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Math Help - Differential eq problem and area problem

  1. #1
    Newbie
    Joined
    May 2009
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    Differential eq problem and area problem

    A:
    State a differential equation that has the solution:
    y=3sin(4x+v)

    B:
    decide the number k that the area beneath the curve y=ke^-x on the first square will be 2e areaunits


    Would be great if you showed how to solve these problems, step by step. thanks
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    408

    Thoughts

    A. y=3\sin(4x+v) \rightarrow y'=12\cos(4x+v) \rightarrow y''=-48\sin(4x+v)=-16y, so the differential equation y''=-16y has the solution you seek, where v is a constant of integration.

    B. Area= \int_a^b ke^{-x}dx=-ke^{-x}|_a^b=-k(e^{-b}-e^{-a}) . I am not sure what is meant by "on the first square" here. You are seeking a k for which the Area equals 2e, but I do not know what this phrase is implying for choosing a and b.
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