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SOLDE_23-32 - Tutorial SOLDE Second-Order Linear...

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Tutorial SOLDE: Second-Order Linear Differential Equations The solution is thus y ( t ) = 2 5 e ¡ t · cos ³ p 5 t ´ + 9 p 5 sin ³ p 5 t ´ ¸ + 3 10 ( t sin2 t ¡ 2 t cos2 t ¡ sin2 t + 2cos2 t ) : Exercise 23 NA Exercise 24 From Newton’s Second Law, the equation of motion for the (unforced) harmonic oscillator is m Ä y = ¡ ¯ _ y ¡ 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 = ¯ and c = k: Exercise 25 The equation for the charge on the capacitor is (for an unforced RLC circuit) L Ä 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( ¡ °t )cos( ! 0 t + ' ) ; we have a £¡ ° 2 ¡ ! 2 0 ¢ cos( ! 0 t + ' ) + 2 ! 0 ° sin( ! 0 t + ' ) ¤ + b ( ¡ ° cos( ! 0 t + ' ) ¡ ! 0 sin( ! 0 t + ' )) + c cos( ! 0 t + ' ) = 0 ; or a ¡ ° 2 ¡ ! 2 0 ¢ ¡ + c = 0 ; 2 a! 0 ° ¡ ! 0 b = 0 : The second of these yields the expression for ° directly. The first then gives that for !
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