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Math Help - Calculus AB exam prep, dy/dt = my

  1. #1
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    Question Calculus AB exam prep, dy/dt = my

    Howdy everyone. I was wondering if someone here could help me out with this test prep question:

    If dy/dt = my and m is a nonzero constant, then y could be

    (A) 4e^(mty)
    (B) 4e^(mt)
    (C) e^(mt) + 4
    (D) mty + 4
    (E) (m/2)y^2 +4

    I thought that the first logical step would be antidifferentiation, which, in my mind, produced:

    y = tm * (y^2/2)

    I then multiplied out and got :

    y = (tmy^2)/2

    multiplying both sides by two yields :

    2y = tmy^2

    dividing both sides by y yields:

    2 = tmy

    Finally, dividing by tm has me believing that y= 2/tm

    However, that is not an answer choice, lol.

    All help is appreciated!

    Have a good one.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by EuptothepiI View Post
    Howdy everyone. I was wondering if someone here could help me out with this test prep question:

    If dy/dt = my and m is a nonzero constant, then y could be

    (A) 4e^(mty)
    (B) 4e^(mt)
    (C) e^(mt) + 4
    (D) mty + 4
    (E) (m/2)y^2 +4
    Differentiate A through E and see which give my for the derivative.

    CB
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by EuptothepiI View Post
    Howdy everyone. I was wondering if someone here could help me out with this test prep question:

    If dy/dt = my and m is a nonzero constant, then y could be

    (A) 4e^(mty)
    (B) 4e^(mt)
    (C) e^(mt) + 4
    (D) mty + 4
    (E) (m/2)y^2 +4
    The type of equation this is rules out D and E already as e is involved.

    You can solve this by separating the variables:

    \frac{dy}{y} = m\,dt

    \int \frac{dy}{y} = \int m\,dt

    Spoiler:
    ln(y) = mt + k

    Let k = ln(C)

    y = e^{mt+ln(C)} = e^{mt}\cdot e^{ln(C)} = Ce^{mt}
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