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Practice Problems Section 8.2 1. Find and classify all equilibria of the autonomous differential equation dy = e−y sin(y), dt y ≥ 0. 2. Consider the differential equation dP = f (P ) = 16P − P 3 . dt (a) Find all equilibria. (b) By graphing f (P ), analyze the stability of the equilibria from (a). (c) Use the method of eigenvalues to check your answers in (b). (d) In the cases, P (0) = 2 P (0) = 5, and P (0) = −1, what would you expect the long-term behavior of P to look like? That is, what do you expect limt→∞ P (t) to be in each case? Explain your answer without solving the differential equation. (e) Solve the differential equation, and find specific solutions in the cases: P (0) = 2, P (0) = 5, P (0) = −1, and P (0) = 0 (you do not need to find explicit solutions). 3. Consider √ dy = e− y , dt y ≥ 0. (a) Find all equilibria, and classify each as either stable or unstable. (b) If y(0) = 4, what do you expect limt→∞ y(t) to be? (c) Solve the differential equation with the initial condition y(0) = 4. 4. Let dy = y ln(y), dt y > 0. (a) Find all equilibria, and analyze the stability of each. (b) If y(0) = e, what do expect limt→∞ y(t) to be? (c) Solve the differential equation in the case of y(0) = e. Find an explicit solution, and take a limit as t → ∞. Compare this with your answer in (b). 1 Solutions: (1) Stable equilibria: kπ, where k ≥ 0 is odd. Unstable: kπ, where k ≥ 0 is even. (2) (a) P = 0, ±4. (b) After graphing f (P ), you should find 0 to be unstable and ±4 to be stable. (c) Since f 0 (P ) = 16 − 3P 2 , we get f 0 (0) = 16, meaning 0 is unstable, and f 0 (4) = f 0 (−4) = −32 < 0, so 4 and −4 are stable. (d) For the cases P (0) = 2 and P (0) = 5, we would expect the limit to be 4 since 4 is a stable equilibrium. In the case P (0) = −1, we would expect the solution to move away from P = 0 and towards P = −4 since P = 0 is unstable and P = −4 is stable. (e) The general solution is 1 1 1 ln |y| − ln |4 + y| − ln |4 − y| = t + C. 16 32 32 1 1 1 In the case P (0) = 2, C = 32 ln(1/3). When P (0) = 5, C = 16 ln(5) − 32 ln(9). When P (0) = −1, 1 C = 32 ln(3/5), and when P (0) = 0, the only solution is P (t) = 0. (3) (a) There are no equilibrium solutions. √ (b) Since e− y > 0, we expect the limit to be ∞. √ √ √ (c) 2( ye y − e y ) = t + 2e2 . (4) (a) The only equilibrium comes at y = 1. Since (y ln(y))0 = 1 + ln(y), and 1 + ln(1) = 1 > 0, it is an unstable equilibrium. (b) 1 is unstable, and so we expect y(t) to grow to ∞. t (c) The explicit solution is y(t) = ee , and the limit as t → ∞ is ∞, as expected. 2