Question

Solve the following initial-value problem.

dp/dt = -kp, p(0) = 10.4, p(2) = 3.7

Answer 1

$$dp/dt=-kp$$

$$\dfrac{dp}{p}=-kdt$$

Integrate

$$\int\dfrac{dp}{p}=-\int kdt$$

$$\ln|p|=-kt+\ln C$$

$$p=Ce^{-kt}$$

The initial conditions

$$p(0)=10.4=>10.4=C$$

$$p=10.4e^{-kt}$$

$$p(2)=3.7=>3.7=10.4e^{-k(2)}$$

$$k=0.5\ln(\dfrac{10.4}{3.7})$$

$$p=10.4(\dfrac{10.4}{3.7})^{-0.5t}$$