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Math Help - Help with exam prep

  1. #1
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    Help with exam prep

    dy/dt = y + 2

    y(0) = 3

    find y(2)

    It's mainly the combination of dy/dt and y in the equation that's causing my confusion. Does this mean that when I take the antiderivative, I need to substitute so that it is with respect to y?
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  2. #2
    Junior Member Ziaris's Avatar
    Joined
    Dec 2008
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    \frac{dy}{dt}=y+2

    then

    \int\frac{dy}{y+2}=\int dt

    after moving differentials and integrating. Taking the antiderivative,

    \ln(y+2)=t+C which, after solving for y, leads to

    y=Ce^t - 2. Plugging in y(0)=3, 3=Ce^0 - 2 = C - 2 and after solving for C, C = 5.

    So our specific solution is y(t)=5e^t-2. Feel free to verify this by differentiating and using initial conditions. Finding y(2),

    y(2)=5e^2 - 2.

    (As a side note, go Ovie!)
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