Student Question

# `dP - kPdt = 0 , P(0) = P_0` Find the particular solution of this differential equation that satisfies the given initial condition.

In the following answer, I assume that k and `P_0` are constants.

Then, the given differential equation can be solved by separation of variables:

`dP - kPdt = 0`

`dP = kPdt`

Dividing by P results in

`(dP)/P = kdt` .

Integrating both sides, we obtain

`lnP = kt + C` , where C is an arbitrary constant. We can now solve for P(t) by rewriting the natural logarithmic equation as an exponential (with the base e) equation:

`P = e^(kt + C)` .

So, the general solution of the equation is `P(t) = e^(kt + C)` . Since the initial condition is `P(0) = P_0` , we can find C:

`P(0) = e^(0 + C) = e^C = P_0` . Therefore,

`P(t) = e^(kt)*e^C = P_0e^(kt)`

The particular solution of the equation with the given initial condition is

`P(t) = P_0e^(kt)`

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