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

Feb 25, 2017

$P = {P}_{0} \setminus {e}^{k t}$

#### Explanation:

This is a first order separable DE, so:

$\setminus \mathrm{dP} - k P \setminus \mathrm{dt} = 0 \implies \mathrm{dP} = k P \setminus \mathrm{dt}$
$\therefore \int \setminus \frac{1}{P} \setminus \mathrm{dP} = \int \setminus k \setminus \mathrm{dt} \setminus \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$

Integrating gives us;

$\ln P = k t + C$

Using the initial Condition $P \left(0\right) = {P}_{0}$ we have:

$\ln {P}_{0} = 0 + C$
$\therefore C = \ln {P}_{0}$

So the solution becomes;

$\setminus \ln P = k t + \ln {P}_{0}$
$\therefore P = {e}^{k t + \ln {P}_{0}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{k t} {e}^{\ln {P}_{0}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {P}_{0} \setminus {e}^{k t}$

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:

$\setminus \setminus \setminus \setminus \setminus \setminus {\int}_{{P}_{0}}^{P} \setminus \frac{1}{\psi} \setminus \mathrm{dp} s i = {\int}_{0}^{t} \setminus k \setminus \eta$

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

$\therefore \setminus \setminus \setminus \setminus \setminus \ln \left(\frac{P}{P} _ 0\right) = k t$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{P}{P} _ 0 = {e}^{k t}$

$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus P = {P}_{0} \setminus {e}^{k t} \setminus \setminus \setminus$, as above