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Initial Value Diff. Equation

  1. Sep 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Solve the Given initial value problem


    2. Relevant equations

    [tex] y' = tan ( x ), y ( pi / 4 ) = 3 [/tex]

    3. The attempt at a solution

    Well I got the answer

    [tex] y = sec^2 ( x ) + 5 / 2 [/tex]

    The books answer is

    [tex] y = 3 - ln (sqrt(2) cos(x) ) [/tex]

    Where did I got wrong? Or is the book screwed up?
     
  2. jcsd
  3. Sep 29, 2007 #2
    you integrated wrong, i don't know how they got sqrt(2) but you need to change tan(x) to sin(x)/cos(x) and do a substitution to get the log
     
  4. Sep 29, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What do YOU believe is the integral of tan(x)??
     
  5. Sep 29, 2007 #4
    I am no expert, but I get the books answer, based on y=- ln (cos x)+C, and C =3 - ln[(sqrt2) /2)] , which = 3+ ln (sqrt 2).

    Or y=-ln (cos x) -ln ( sqrt 2 ) +3

    y= - ln [ (cos x) sqrt 2] +3

    Dont bet your life on it. Jim
     
  6. Sep 30, 2007 #5
    sheesh... I differentiated instead of integrated. It's been a long summer. Thanks for your time guys, appreciate it.
     
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