Comment on Activity 4

  1. We start by writing the differential equation in differential form with the variables separated:

    `(dP)/P^2=k dt`

    Now we antidifferentiate (formally) on both sides:

    An antidifferential of `(dP)/P^2=`some antidifferential of `k dt`.

    On the right, as before, we get `kt+C`. On the left, we use the Power Rule to find an antidifferential. We know that `d/(dP)P^-1=-P^-2`, so

    `d/(dP)(-1/P)=d/(dP)(-P^(-1))=1/P^2`.

    Thus

    `d(-1/P)=1/P^2 dP`.

    So now we know

    `-1/P=kt+C`.

    Next, we determine `C` from the initial condition. When we set `t=0` and `P=P_0`, we find that `C=-1//P_0`. Thus

    `-1/P=kt-1/P_0`.

    The rest of the problem is algebra; we have to solve for `P` as a function of `t`. If we take reciprocals of both sides of our equation, we get

    `-P=1/(kt-1/P_0)=P_0/(kP_0t-1)`.

  2. To differentiate `P`, we first write it in the form `P=P_0(1-kP_0t)^-1`. Then, using the Power Rule and the Chain Rule, we find

    3. `(dP)/(dt)=-P_0(1-kP_0t)^(-2)(-kP_0)=(kP_0^2)/(1-kP_0t)^2=kP^2`,

    so our solution function does indeed satisfy the differential equation. Finally, substituting `t=0` in the formula for the solution function, we see that `P=P_0` when `t=0`.