MATH 316, DIFFERENTIAL EQUATIONS, WINTER 2000

Answers to the First Midterm

Problem 1 (12 points) Suppose the temperature of a cup of coffee obeys Newton’s law of cooling. A cup of coffee is freshly poured and placed outdoors on a cold day. Suppose the coffee has a temperature of 200^{∘}F when it is first poured, and after ten minutes is has cooled down to 146^{∘}F. If the outside temperature is 38^{∘}F, then at what time does the coffee reach 86^{∘}F? (You may use the facts that ln \frac{2}{3} ≈ −0.4 and that ln \frac{8}{27} ≈ −1.2).

Answer: Let y be the temperature, in^{∘}F, as a function of time t, in minutes. Newton’s law of cooling states that y′ is proportional to the temperature difference, so y′ = k(y − 38). Separating, we get dy/(y − 38) = k dt. Integrating, we obtain ln |y − 38| = kt + C, or y − 38 = Ae^{kt}, where A is any constant. We plug in the first initial condition t = 0 and y = 200, to obtain A = 162. Next, we plug in the second “initial” condition t = 10 and y = 146 to obtain 146 − 38 = 162e^{10k}, or k = \frac{1}{10} ln \frac{108}{162} = \frac{1}{10} ln \frac{2}{3}. Finally, we set

y = 86 and solve for t. We get 86 − 38 = 162e^{kt}, or t = \frac{1}{k} ln \frac{48}{162} = 10 ln \frac{8}{27} ln \frac{2}{3} = 10 · 3 = 30. Thus the coffee reaches 86^{∘}F thirty minutes after it is poured.

Problem 2 (12 points) For each of the following differential equations, state its type (for instance linear, non-linear, separable, exact, homogeneous) and find the general solution to each equation. (You do not need to solve your answer for y).

(a) \frac{dy}{dx} = \frac{x^{2}}{y^{2}} + \frac{y}{x}.

Answer: This equation is homogeneous. If we do the substitution y = vx (thus y′ = v′x + v), we get v′x + v = 1/v^{2} + v, or v^{2}dv = dx/x. Integrating, we obtain v^{3}/3 = ln |x| + C. Resubstituting v = y/x, we get \frac{y^{3}}{3x^{3}} = ln |x| + C.

(b) y′ + y = t.

Answer: This is a linear equation of the form y′ + p(t)y = g(t), where p(t) = 1, g(t) = t. We find an integrating factor μ(t) = exp ∫ p(t)dt = e^{t}. We calculate ∫ μ(t)g(t)dt = ∫ te^{t}dt = te^{t} − e^{t} + C (using integration by parts). Thus the general solution is

y = \frac{∫ μ(t)g(t)dt + C}{μ(t)} = \frac{te^{t} − e^{t} + C}{e^{t}} = t − 1 + Ce^{−t}.

(c) (x^{2} + 2x^{3}y)\frac{dy}{dx} + 2xy + 3x^{2}y^{2} = 0.

Answer: This is an exact equation, because \frac{d}{dx}(x^{2} + 2x^{3}y) = 2x + 6x^{2}y = \frac{d}{dy} (2xy + 3x^{2}y^{2}). We find the potential ϕ(x, y) = x^{2}y + x^{3}y^{2}. The solution is ϕ(x, y) = C, or x^{2}y + x^{3}y^{2} = C.

Problem 3 (12 points) Suppose that the size y of a certain fish population in a pond is governed by the differential equation

y′ = y(1 − \frac{y}{400}) − R.

Here R is the rate at which fish are being removed from the pond (and turned into cat food by a nearby cannery).

(a) Suppose the cannery harvests fish at a rate of R = 64. Sketch a slope field for the differential equation. Make sure you indicate the scale on the y-axis. In your slope field, draw all equilibrium solutions, and at least two increasing and two decreasing solution curves.

Answer: We first determine the equilibrium solutions by setting y′ = 0:

y(1 − y/400) − 64 = 0 ⟺ −y^{2}/400 + y − 64 = 0
⟺ y = ^{\frac{−1±\sqrt{1−64/100}}{−2/400}}
⟺ y = 80 or y = 320.

Moreover, y′ is positive for 80 < y < 320, and negative for y < 80 and 320 < y. Thus we get the following sketch:

(b) With R = 64, what will be the size of the fish population in the long run (as t → ∞) if y(0) = 100? What if y(0) = 50?

Answer: From the slope field, you can see that y will tend towards the stable equilibrium (y=320) if the initial condition is y(0) = 100. If the initial condition is y(0) = 50, the fish population will go to zero.

(c) What is the maximal rate R at which the cannery can harvest fish without risking the extinction of the fish population?

Answer: If the harvest rate is R, then the equilibrium solutions are determined by y = ^{\frac{−1±\sqrt{1−R/100}}{−2/400}} as in part (a). By looking at the term under the square root, we find that when R > 100, then there is no equilibrium solution and y′ is always negative, whereas when R ⩽ 100, there are some equilibrium solutions. Thus R = 100 is the maximum harvest rate at which the fish population has a chance to survive.