
  4. Sep 25, 2016 #3
    So all of my work is fine up until the last step?
     
  5. Sep 26, 2016 #4

    Mark44

    Staff: Mentor

    I don't see anything wrong, but you shouldn't use u for both of your substitutions -- maybe v or w for the other one.
     
  6. Sep 26, 2016 #5
    yea, I figured that was bad notation. Anyways, for the next step, is the left side of the equation is right? sin(y) +1

    as far as the right side and applying the rules you gave me. The right side would look something like this before the exponentiating:

    -(x+1)ln|e| +c
     
  7. Sep 27, 2016 #6

    Mark44

    Staff: Mentor

    You have a mistake that I didn't notice before.
    On the right side it should be ##-\ln(e^x + 1) + C##. You had ##-\ln|e^{x + 1}| + C##.
    Also note that I didn't use absolute values, since ##e^x > 0## for all real x, so ##e^x + 1 > 1 > 0## for all real x.

    You don't need absolute values on the left side, either, since ##\sin(y) + 1 \ge 0## for all real y.
     
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