How do you find the particular solution to dP-kPdt=0 that satisfies P(0)=P_0?

1 Answer
Feb 25, 2017

P = P_0 \ e^(kt)

Explanation:

This is a first order separable DE, so:

\ dP - kP \ dt = 0=> dP = kP \ dt
:. int \ 1/P \ dP = int \ k \ dt \ \ \ \ \ ..... [1]

Integrating gives us;

lnP = kt + C

Using the initial Condition P(0)=P_0 we have:

lnP_0 = 0 + C
:. C = lnP_0

So the solution becomes;

\ lnP = kt + lnP_0
:. P = e^(kt + lnP_0)
\ \ \ \ \ \ \ \ = e^(kt)e^(lnP_0)
\ \ \ \ \ \ \ \ = P_0 \ e^(kt)

We can also take an approach used by some texts/tutors where the initial conditions are incorporated directly in a definite integral.

Here we apply integration limits (using the initial condition) and arbitrarily change the dummy variable of integration to the integral [1] to get:

\ \ \ \ \ \ int_(P_0)^P \ 1/psi \ dpsi = int_0^t \ k \ eta

:. \ \ \ \ [ color(white)(""/"") ln psi \ ]_(P_0)^P = [color(white)(""/"") k \ eta \ ]_0^t
:. ln P - ln P_0 = kt - 0

:. \ \ \ \ \ ln (P/P_0) = kt
:. \ \ \ \ \ \ \ \ \ \ \ \ \ P/P_0 = e^(kt)

:. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P = P_0 \ e^(kt) \ \ \ , as above