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Thread: Hi. I'm struggling with this problem. Someone helps me please. Thank you.

  1. #1
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    Hi. I'm struggling with this problem. Someone helps me please. Thank you.

    Hi. I'm struggling with this problem. Someone helps me please. Thank you.-img_5847.jpg
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  2. #2
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    Re: Hi. I'm struggling with this problem. Someone helps me please. Thank you.

    $a=0.2,~b=0.1$

    $\dfrac {dV}{dt} = a V e^{-bt}$

    $\dfrac{dV}{V} = a e^{-b t} ~dt$

    $\ln(V) = -\dfrac a b e^{-b t} + C$

    $V(t) = e^{-\frac a b e^{-b t}}e^C = e^C e^{-2e^{-0.1 t}}$

    $V(0) = e^C e^{-2} = 1.86$

    $e^C \approx 13.7436$

    $V(t) \approx 13.7436 e^{-2e^{-0.1 t}}$
    Thanks from jackson990 and MarkFL
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Hi. I'm struggling with this problem. Someone helps me please. Thank you.

    Another approach is to write the given ODE in standard linear form:

    \frac{dV}{dt}-0.2e^{-0.1t}V=0

    Compute the integrating factor:

    \mu(t)=\exp\left(-0.2\int e^{-0.1t}\,dt\right)=e^{2e^{-0.1t}}

    Multiply through by the integrating factor:

    e^{2e^{-0.1t}}\frac{dV}{dt}-0.2e^{-0.1t}e^{2e^{-0.1t}}V=0

    Observe that the LHS can be rewritten as the derivative of a product:

    \frac{d}{dt}\left(e^{2e^{-0.1t}}V\right)=0

    Integrate through with respect to t:

    e^{2e^{-0.1t}}V=C

    Thus:

    V(t)=Ce^{-2e^{-0.1t}}

    Use the initial conditions to find C

    V(0)=Ce^{-2e^{-0.1(0)}}=Ce^{-2}=1.86\implies C=1.86e^2

    And so the solution to the IVP is:

    V(t)=1.86e^2e^{-2e^{-0.1t}}=1.86e^{2-2e^{-0.1t}}

    This is equivalent to the solution posted by romsek, just written in a slightly different form.
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