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Unformatted text preview: Tutorial SOLDE: Second-Order Linear Differential Equations The solution is thus y ( t ) = 2 5 e Â¡ t Â· cos Â³ p 5 t Â´ + 9 p 5 sin Â³ p 5 t Â´ Â¸ + 3 10 ( t sin 2 t Â¡ 2 t cos2 t Â¡ sin2 t + 2cos2 t ) : Exercise 23 NA Exercise 24 From Newtonâ€™s Second Law, the equation of motion for the (unforced) harmonic oscillator is m Ã„ y = Â¡ Â¯ _ y Â¡ ky; where y is the displacement from equilibrium. The force of gravity, mg; is cancelled by the elastic restoring force from the original stretch in reaching equilibrium. We have, therefore, a = m; b = Â¯ and c = k: Exercise 25 The equation for the charge on the capacitor is (for an unforced RLC circuit) L Ã„ Q + R _ Q + Q=C = 0 : Thus, a = L; b = R and c = 1 =C: Exercise 26 Using the form for the solution, A exp( Â¡ Â°t )cos( ! t + ' ) ; we have a Â£Â¡ Â° 2 Â¡ ! 2 Â¢ cos( ! t + ' ) + 2 ! Â° sin ( ! t + ' ) Â¤ + b ( Â¡ Â° cos( ! t + ' ) Â¡ ! sin( ! t + ' )) + c cos( ! t + ' ) = 0 ; or a Â¡ Â° 2 Â¡ !...
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This note was uploaded on 07/13/2008 for the course PHY 201 taught by Professor Covatto during the Spring '08 term at ASU.
- Spring '08