DIFFERENTIAL EQUATIONS - EXAMPLE 20.3 : A differential equation is given as y&rd

DIFFERENTIAL EQUATIONS - EXAMPLE 20.3 : A differential equation is given as y” + 5y’ + 6y = 0, y(0) = 2 and y’(0) = 3. By using Laplace transform, an engineer has correctly produced the equation L {y} = (2s + 13) / [(s + 2)(s + 3)] = A / (s + 2) + B (s + 3). (a) Find the values of A and B. (b) The inversed Laplace transform of 1 / (s + a) is given by exp (-at) where a is a constant. If the statement : L {y} = 9 L { exp (-2t) } - 7 L { exp (-3t) } is correct, find the equation of y as a function of t as a solution to the differential equation stated in the beginning of this question. When L {d} = 9 L {b} - 7 L {c}, then d = 9b - 7c with b, c and d are unknowns.

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DIFFERENTIAL EQUATIONS - EXAMPLE 20.3 : A differential equation is given as y” + 5y’ + 6..

Answer / kangchuentat

DIFFERENTIAL EQUATIONS - ANSWER 20.3 : (a) Let (2s + 13) / [(s + 2)(s + 3)] = [A (s + 3) + B (s + 2)] / [(s + 2)(s + 3)]. Then 2s + 13 = As + 3A + Bs + 2B = (A + B)s + (3A + 2B), then A + B = 2 and 3A + 2B = 13. Let 2A + 2B = 4 as first equation by doubling A + B = 2 and 3A + 2B = 13 as second equation, the difference of first and second equations will produce A = 9. When A = 9, B = 2 - A = 2 - 9 = -7. (b) Let d = y, b = exp (-2t) and c = exp (-3t), then y = 9 exp (-2t) - 7 exp (-3t). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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