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    The voluem of liquid Vcm^3 at time t seconds satisfies

    -15DV/dt = 2V - 450

    Given that initially the volume is 300cm^3, find tot he nearest cm^3 the volume after 15 seconds.
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    (Original post by Claxxy)
    The voluem of liquid Vcm^3 at time t seconds satisfies

    -15DV/dt = 2V - 450

    Given that initially the volume is 300cm^3, find tot he nearest cm^3 the volume after 15 seconds.
    What are you stuck with in particular? Do you know the separation of variables method?
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    Not really
    We haven't been taught this section properly
    I just have no idea how to go about this
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    (Original post by Claxxy)
    Not really
    We haven't been taught this section properly
    I just have no idea how to go about this
    The way the method works is:

    First you rearrange the equation so that (roughly speaking) everything involving V appears on the same side as \frac{dV}{dt}, and everything involving t appears on the other side. So what you you get should look something like

    f(V)\frac{dV}{dt}=g(t)

    For some functions f and g.

    Once you have this, the next step is:

    \int f(v) dV = \int g(t) dt

    To get to this step, you can visualise bringing the dt over to the right-hand side of the equation and integrating both sides. This isn't acutally what happens, but it's what I was taught at A level and seems like a good way to make sense of the method.*

    Now you can just evaluate these two integrals as usual (remembering a constant of integration), and apply the initial conditions.

    *If you're interested, at this point we're actually integrating both sides with respect to t.
 
 
 
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