변수분리형 미분방정식

(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.

(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.