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.