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    I just wanted to check this, I'm trying to get the general solution:

     sec x \frac {dy}{dx} = e^y

    I separate the variables to get
     (e^y)^{-1} \:dy = \frac {1}{sec x}\:dx

      (e^y)^{-1} =  e^{-y}

    because of the rule that  (a^n)^m = a^{nm}

    So  \int (e^y)^{-1} =  e^{-y}

    My answer is  e^{-y} = sin x + A

    The answer I have on the sheet is  -e^{-y} = sin x + A
    why is the e minus?

    thanks!
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    \displaystyle\int e^{ax+b}\ dx=\frac{1}{a}e^{ax+b}+c
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    You've separated the variables correctly. If you think about it:

    if y = e^f(x)
    then dy/dx = f'(x)e^f(x).

    Try differentiating e^-y and -e^-y.

    The first will give you -e^-y (dy/dx) and the latter will give you e^-y (dy/dx)
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    integral of e^(-y) is -e^(-y).
    Since integral of e^(ax+b) is (1/a)*e^(ax+b)
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    As you may well imagine, it's a really simple error:

    when you integrate e^-y with respect to y, you get -e^-y, because of the chain rule. The derivative of -y is -1, which multiples with e^-y to give -e^-y.
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    (Original post by mh1985)
    So  \int (e^y)^{-1} =  e^{-y}
    This bit is wrong.

    In general, for linear powers, we know that:

    \displaystyle\int ae^{bx+c}dx = \dfrac{a}{b}e^{bx+c} + K

    Where a,b and c are constants, and b\not=0

    Remember that this won't work in general, it has to be in the form I've highlight above for this exact 'rule' to follow.
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    Thanks a lot everyone! Will rep when on recharge
 
 
 
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