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

Question

Elijah Abram · Accepted Answer

# P = P_0 \ e^(kt) #

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

Emilia Aguilar · Answer

undefined To find the particular solution to the differential equation $\frac{{dP}}{{dt}} - kP = 0$ that satisfies the initial condition $P(0) = P_0$, we can separate variables and integrate both sides.

First, rearrange the equation to isolate the variables:
$$\frac{{dP}}{{dt}} = kP$$

Next, divide both sides by $P$ and multiply both sides by $dt$:
$$\frac{{dP}}{{P}} = k \, dt$$

Now, integrate both sides with respect to their respective variables:
$$\int \frac{{dP}}{{P}} = \int k \, dt$$

This yields:
$$\ln|P| = kt + C$$

where $C$ is the constant of integration.

To find the value of $C$, use the initial condition $P(0) = P_0$. Substitute $t = 0$ and $P = P_0$ into the equation:
$$\ln|P_0| = 0 + C$$

Solving for $C$, we get $C = \ln|P_0|$.

Substitute the value of $C$ back into the equation to find the particular solution:
$$\ln|P| = kt + \ln|P_0|$$

Exponentiate both sides to eliminate the natural logarithm:
$$|P| = e^{kt} \cdot |P_0|$$

Since the absolute value can be removed by considering the sign of $P_0$, we have:
$$P = P_0 \cdot e^{kt}$$

Therefore, the particular solution to the differential equation $\frac{{dP}}{{dt}} - kP = 0$ that satisfies the initial condition $P(0) = P_0$ is $P = P_0 \cdot e^{kt}$.