**Example 1:** Find the orthogonal trajectories of the family of curves y=k/x. Use a graphing device to draw several members of each family on a common screen.

**Hint:** Differentiate xy=k and get xdy+ydx=0. Therefore the orthogonal trajectories satisfy xdy - ydx=0.

**Example 2:** We formulate a model for learning in the form of the differential equation dP / dt =k(M-P), where P(t) measures the performance of someone learning a skill after a training time t, M is the maximum level of performance, and k is a positive constant. Solve this differential equation to find an expression for P(t). What’s the limit of this expression?

**Hint:** Use separation of variables and get dP / M-P = kdt.

**Example 3:** A model for the concentration C=C(t) of the glucose solution in the bloodstream is dC / dt = r-kC, where k is a positive constant. **(a)** Suppose that the concentration at t=0 is C_0. Determine the concentration at any time t by solving the differential equation. **(b)** Assuming that C_0 < r/k, find \lim_{t\to\infty} C(t) and interpret your answer.

**Hint: (a)** Use separation of variables and get dC / r-kC = dt. **(b)** In the long run, C(t) tends to r/k.

**Example 4:** One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor. **(a)** Write a differential equation that is satisfied by y. **(b)** Solve the differential equation. (c) A small town has 1000 inhabitants. At 8 AM, 80 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor?

**Hint: (a)** According to the description of the model, dy / dt = ky(1-y). **(b)** Use separation of variables and get dy/ y(1-y) = kdt. **(c)** Set 8AM as t=0. Then noon stands for t=4. Use the data and find A = 2/23, k = \frac{1}{4}\ln(23/2). Solve t for y(t) = 0.9.

**Example 5:** Let c be a positive number. A differential equation of the form dy /dt = ky^{1+c}, where k is a positive constant, is called a doomsday equation because the exponent in the expression ky^{1+c} is larger than the exponent 1 for natural growth. **(a)** Determine the solution that satisfies the initial condition y(0)=y_0. **(b)** Show that there is a time t=T (doomsday) such that \lim_{t\to T^{-}}y(t)=\infty. **(c)** An especially prolific bread of rabbits has the growth term My^{1.01}. If 2 such rabbits bread bread initially and the warren has 16 rabbits after 3 months, then when is doomsday?

**Solution:** **(a)** Use separation of variables and get dy / y^{1+c} = kdt. Integrate, use the initial value, and get y = (-ckt + y_0^{-c})^{-1/c}. **(b)** As -ckt + y_0^{-c} tends to 0^+, y(t) tends to infinity. Therefore the doomsday T = y_0^{c}/ck. **(c)** According to the data, y_0 = 2, c = 0.01 and 16 = (-0.01\cdot k\cdot 3 + 2^{-0.01})^{-1/0.01}, hence k=0.681252\ldots. The doomsday is then 145.775 months, or 12 years plus 2 months.

**Disclaimer:** The author disclaims responsibility for any adverse effects resulting, directly or indirectly, from information contained in the previous problem. In particular, when you see 16 rabbits, don’t try killing them.

**Remark:** See *A Mole of Moles* by xkcd for a hypothetical analysis of the moment before the doomsday.