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