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Math Help - help plz

  1. #1
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    help plz

    How would you solve this and find y?
    (e^(2x)(ydoubleprime) = 4(e^4x + 1)
    I did this:
    ydoubleprime = 4(e^4x + 1)/(e^(2x)
    ..but that doesnt seem to help, taking the integration of that looks difficult. Is there a different approach I could try? Thanks a lot!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    How would you solve this and find y?
    (e^(2x)(ydoubleprime) = 4(e^4x + 1)
    I did this:
    ydoubleprime = 4(e^4x + 1)/(e^(2x)
    ..but that doesnt seem to help, taking the integration of that looks difficult. Is there a different approach I could try? Thanks a lot!
    What's so hard?

    e^{2x}y^{\prime \prime} = 4(e^{4x} + 1)

    y^{\prime \prime} = 4\frac{e^{4x} + 1}{e^{2x}}

    y^{\prime \prime} = 4(e^{2x} + e^{-2x})

    You can either do this directly, or observe that the RHS can also be considered as:
    y^{\prime \prime} = 8 \cdot cosh(2x)

    -Dan
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