Chapter4diff eq - CHAPTER 4 Higher-Order Linear...

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294 CHAPTER 4 4.1 The Harmonic Oscillator ± The Undamped Oscillator 1. 0 xx += ±± , ( ) 01 x = , ( ) 00 x = ± The general solution of the harmonic oscillator equation 0 + = is given by ( ) () 12 cos sin sin cos . x tc t c t x t ct =+ =− + ± Substituting the initial conditions ( ) x = , ( ) x = ± , gives ( ) 1 2 xc = = = = ± so 1 1 c = , 2 0 c = . Hence, the IVP has the solution ( ) cos x tt = . 2. 0 , ( ) x = , ( ) x = ± The general solution of the harmonic oscillator equation 0 + = is given by ( ) cos sin sin cos . x t c t x t + ± Substituting the initial conditions ( ) x = , ( ) x = ± , gives ( ) 1 2 = = = = ± or 1 cc == . Hence, the IVP has the solution ( ) cos sin x t . Higher-Order Linear Differential Equations
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SECTION 4.1 The Harmonic Oscillator 295 In polar form, this would be () 2cos 4 xt t π ⎛⎞ =− ⎜⎟ ⎝⎠ . 3. 90 xx += ±± , ( ) 01 x = , ( ) x = ± The general solution of the harmonic oscillator equation + = is given by ( ) 12 cos3 sin3 3s i n 3 3c o s 3 . c t c t x tc t ct =+ + ± Substituting the initial conditions ( ) x = , ( ) x = ± , gives ( ) 1 2 03 1 xc = = = = ± so 1 1 c = , 2 1 3 c = . Hence, the IVP has the solution 1 cos3 3 x tt t . In polar form, this would be 10 cos 3 3 t δ = where 1 1 tan 3 = . This would be in the first quadrant. 4. 40 , ( ) x = , ( ) 02 x = − ± The general solution of the harmonic oscillator equation + = is given by ( ) cos2 sin 2 2s i n 2 2c o s 2 . c t x t + ± Substituting the initial conditions ( ) x = , ( ) x = − ± , gives ( ) 1 2 2 == = ± so 1 1 c = , 2 1 c = − . Hence, the IVP has the solution ( ) sin 2 x t . In polar form, this would be 2cos 2 4 t .
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296 CHAPTER 4 Higher-Order Linear Differential Equations 5. 16 0 xx += ±± , ( ) 01 x =− , ( ) 00 x = ± The general solution of the harmonic oscillator equation 16 0 + = is given by ( ) () 12 cos4 sin 4 4s i n 4 4c o s 4 . xt c t c t x tc t ct =+ + ± Substituting the initial conditions ( ) x = − , ( ) x = ± , gives ( ) 1 2 04 0 xc = = = ± so 1 1 c , 2 0 c = . Hence, the IVP has the solution ( ) x tt . 6. 16 0 , ( ) x = , ( ) x = ± The general solution of the harmonic oscillator equation 16 0 + = is given by ( ) sin 4 i n o s 4 . c t x t + ± Substituting the initial conditions ( ) x = , ( ) x = ± , we get ( ) 1 2 4 == = = ± so 1 0 c = , 2 1 c = . The IVP has the solution ( ) sin 4 x = . 7. 2 16 0 π , x (0) = 0, ( ) 0 x = ± 2 0 16 1 ω = = 4 x = c 1 cos 4 t + c 2 sin 4 t x ± = 4 c 1 sin 4 t + 4 c 2 cos 4 t x (0) = 0 = c 1 (0) x ± = = 4 c 2 c 2 = 1 4 x = 1 sin 4 4 t
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SECTION 4.1 The Harmonic Oscillator 297 8. 2 40 xx π += ±± , x (0) = 1, ( ) 0 x = ± 2 0 42 ω == x = c 1 cos 4 t + c 2 sin 4 t x ± = 12 sin cos 22 ct c t ππ + x (0) = 1 = c 1 (0) x ± = = 2 2 c , c 2 = 2 x = cos 2sin tt + ± Graphing by Calculator 9. cos sin ytt =+ The equation tells us 2 T = and because 0 2 T = , 0 1 = . We then measure the delay 0 0.8 δ which we can compute as the phase angle ( ) 0.8 1 0.8 ≈= . The amplitude A can be measured directly giving 1.4 A . Hence, ( ) cos sin 1.4cos 0.8 t +≈ . Compare with the algebraic form in Problem 15. Š1.5 1.5 t T = 2 A 1.4 0.8 y 0 10. 2cos sin yt t The equation tells us 2 T = and because 0 2 T = , 0 1 = . We then measure the delay 0 0.5 , which we can compute as the phase angle ( ) 0.5 1 0.5 . The amplitude A can be measured directly giving 2.2 A . Hence, ( ) 2cos sin 2.2cos 0.5 t .
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This note was uploaded on 04/07/2008 for the course MA 232 taught by Professor Toland during the Fall '08 term at Clarkson University .

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Chapter4diff eq - CHAPTER 4 Higher-Order Linear...

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