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Section 8.6

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1. (a) x = c_1\cos(t) + c_2\cos(t) , (0, 0) is a center

(b) x = c_1e^{-t} + c_2e^{-2t} , (0, 0) is a stable equilibrium

(c) x = c_1e^t + c_2e^{-t} , (0, 0) is an unstable equilibirum

(d) x = e^{-t}(c_1\cos(t) + c_2\sin(t)) , (0, 0) is a stable equilibrium

(e) x = e^t(c_1\cos(t) + c_2\sin(t)) , (0, 0) is an unstable equilibrium

4. (a) Equations:

\dot{x} = y
\dot{y} = -x^2

Graph of x(t) :
Section 8.6 - Problem 4. (a)
Phase curve:
Section 8.6 - Problem 4. (a)

(c) Equations:

\dot{x} = y
\dot{y} = -x^3

Graph of x(t) :
Section 8.6 - Problem 4. (c)
Phase curve:
Section 8.6 - Problem 4. (c)

(e) Equations:

\dot{x} = y
\dot{y} = (x^2 - 1)y - x

Graph of x(t) :
Section 8.6 - Problem 4. (e)
Phase curve:
Section 8.6 - Problem 4. (e)

5. (a) Graph of x(t) (blue) and y(t) (red):
Section 8.6 - Problem 5. (a)
Phase curve:
Section 8.6 - Problem 5. (a)

(c) Graph of x(t) (blue) and y(t) (red):
Section 8.6 - Problem 5. (c)
Phase curve:
Section 8.6 - Problem 5. (c)

7. Graph of x(t) :
Section 8.6 - Problem 7
The period is approximately 3.00 seconds. The period of the linearized system is 2.84 seconds. The exact period (see Problem 2 in Section 8.5) is 3.03 seconds.

9. (a) Equations:

\dot{x} = y
\dot{y} = -\alpha y + x(1 - x^2)

Stationary points: (-1, 0) , (0, 0) , and (1, 0)

(b) (-1, 0) and (1, 0) are stable equilibrium points and (0, 0) is an unstable equilibrium point.

(c) (-1, 0) , (0, 0) , and (1, 0) are all unstable equilibrium points.

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