Image

CHAPTER 1

Section 1.1 page 14

1. yImage as t → ∞

2. y diverges from −Image as t → ∞

3. y diverges from − Image as t → ∞

4. y → − Image as t → ∞

5. y diverges from − Image as t → ∞

6. y diverges from −2 as t → ∞

7. y′ = 3 − y

8. y′ = 2 − 3y

9. y′ = y − 2

10. y′ = 3y − 1

11. y = 0 and y = 4 are equilibrium solutions; y → 4 if initial value is positive; y diverges from 0 if initial value is negative.

Image     Direction field for Problem 11.

12. y = 0 and y = 5 are equilibrium solutions; y diverges from 5 if initial value is greater than 5; y → 0 if initial value is less than 5.

13. y = 0 is equilibrium solution; y → 0 if initial value is negative; y diverges from 0 if initial value is positive.

14. y = 0 and y = 2 are equilibrium solutions; y diverges from 0 if initial value is negative; y → 2 if initial value is between ...

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