Chapter 6 - CHAPTER 6 Linear Systems of Differential...

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CHAPTER 6 Linear Systems of Differential Equations 6.1 Theory of Linear DE Systems ! Nullcline Sketching 1. ′ = ′ = xx yy Equilibrium (unstable) at (0, 0) hy x −= nullcline nullcline 0 0 υ (See figure.) 2 –2 –2 2 y x h -nullcline -nullcline 2. =+ ′ = y yx 32 2 Equilibrium (unstable) at (0, 0) hx xy −+ = nullcline nullcline 0 0 (See figure.) 2 –2 –2 2 y x 3. ′ = ′=− − y 52 Equilibrium (stable) at (0, 0) y y = nullcline nullcline 0 0 (See figure.) 2 –2 –2 2 y x 435
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436 CHAPTER 6 Linear Systems of Differential Equations 4. ′ = ′=− − + xy yx y 2 22 Equilibrium (stable) at 1 0 , () hx y y −+ = −= nullcline nullcline 0 υ (See figure.) 3 –1 –2 2 y x 5. =+− =− xx y yy 2 1 Equilibrium (unstable) at 1 1 hy = nullcline nullcline 1 2 (See figure.) 2 –1 –1 2 y x 6. ′=− − + y 1 522 Equilibrium (stable) at 0 1 y y = nullcline nullcline 52 2 1 (See figure.) –2 2 y 2 –1 x ! Sketching Second-Order DEs 7. ′′ + += x 0 (a) Letting = , we write the equation as the first-order system ′ = y . (b) The equilibrium point is ,, = 00.
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SECTION 6.1 Theory of Linear DE Systems 437 (c) hx y y −+ = −= nullcline nullcline 0 0 υ (See figure.) (d) From the direction field, the equilibrium point xy ,, () = 0 0 is stable. (e) A mass-spring system with this equation shows damped oscillatory motion about xt () ≡ 0 is stable. 2 –2 –2 2 y x 8. ′′ − ′ += xx x 0 (a) Letting yx = , we write the equation as the first-order system ′ = ′=− + y . (b) The equilibrium point is = 00. (c) y y −− = nullcline nullcline 0 0 (See figure.) (d) From the direction field, the equilibrium point = 0 0 is unstable. (e) A mass-spring system with this equation tends to fly apart. Hence, 0 is unstable. 2 –2 –2 2 y x 9. ′′ 1 (a) Letting = , we write the equation as the first-order system ′ = 1. (b) The equilibrium point is = 10.
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438 CHAPTER 6 Linear Systems of Differential Equations (c) hx y −= nullcline nullcline 1 0 υ (See figure.) (d) From the direction field, the equilibrium point xy ,, () = 1 0 is stable. (e) A mass-spring system with this equation shows no damping and steady forcing; hence, periodic motion about an equilib- rium is to the right of the origin. Hence, xt () ≡ 1 is stable. 3 –1 –2 2 y x 10. ′′ + += xx x 22 (a) Letting yx = , we write the equation as the first-order system ′ = ′=− − + y . (b) The equilibrium point is , = 2, 0 . (c) y y −+ = nullcline nullcline 0 (See figure.) (d) From the direction field, the equilibrium point = 2, 0 is stable. (e) A mass-spring system with this equation shows heavy damping. The force moves the equilibrium two units to the right of the origin. Hence, 2 is stable. 3 –1 –2 2 y x ! Breaking Out Systems 11. =+ =− x x 11 2 21 2 2 4 12. ′ = ′=− + 1 13. =++ xxx e x t 112 2 43 14. ′ = ′ = ′=− + + + x 12 23 31 2 3 s i n
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SECTION 6.1 Theory of Linear DE Systems 439 ! Checking It Out 15. ′ = L N M O Q P xx 13 31 Plugging u t e e t t () = L N M O Q P 4 4 and v t e e t t L N M O Q P 2 2 into the given system easily verifies: 4 4 4 4 4 4 e e e e t t t t L N M O Q P = L N M O Q P L N M O Q P and L N M O Q P = L N M O Q P L N M O Q P 2 2 2 2 2 2 e e e e t t t t .
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Chapter 6 - CHAPTER 6 Linear Systems of Differential...

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