Each deserves a separate constant so redefining

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Each deserves a separate constant. So redefining constants we guess y P ( t ) = A cos(2 t ) + B sin(2 t ) + Ct cos(2 t ) + Dt sin(2 t ) Plugging into the equation we get - 4 A cos(2 t ) - 4 B sin(2 t ) - 4 C sin(2 t ) - 4 Ct cos(2 t ) + 4 D cos(2 t ) - 4 Dt sin(2 t ) + A cos(2 t ) + B sin(2 t ) + Ct cos(2 t ) + Dt sin(2 t ) = sin(2 t ) + t cos(2 t ) Rewrite to obtain ( - 4 A +4 D + A ) cos(2 t )+( - 4 B - 4 C + B ) sin(2 t )+( - 4 C + C ) t cos(2 t )+(4 D + D ) t sin(2 t ) = sin(2 t )+ t cos(2 t ) Comparing coefficients of similar functions on either side t sin(2 t ) : 4 D + D = 0 D = 0 t cos(2 t ) : - 4 C + C = 1 C = - 1 3 cos(2 t ) : - 4 A + 4 D + A = 0 A = 0 sin(2 t ) : - 4 B - 4 C + B = 1 B = 1 9 Thus y P ( t ) = 1 9 sin(2 t ) - 1 3 t cos(2 t ) and so the general solution is y ( t ) = c 1 cos( t ) + c 2 sin( t ) + 1 9 sin(2 t ) - 1 3 t cos(2 t )
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8. 8 y 00 + 4 y 0 + y = te t + 6 Solution : Step 1 Homogeneous solution y H ( t ): Characteristic equation 8 m 2 + 4 m + 1 = 0 which has roots m = - 1 / 4 ± i/ 4 and thus the homogeneous solution is y H ( t ) = c 1 e - t/ 4 cos( t/ 4) + c 2 e - t/ 4 sin( t/ 4) Step 2 Particular solution y P ( t ): Our guess based on the right-hand side consists of two functions y P 1 ( t ) = ( At + B ) e t , y P 2 ( t ) = C and so y P ( t ) = ( At + B ) e t + C Plugging into the equation we get 8(2 Ae t + ( At + B ) e t ) + 4( Ae t + ( At + B ) e t ) + ( At + B ) e t + C = te t + 6 Rewrite to obtain (8 A + 4 A + A ) te t + (16 A + 8 B + 4 A + 4 B + B ) e t + C = te t + 6 Comparing coefficients of similar functions on either side te t : 8 A + 4 A + A = 1 A = 1 13 e t : 16 A + 8 B + 4 A + 4 B + B = 0 B = - 20 169 e 0 = 1 : C = 6 Thus y P ( t ) = 1 13 te t - 20 169 e t + 6 and so the general solution is y ( t ) = c 1 e - t/ 4 cos( t/ 4) + c 2 e - t/ 4 sin( t/ 4) + 1 13 te t - 20 169 e t + 6
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Exercise 2 Consider the oscillator given by x 00 + γx 0 + 2 x = 0 . (a) Find the value of γ so that the system is critically damped. Solution : The characteristic equation is λ 2 + γλ + 2 = 0. This gives λ = - γ ± p γ 2 - 8 2 . Therefore the system will be criticaly damped for γ 2 = 8 or, since γ > 0, γ = 2 2. (b) For this value of γ , solve the initial value problem with x (0) = 0 , x 0 (0) = 1. Solution : The general solution is given by x = c 1 e - γt/ 2 + c 2 te - γt/ 2 . Imposing initial conditions x (0) = c 1 = 0 , x 0 ( t ) = - γ 2 c 1 e - γt/ 2 + c 2 e - γt/ 2 - γ 2 c 2 te - γt/ 2 , x 0 (0) = c 2 = 1 . The solution is then given by x = te - γt/ 2 = te - 2 t
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Exercise 3 A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 10 cm/s, and if there is no damping, determine the position u of the mass at any time t . Solution : The weight is given in grams and so the mass m = 100 The spring constant is k = Weight L , = 100 × 980 5 , = 19600 No damping and hence γ = 0 The initial conditions are u (0) = 0 , u 0 (0) = 10 Hence the equation is 100 u 00 + 19600 u = 0 which has characteristic equation 100 m 2 + 19600 = 0 and so roots are m = ± 14 i . The general solution is given by
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