미분방정식의 기초

$y=4{e}^{3t}$

$y=3-2t+{t}^{2}$

$y=\frac{1}{k}({e}^{kt}-1)$ and $y=x$

(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.

(a) (i) To see if $y(t) = t + 1 + 2e^t$ is a solution to $\frac{dy}{dt} = y - t$, we compute $\frac{dy}{dt}$ and $y - t$ separately and see if they are equal. First, using $y(t) = t + 1 + 2e^t$, $$\frac{dy}{dt} = 1 + 2e^t\,.$$ Next, $$y - t = ( t + 1 + 2e^t ) - t = 1 + 2e^t\,.$$ Thus, $y(t) = t + 1 + 2e^t$ satisfies the DE and is indeed a solution. (ii) For $y(t) = t + 1$, observe that $$\frac{dy}{dt} = 1$$ and $$y - t = 1$$ as well, so $y(t) = t + 1$ is a solution to the DE. (iii) If $y(t) = t + 2$, then $$\frac{dy}{dt} = 1$$ but $$y - t = 2$$ as well, so $\frac{dy}{dt} \ne y-t$, and therefore $y(t) = t + 2$ is a not solution to the DE. (b) In order for $h(t) = 4\sin(3t)$ to be a solution to $$\frac{d^2h}{dt^2} = -kh\,,$$ it must make the equation true. To find $k$, we substitute the formula for $h$ into the DE and see what is required of $k$. We observe that for $h(t) = 4\sin(3t)$, $$\frac{dh}{dt} = 12\cos(3t)$$ and $$\frac{d^2h}{dt^2} = -36\sin(3t)\,.$$ To satisfy the DE $$\frac{d^2h}{dt^2} = -kh\,,$$ it follows that we must have $-36\sin(3t) = -kh = -k \cdot 4\sin(3t)$. Thus, we need $-36 = -4k$, and hence $k = 9$.

(a) $P$ increases for $1 < P < 3$ since $\frac{dP}{dt} > 0$ for these $P$-values. (b) $P$ decreases for $P < 1$ and $3 < P < 4$ since $\frac{dP}{dt} < 0$ for these $P$-values. (c) If $P(0) = 3$, $\frac{dP}{dt}\vert_{P=3} = 0$, so $P$ will not change at all with time since its derivative is $0$. (d) If the initial population satisfies $0< P(0)< 1$, we expect that the population will decrease toward $P = 0$ with $P$ always being positive. Since $\frac{dP}{dt}$ is always negative for $0< P < 1$, $P$ is always decreasing on this interval. The population can't drop below $P = 0$ because at that value of $P$, $\frac{dP}{dt} = 0$ and at that value the population wouldn't change. (e) If the initial population satisfies $1< P(0)< 3$, we expect that the population will increase toward $P = 3$ with $P$ always being between $1$ and $3$. Since $\frac{dP}{dt}$ is always positive for $1 < P < 3$, $P$ is always increasing on this interval. The population can't increase above $P = 3$ because at that value of $P$, $\frac{dP}{dt} = 0$ and at that value the population wouldn't change. (f) If the initial population satisfies $3< P(0)$, we expect that the population will decrease toward $P = 3$ with $P$ always being above $3$. Since $\frac{dP}{dt}$ is always negative for $P > 3$, $P$ is always decreasing on this interval. The population can't drop below $P = 3$ because at that value of $P$, $\frac{dP}{dt} = 0$ and at that value the population wouldn't change. (g) It looks like there are two thresholds for the population: there's a minimum threshold of $P = 1$, since below that population, the population decreases to $0$, and there's a maximum threshold of $P = 3$, where if we are above the minimum but below the maximum, the population works to increase toward the maximum threshold, while if the population is above the maximum, its value will fall back toward it.