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    Hi!
    I have become a bit stuck on this question the C3C4 Elmwood textbook.

    Given x and y are positive, dy/dx +y/x^2 =0 and y= e when x=1, show that the solution may be written in the form y=e^(x-1)

    My attempt:
    dy/dx = -y/x^2
    integral of 1/y with respect to y = integral of -1/x^2 with respect to x
    lny = 1/3x^(-3) +c
    y= Ae^(1/3x^(-3))
    When y=e and x=1
    e = Ae^(1/3)
    A=e / e^(1/3)
    A=e^(1-1/3)
    A=e^(2/3)

    Therefore
    y=e^(1/3)*e^(1/3x^(-3))
    y=e^(2/9x^3)
    Which is not what I am supposed to get :/

    Any suggestions to where I have gone wrong?
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    (Original post by Ruineth)
    Hi!
    I have become a bit stuck on this question the C3C4 Elmwood textbook.

    Given x and y are positive, dy/dx +y/x^2 =0 and y= e when x=1, show that the solution may be written in the form y=e^(x-1)

    My attempt:
    dy/dx = -y/x^2
    integral of 1/y with respect to y = integral of -1/x^2 with respect to x
    lny = 1/3x^(-3) +c
    y= Ae^(1/3x^(-3))
    When y=e and x=1
    e = Ae^(1/3)
    A=e / e^(1/3)
    A=e^(1-1/3)
    A=e^(2/3)

    Therefore
    y=e^(1/3)*e^(1/3x^(-3))
    y=e^(2/9x^3)
    Which is not what I am supposed to get :/

    Any suggestions to where I have gone wrong?
    please post a photo of the workings
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    (Original post by TeeEm)
    please post a photo of the workings
    Of course
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    (Original post by Ruineth)
    Of course
    I am off to work now but someone will help you real soon

    All the best !!
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    (Original post by Ruineth)
    Of course
    When integrating you need to raise the power by one and then divide, careful with the negative power - see your third and fourth lines of working.
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    (Original post by GingerGnu)
    When integrating you need to raise the power by one and then divide, careful with the negative power - see your third and fourth lines of working.
    Ah yes! Got it, stupid mistakes...

    Thank you so much
 
 
 
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