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6.4 - Math 334 Lecture#26 6.4 Discontinuous Forcing Example...

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Math 334 Lecture #26 § 6.4: Discontinuous Forcing Example. Use the Laplace transform to solve the IVP y + 4 y + 4 y = u 1 ( t ) - u 2 ( t ) , y (0) = 0 , y (0) = 0 . Applying the Laplace tranform (by its rules) gives L{ y + 4 y + 4 y } = L{ u 1 ( t ) - u 2 ( t ) } use linearity ⇒ L{ y } + 4 L{ y } + L{ y } = L{ u 1 ( t ) } - L{ u 2 ( t ) } use derivative rules , and step function rules s 2 L{ y } - sy (0) - y (0) + 4 s L{ y } - y (0) + 4 L{ y } = e - s s - e - 2 s s use initial conditions ( s 2 + 4 s + 4) L{ y } = e - s s - e - 2 s s solve for L{ y } ⇒ L{ y } = e - s 1 s ( s 2 + 4 s + 4) - e - 2 s 1 s ( s 2 + 4 s + 4) . To invert the Laplace transform, find the partial fraction decomposition of the rational function common to both terms: 1 s ( s 2 + 4 s + 4) = 1 s ( s + 2) 2 = A s + B s + 2 + C ( s + 2) 2 , where 1 = A ( s + 2) 2 + Bs ( s + 2) + Cs implies that s = 0 : 1 = 4 A A = 1 / 4 , s = - 2 : 1 = - 2 C C = - 1 / 2 , s = - 1 : 1 = A - B - C B = A - C - 1 = 1 / 4 + 1 / 2 - 1 = - 1 / 4 . Check the partial fraction decomposition: 1 / 4 s - 1 / 4 s + 2 - 1 / 2 ( s + 2) 2 = (1 / 4)( s + 2) 2 - (1 / 4) s ( s + 2) - (1 / 2) s s ( s + 2) 2 = s 2 / 4 + s + 1 - s 2 / 4 - s/ 2 - s/ 2 s ( s + 2) 2 = 1 s ( s + 2) 2 .

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So the Laplace transform of the solution of the initial value problem is L{ y } = e - s 1 / 4 s - 1 / 4 s + 2 - 1 / 2 ( s + 2) 2 - e - 2 s 1 / 4 s
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