변수분리형 미분방정식

$y={e}^{t}-1$

$y=1+C{e}^{-t}$

(a) Separating the variables, we find that $$y \frac{dy}{dt} = -t$$ and integrating with respect to $t$ $$\int y \, dy = -\int t \,dt\,.$$ Evaluating the integrals and solving for $y$, we see that $\frac{1}{2}y^2 = -\frac{1}{2}t^2 + C$ and therefore $$y^2 = C - t^2\,,$$ so $y = \pm \sqrt{C - t^2}$. Applying the initial condition $y(0) = 8$, we find that $8 = \sqrt{C}$ (we choose the $+$ from $\pm$ since $8 > 0$), so $C = 64$. Hence the solution to the IVP is $$y = \sqrt{64 - t^2}\,.$$ (b) The solution to the IVP is defined for $-8 \le t \le 8$. (c) The last time that the solution is defined is $t = 8$ and the corresponding value of $y$ at this point is $y(8) = 0$. (d) Recall that the differential equation is $\frac{dy}{dt} = -\frac ty$, and the righthand expression is undefined when $y = 0$. Hence the solution value is technically not defined at the point we noted in (c); this is connected to there being a vertical tangent line at this point in the direction field.

(a) Since $\frac{dh}{dt}$ is proportional to the square root of the height of the water, we have $$\frac{dh}{dt} = k \sqrt{h}\,.$$ (b) If one tank has $k=-1$ and the other has $k=-10$, the tank with $k = -10$ will have water leaving the tank much more rapidly, which should correspond to that tank having a larger hole. (c) Since height decreases at $20$ inches per minute when the water is filled to a depth of $100$ inches, this tells us that $\frac{dh}{dt}\vert_{h=100} = -20$. Applying this information to the differential equation, $$-20 = k \sqrt{100}$$ and thus $k = -2$. (d) To solve the differential equation $\frac{dh}{dt} = -2 \sqrt{h}$ with initial condition $h(0) = 100$, we first separate the variables and write $$\frac{1}{\sqrt{h}} \frac{dh}{dt} = -2\,.$$ Rewriting the first fraction and integrating with respect to $t$, we have $$\int h^{-1/2} \, dh = \int -2 \, dt$$ and thus $$2h^{1/2} = -2t + C\,,$$ or $h = \left( C - t \right)^2$. Knowing that $h(0) = 100$, we see that $100 = C^2$, and thus $C = 10$. This shows that the solution to the IVP is $$h(t) = \left(10 - t \right)^2\,.$$ A plot of this function shows it to be a concave up parabola with $y$-intercept $(0,100)$ and a repeated zero at $(10,0)$. (e) It takes $10$ minutes for the tank to empty since $h(10) = 0$. (f) No, this solution is not valid for all time $t$. The physical constraints of the problem tell us that the solution is only valid for $0 < t < 10$ since at $t = 10$ the tank is empty.