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Thread: exponential differentiation

  1. May 12th 2009, 01:16 AM #1
    slaypullingcat
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    exponential differentiation

    show that y= Ae^kx+C is a solution of y'=k(y-C)

    Thank you
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  2. May 12th 2009, 01:52 AM #2
    Spec
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    \frac{d}{dx}\left(Ae^{kx}+C \right)=k\cdot Ae^{kx}
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