미분방정식의 기초

(a) Since $\frac{dT}{dt}= -\frac1{15}T+5$ and $T(0)=105$, $\frac{dT}{dt}\vert_{T=105} = -\frac{1}{15}(105) + 5 = -7 + 5 = -2$. The fact that $T(0) = 105$ tells us the coffee's temperature at time $t = 0$ is $T = 105$ degrees F, and $\frac{dT}{dt}\vert_{T=105} = -2$ tells us that at the instant $T = 105$ (i.e. when $t = 0$), the coffee's temperature is decreasing at an instantaneous rate of $-2$ degrees F per minute. (b) At $t = 0$, $T = 105$, and since $\frac{dT}{dt}\vert_{T=105}$ is negative, $T$ decreasing at $t=0$. (c) Using the linear approximation $T(1) \approx T(0) + \frac{dT}{dt}\vert_{t=0} (1-0)$ along with the given information, it follows that $$T(1) \approx 105 + (-2)(1) = 103$$ degrees F. (d) We note that $\frac{dT}{dt}$ is a linear function of $T$ with slope $-\frac{1}{15}$ and vertical intercept $(0,5)$, so its graph is as shown below. Note particularly that $\frac{dT}{dt}\vert_{T = 75} = 0$. (e) For $T < 75$, $\frac{dT}{dt} > 0$, so for these $T$-values, $T$ increases. For $T > 75$, $\frac{dT}{dt} < 0$, so for these $T$-values, $T$ decreases. (f) In the previous question, we observed that $75$ degrees F is the critical temperature that changes the behavior of $T$. There are several reasons to think that $75$ degrees F is room temperature: if the coffee's temperature is above room temperature ($75$), its temperature will decrease to room temperature, while if the coffee is below room temperature, its temperature will increase to room temperature. Moreover, if the coffee's temperature is $75$ (which would match the room temperature), then its rate of change is $0$ and the coffee's temperature doesn't change. (g) Verify that $T(t) = 75 + 30e^{-t/15}$ is the solution to the differential equation with initial value $T(0) = 105$. What happens to this solution after a long time? Assuming that $T(t) = 75 + 30e^{-t/15}$, we observe that $$\frac{dT}{dt} = 30e^{-t/15} \left(-\frac{1}{15} \right) = -2e^{-t/15}\,.$$ In addition, $$-\frac{1}{15}T + 5 = -\frac{1}{15}(75 + 30e^{-t/15}) + 5 = -5 - 2e^{-t/15} + 5 = -2e^{-t/15}\,.$$ The two preceding equations show that $$\frac{dT}{dt}= -\frac1{15}T+5$$ when we substitute $T(t) = 75 + 30e^{-t/15}$ in for $T$, and thus this function is a solution to the differential equation. Moreover, $T(0) = 75 + 30e^0 = 75 + 30 = 105$, so the function satisfies the initial condition, too. Finally, because $e^{-t/15} \to 0$ as $t \to \infty$, it follows that $T(t) = 75 + 30e^{-t/15} \to 75$ as $t \to \infty$, which shows that the solution function's value tends to room temperature as time increases without bound, as we expect.