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