6.4 - Math 334 Lecture #26 § 6.4: Discontinuous Forcing...

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Unformatted text preview: Math 334 Lecture #26 § 6.4: Discontinuous Forcing Example. Use the Laplace transform to solve the IVP y 00 + 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 00 + 4 y + 4 y } = L{ u 1 ( t )- u 2 ( t ) } use linearity ⇒ L{ y 00 } + 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...
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This note was uploaded on 03/08/2011 for the course MATH 334 taught by Professor Smith during the Spring '11 term at Vanderbilt.

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6.4 - Math 334 Lecture #26 § 6.4: Discontinuous Forcing...

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