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    Hi I'm having some trouble with this question - I can't see what I've done wrong but I'm not coming out with the same answer as the book. Any help?

    Nb: there is the possibility that the answer in the book might be wrong - it's done that before.

    Q: solve the differential equation

    dy/dx = 2ysec(^2)2x, given that y=2 when x=0

    My answer:

    dy/dx = 2ysec(^2)2x
    ?2y dy = ?sec²2x dx
    ln(2y) = 1/2 tan2x+c

    @ x=o,y=2

    ln(4)=1/2 tan(0) + c
    c = ln(4)

    =>ln(2y) = 1/2 tan2x+ln(4)

    2y = e^(1/2 tan2x + ln(4))
    2y = e^1/2 tan2x x e^ln(4)
    2y = 4e^1/2 tan2x
    y = 2e^1/2 tan2x

    answer is supposed to be:

    y = 2e^tan2x

    what happened to the 1/2 ?

    Thanks!
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    (Original post by SparksInTheSky)
    ?2y dy = ?sec²2x dx
    ln(2y) = 1/2 tan2x+c
    Not sure what that first line is meant to be, but:

    \int\frac{1}{2y}\not=\ln (2y)

    and your error is propagated through the rest of your working; and by the looks of things doesn't show up until the end.

    PS: Book's right.
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    (Original post by Farhan.Hanif93)
    The book right, you're wrong.
    Corrected :p: As per ghostwalker's post.
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    (Original post by nuodai)
    Corrected :p: As per ghostwalker's post.
    Oops, didn't notice that. :p:
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    Yeah should be 1/2ln(2y), or keep the 2 with the x
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    (Original post by ghostwalker)
    Not sure what that first line is meant to be, but:

    \int\frac{1}{2y}\not=\ln (2y)

    and your error is propagated through the rest of your working; and by the looks of things doesn't show up until the end.

    PS: Book's right.
    Yeah that was meant to be an intergral sign, dunno why it didn't come out right.

    Yeah I see it now, integral 1/2y = 1/2 integral 1/y = 1/2ln(y)

    then it all works out. Thanks! (feel a bit stupid now...!)
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    (Original post by SparksInTheSky)
    Yeah that was meant to be an intergral sign, dunno why it didn't come out right.
    Have a look at using LaTex; there's a link on the forum page, and at the top of this thread.
 
 
 
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