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General solution

  1. Mar 12, 2008 #1
    im asked to find the general solution to the equation below and after find the implicit form ive done some work on it and just wanted to see if im going in the right direction.

    the equation

    dy[tex]/dx = e^-3y cos x (1+sin x )^2[/tex]



    My attempt at a general solution


    1[tex]/3 e^3 = (sin x +1)^3 = y =e^3/(sin + 1)^3 +c[/tex]

    and the implicit form as

    1[tex]/12 (1+sin x )^3/exp (3y)[/tex]

    this is harder maths than ive done so not as sure as with most.
     
  2. jcsd
  3. Mar 12, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Do you mean
    [tex]\frac{dy}{dx}= e^{-3y}cos x(1+ sin x)^2[/tex]?




    Surely you don't mean this- it makes no sense. That says that y is a constant, that it is equal to (sin x+ 1)3, and that it is equal to e3 over a constant!

    That last is not even an equation!
    If you meant
    [tex]\frac{dy}{dx}= e^{-3y}cos x(1+ sin x)^2[/tex]
    then it separates into
    [tex]e^{3y}dy= cos x (1+ sin x)^2 dx[/itex]
    The left side integrates easily and the substitution u= 1+ sin x makes the right side simple.
     
  4. Mar 12, 2008 #3
    no what i ment that y was equaled to

    e^3/(sinx+1)^3 +c
     
    Last edited: Mar 12, 2008
  5. Mar 12, 2008 #4

    Shooting Star

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    Homework Helper

    The second eqn doesn't follow from the first. How did you get it? Show the calculation.

    Look again at where you got the problem from. What HallsofIvy says makes the most sense.
     
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