미분방정식 전체 문제

Answers will vary.

Answers will vary.

Answers will vary.

Answers will vary.

$C = 2$

$C = 6$

(a) Since $|y| \geq 0$ no matter what $y$ is, we see $Cx \ge 0$ for all $x$ in the domain of $f(x)$. Since $C$ is positive, that means the domain of $f(x)$ only includes nonnegative numbers. So, the largest possible domain of $f(x)$ is $[0,\infty)$. (b) None exists.

The graph of $Cx$ is given below for some positive constant $C$, also with the graph of $-Cx$. If $y=f(x)$ were sometimes the top function, and other times the bottom function, then there would be a jump discontinuity where it switched. Then the derivative of $f(x)$ would not exist, violating the second property.

A tiny technical note is that it's possible that $f(x)=Cx$ when $x=0$ and $f(x)=-Cx$ when $x>0$ (or vice-versa). This would not introduce a jump discontinuity, but it also does not satisfy that $f(x)>0$ for some values of $x$.

Remark: in several instances below, solving a differential equation will lead us to conclude something like $|y|=g(x)$. In these cases, we choose either $y=g(x)$, or $y=-g(x)$, but not $y=\pm g(x)$ (which is not a function) or that $y$ is sometimes $g(x)$, and other times $-g(x)$. The reasoning above somewhat explains this choice: if $y$ were sometimes positive and sometimes negative, then $\frac{\mathrm{d}y}{\mathrm{d}x}$ would not exist at the values of $x$ where the sign of $y$ switches, unless that switch occurrs at a root of $g(x)$. Since that's a pretty specific occurrence, we usually feel safe ignoring it to avoid getting bogged down in technical details.

(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) Sketching the slopes at individual points by computing $\frac{dy}{dt}\vert_{(t,y)}$ at various values of $(t,y)$, we see the following slope field. (b) By following the slope field so that we pass through the points $(0,-4)$, $(0,-3)$, $\ldots$, $(0,4)$, we see the collection of curves in the figure above. (c) It looks like the solution that passes through $(0,-1)$ is a straight line with slope $m = 1$, and thus has the formula $y(t) = t-1$. To verify this function is a solution, we compute both $\frac{dy}{dt}$ and $t-y$. Doing so, we see $$\frac{dy}{dt} = 1$$ and $$t-y = t - (t-1) = 1\,.$$ Thus, $y(t) = t-1$ is indeed a solution to the differential equation. (d) From both the differential equation and the graphs we have sketched, it appears that $t$ and $y$ are equal at any point where a solution has a local minimum. This is because when $t = y$, $\frac{dy}{dt} = t - y = 0$, and having the derivative be zero is required for any local minimum on a differentiable function.

(1) (2) The only equilibrium solution is $P = 3$, since $\ln(1) = 0$ (note: $\ln(0)$ is undefined, so the differential equation is not defined for $P = 0$). From the slope field pictured in (a), the equilibrium solution $P = 3$ is stable. (3) To solve the IVP with $\frac{dP}{dt} = -P\ln\left(\frac P3\right)$ and $P(0) = 1$, we use separation of variables. Observe that $$\frac{1}{P\ln\left(\frac P3\right)}\frac{dP}{dt} = -1$$ and thus $$\int \frac{1}{P\ln\left(\frac P3\right)} \, dP = - \int 1 \, dt$$ On the left, using the substitution $u = \ln (P/3)$, it follows $du = \frac{3}{P} \cdot \frac{1}{3} \, dP$, we can rewrite the most recent equation as $$\int \frac{du}{u} = - \int 1 \, dt\,.$$ Therefore, $$\ln |u| = -t + C\,,$$ so $|u| = e^{C-t}$, or $u = Ke^{-t}$. Next, we recall that $u = \ln \left(\frac P3\right)$, so $$\frac{P}{3} = e^{Ke^{-t}}$$ and $P = 3e^{Ke^{-t}}$. Applying the initial condition, $1 = 3e^K$, so $K = \ln \left(\frac{1}{3} \right)$, and thus $$P(t) = 3e^{\ln \left(\frac{1}{3} \right) e^{-t}}\,.$$ This function's plot is shown in blue in the image above in (a) and we see (both graphically and algebraically) that $P(t) \to 3$ as $t \to \infty$ since $e^{-t} \to 0$. (4) Changing the initial condition to $P(0) = 6$, we can use all of our preceding work in (c) up to where we established $P = 3e^{Ke^{-t}}$. Applying this initial condition, $6 = 3e^K$, so $K = \ln \left(2 \right)$, and thus $$P(t) = 3e^{\ln \left( 2 \right) e^{-t}}\,.$$ This function's plot is shown in red in the image above in (a) and we see (both graphically and algebraically) that $P(t) \to 3$ as $t \to \infty$ since $e^{-t} \to 0$. (5) Our work in (c) and (d) suggests that regardless of the initial condition with $P(0) > 0$, the long-term behavior of $P$ will be for it to tend to $3$ as $t$ increases without bound. This matches our initial observations based on the slope field.