변수분리형 미분방정식

(a) A differential equation that represents the given description is $$\frac{dM}{dt} = kM\,.$$ (b) Separating the variables to solve the differential equation yields $$\frac{1}{M} \, dM = k \, dt$$ so $$\int \frac{1}{M} \, dM = \int k \, dt\,.$$ Integrating and solving for $M$, we find $\ln|M| = kt + C$ and thus $|M| = e^{kt+C} = e^{kt}e^C = Ke^{kt}$ for some constant $K$. By absorbing the absolute value into the constant $K$, we have $M = Ke^{kt}$. Using the initial condition $M(0) = M_0$, it follows that $K = M_0$. Therefore, the mass at time $t$ is given by $$M(t) = M_0e^{kt}\,.$$ (c) If we let $5730$ years pass from time $0$, the remaining mass of the radioactive sample will be $\frac{1}{2}M_0$. This allows us to find the value of $k$ by setting $M(5730)$ equal to $\frac{1}{2}M_0$ and solving for $t$. Doing so, we have $\frac{1}{2}M_0 = M_0e^{5730k}$, and thus $\frac{1}{2} = e^{5730k}$. It follows that $$\ln\left(\frac{1}{2}\right) = 5730k$$ from which we find $k = \frac{1}{5730}\ln\left(\frac{1}{2}\right) = -\frac{\ln(2)}{5730}$. Hence, $$M(t) = M_0e^{-\frac{\ln(2)}{5730}t} \approx M_0e^{-0.000121t}$$ (d) Now we want to find the time $t$ so that $M(t) = \frac{1}{4}M_0$. Similar work to our recent efforts in (c) shows that $t = \frac{5730\ln(4)}{\ln(2)} \approx 11460$ years. This number makes sense because it takes $5730$ years for half of the sample to decay, then another $5730$ years for half to decay again, for a total of $(2)(5730)$ years for the substance to decay to one quarter of its original amount. (e) We want to find the time it takes for $M(t)$ to be $30$% of $M_0$, so we need to solve $0.3M_0 = M_0e^{-\frac{\ln(2)}{5730}t}$. Doing so in the usual way, we find that $t = -\frac{5730\ln(0.3)}{\ln(2)} \approx 9952.8$ years, so we estimate that the sample is about $9973$ years old.