Chem Differential Eq HW Solutions Fall 2011 21

# Chem Differential Eq HW Solutions Fall 2011 21 - Section...

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Unformatted text preview: Section 2.7 Forced Oscillations 21 Solutions to Exercises 2.7 1. (a) General solution of y + 2 y + y = 0. The characteristic equation is λ 2 +2 λ +1 = 0 or ( λ +1) 2 = 0. It has one double characteristic root λ =- 1. Thus the general solution of the homogeneous equation y + 2 y + y = 0 is y = c 1 e- t + c 2 te- t . To find a particular solution of y + 2 y + y = 25 cos 2 t , we apply Theorem 1 with μ = 1 , c = 2, and k = 1. The driving force is already given by its Fourier series: We have b n = a n = 0 for all n , except a 2 = 25. So α n = β n = 0 for all n , except α 2 = A 2 a 2 A 2 2 + B 2 2 and β 2 = B 2 a 2 A 2 2 + B 2 2 , where A 2 = 1- 2 2 =- 3 and B 2 = 4. Thus α 2 =- 75 25 =- 3 and β 2 = 100 25 = 4, and hence a particular solution is y p =- 3 cos 2 t + 4 sin 2 t . Adding the general solution of the homogeneous equation to the particular solution, we obtain the general solution of the differential equation y + 2 y + y = 25 cos 2 t y = c 1 e- t + c 2 te- t- 3 cos 2 t + 4 sin 2...
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